Category: CBSE Class 11

Here we are providing Class 11 Biology Important Extra Questions and Answers Chapter 3 Plant Kingdom. Important Questions for Class 11 Biology are the best resource for students which helps in Class 11 board exams.

Class 11 Biology Chapter 3 Important Extra Questions Plant Kingdom

Plant Kingdom Important Extra Questions Very Short Answer Type

Question 1.
Define pyrenoid.
Answer:
Pyrenoid is a starch storing organelle present in green algae.

Question 2.
Define ramenta.
Answer:
The hair-like structure present on the younger stem, petiole, and rachis of mature leaves is known as ramenta.

Question 3.
What is the function of mucilage in aquatic plants?
Answer:
Mucilage protects the algae from epiphytic growth and the decaying effect of water.

Question 4.
How much water can Sphagnum absorb?
Answer:
Sphagnum can absorb water up to 18 times its weight.

Question 5.
What is the function of air vesicles in brown algae?
Answer:
The air vesicles in brown algae maintain buoyancy.

Question 6.
Why is Adiantum called a ‘walking fern’?
Answer:
Adiantum is known as ‘walking fern’ because the leaf tips come in contact with the soil, They strike adventitious roots and develop into a new plant.

Question 7.
Give one example of the monocarpic plant.
Answer:
Bamboo.

Question 8.
What are sori?
Answer:
They are groups of separation found in Dryopteris fern.

Question 9.
What are rhizoids?
Answer:
They are slender unicellular or multicellular hair-like structures that penetrate in the moist soil and absorb the water for the plants.

Question 10.
Which pigments are found in green algae?
Answer:
Chlorophyll ‘a’ and ‘b’ and ‘Beta’ carotene.

Question 11.
Define a fruit.
Answer:
Fruit is a developed ovary of the flower that encloses seeds and may be associated with other parts of the flower.

Question 12.
Name the group of vascular plants with naked seed.
Answer:
Gymnosperms.

Question 13.
Name the green algae used as food.
Answer:
Chlorella, Ulva.

Question 14.
Name the following:
(i) Photosynthetic pigments of brown algae
Answer:
Chlorophyll a and c and fucoxanthin,

(ii) Unicellular, biflagellate, pear-shaped green algae.
Answer:
Chlamydomonas,

Question 15.
What are coralloid roots?
Answer:
Coralloid roots are irregular, negatively geotropic,

Question 16.
What is triple fusion?
Answer:
The fusion of the diploid secondary nucleus and one male gamete is called triple fusion.

Question 17.
Define a seed.
Answer:
It is a ripened ovule and capable of forming a new plant.

Question 18.
Define a fruit.
Answer:
Fruit is a ripened ovary.

Question 19.
Give one example of a dicot seed and one of a monocot seed.
Answer:
Dicot: Gram, Monocot: Maize Grain.

Plant Kingdom Biology Important Extra Questions  Short Answer Type

Question 1.
Why are red algae able to survive in the deep-sea?
Answer:
Red algae contain phycoerythrin and phycocyanin pigments. Phycoerythrin is able to absorb the blue wavelengths of light and thus can photosynthesize. Since red algae can utilize blue and green rays they can live at greater depths.

Question 2.
What are the features that have led to the success and dominance of vascular plants?
Answer:

1. Development of deep penetrating roots to anchor the plant in soil and absorb water and minerals for the plants from the deeper layers of the soil.
2. Development of cutin as a waterproof layer on leaves to reduce transpiration.
3. Development of mechanical tissue to provide support.
4. Development of a well developed vascular system.

Question 3.
Define monopodial growth?
Answer:
When the main axis of the trunk rises straight from the base and reaches up to the tip, this type of growth is known as monopodial growth.

Question 4.
Why do marine algae have no mechanical tissue?
Answer:
Marine algae have no mechanical tissue because buoyancy holds them erect under the sea surface.^

Question 5.
Explain the different types of sexual reproduction in green algae.
Answer:
Sexual reproduction in green algae can be of three different types:

1. Isogamy: Both the fusing gametes are morphologically and physiologically similar.
2. Anisogamy: The fusing gametes are structurally similar but differ in size and behaviour.
3. Oogamy: The female gamete is bigger, food-laden and non-motile, whereas the male gamete is smaller, without food reserve and motile.

Question 6.
Why are seed plants considered the most successful land plants?
Answer:
Seed plant is considered as the most successful land plants because:

1. Fertilization is not water-dependent.
2. Seed enclosing the future embryo is well protected within the ovary.
3. The extensive root system for anchoring and absorption of water.
4. Well developed mechanical tissue
5. Formation of bark during secondary growth for protection.

Question 7.
Give one example of each
(i) Liverworts
Answer:
Riccia

(ii) Mosses
Answer:
Funaria

(iii) A pteridophyte having bipinnate leaves.
Answer:
Dryopteris

(iv) A pteridophyte having Unipinnate leaves.
Answer:
Adiantum caudatum

Question 8.
Give five distinguishing characteristics of red algae.
Answer:
Five distinguishing characteristics of red algae are:

1. Most of the red algae are marine.
2. The motile stage is absent in the life cycle.
3. The plant body varies from unicellular filamentous to parenchymatous form.
4. The cell wall possesses cellulose and hydrocolloids.
5. Photosynthetic pigments include chlorophyll a, carotenoids and phycobilins.

Question 9.
Discuss the development of seed habit.
Answer:
The development of seed habit takes place due to the

1. Development of heterospory.
2. The megasporangium developed an intelligent covering with a micropyle.
3. The development of female gametophyte takes place from functional-megaspore.
4. Development of pollen tube.
5. The fertilized ovule developed into a seed.

Question 10.
Draw a neat diagram of Chlamydomonas.
Answer:

Chlamydomonas

Question 11.
Draw a neat diagram of Spirogyra.
Answer:

Spirogyra

Question 12.
Describe the fern sporophyte.
Answer:
Fem sporophyte is differentiated into root, the stem is a rhizome with adventitious roots, the young part of the rhizome has ramenta. the leaves are bipinnately compound.

Question 13.
Distinguish between Antherida and Archaegonia.
Answer:

Antheridia	Archaegonia
1. It is the male reproductive organ.	1.            It is the female reproductive organ.
2. It is racket-shaped	2.            It is usually flask-shaped.
3. Sterile cells are absent inside antheridia.	3. Neck canal cells, venter canal cells form the sterile cells.
4. It produces many male gametes	4. It produces only one female gamete.
5. The male gamete is usually motile	5. The female gamete is usually non-motile.

Question 14.
How do red algae differ from brown algae?
Answer:
Differences between red algae and brown algae:

Red algae	Brown algae
1. Unicellular and microscopic only a few are filamentous and Heterotrichous.	1. Filaments and heterotrichous.
2. Phycoerythrin, phycocyanin and phycobilins pigments are present.	2. Fucoxanthin pigment is present.
3. Reserve food material is Floridian starch.	3. Reserve food material is Laminarian starch.
4. Chlorophyll ‘a’ present.	4. Chlorophyll ‘a’ + ‘c’ present.
5. e.g. Gelidium Polysiphonia.	5. e.g. Laminaria, Fucus and Surgaassum.

Red algae

Porphyra or Polysiphonia

Question 15.
Distinguish the reproductive organs of gymnosperms and angiosperms.
Answer:

Gymnosperms	Angiosperm
1. The sporophylls are aggregated to form cones.	1. The sporophylls are aggregated to produce flowers.
2. Microporosphyll consists of micro-sporangia bearing microspores.	2. Microsporophyll is differentiated into anther and filament.
3. Stigma and style are absent.	3. Stigma and style are present.
4. Ovule lies on the lower side and exposed to the megasporophyll.	4. Ovules are attached to the funic-ulus and are present inside the ovary.
5. One too many cotyledons are present.	5. One or two cotyledons are present.
6. Fruit formation doesn’t take place.	6. Fruit formation takes place.
7. Double fertilization is absent.	7. Double fertilization is present.
8. Archaegonia are present.	8. Archaegonia are present.

Question 16.
Mention the changes that take place when the fruit ripens.
Answer:

1. Starch is converted into sugar.
2. The production of various organic substances gives it a texture, taste and flavour.
3. The breakdown of chlorophyll leads to changes in the colour of the skin of the fruit.

Question 17.
What is the importance of seed?
Answer:

1. The seed contains the young embryo which develops into a new plant.
2. Seed can be dispersed and carried to faraway places without losing viability.
3. Seed is neither the beginning nor the end of life. It is a state of suspended animation.

Question 18.
Describe the important characteristics of gymnosperms.
Answer:
Characteristic features of gymnosperms:

1. Gymnosperms are evergreen woody, perennial plants.
2. Plants are heterosporous.
3. Reduction of gametophytic generation.
4. The enclosure of the female gametophyte by the megasporangium.
5. Ovules are exposed to receive pollen grains.
6. Gymnosperms possess exposed or naked seeds.
7. Polyembryony is a common occurrence.
8. Xylem lacks vessels and phloem lacks companion cells. Example Cycas, Pinus and Cedms.

Question 19.
Name two characters Used for the classification of dicotyledons in 3 sub-classes.
Answer:
Number and nature of floral whorls. Sub-classes are divided into series mainly on the position of the ovary with respect to other floral parts.

Question 20.
Explain briefly the alternation of generation in bryophytes.
Answer:
Alternation of generations: Moss plants are a gametophyte. Spore is the beginning of the gametophytic generation. It develops into protonema which rises to male and female gametes produced in them. Club-shaped antheridium bears biflagellate sperms or antherozoids. Flask-shaped archegonium encloses the female egg. a zygote is formed after the fertilization (syngamy) of male and female gametes with the help of water.

Repeated divisions of the zygote give rise to the embryo (2n) which soon develop into a sporophyte. The sporophyte of moss gets differentiated into three parts -foot, seta and capsule. Inside the capsule, single-celled spores are produced. After the dehiscences, they begin to germinate and give rise to the protonema to start the cycle again. Gametophytic generation al¬ternates the sporophytic generation.

Question 21.
Draw the haplontic life-cycle.
Answer:

Haplontic life-cycle

Question 22.
Draw the diplontic and haplo-diplontic cycles.
Answer:

Diplontic cycles

Haplo-diplontic cycles

Plant Kingdom Biology Important Extra Questions Long Answer Type

Question 1.
What are angiosperms? Give their characteristic features.
Answer:
Angiosperms are a group of flowering plants where seeds are embedded in the fruits.

They show the following characters:

1. The ovules/seeds are enclosed within the ovary, or we may say that after fertilization seeds are located in the fruit.
2. Male and female gametes i.e. pollen grains and egg nucleus are borne by the flowers.
3. During pollination pollen grains fall on the stigma, they develop on the stigma of the ovary and male gametes enter the egg nucleus through Onicropyh.
4. Male gametophyte is a three-celled structure when dehisced.
5. Embryosac or female gametophyte is eight celled when young and becomes seven celled at the time of fertilization.
6. There is double fertilization wherein one male gamete fuses with the egg nucleus to form a diploid zygote and another fuse with the secondary nucleus to form a triploid endosperm.
7. Xylem consists of tracheids, vessels fibres and parenchyma while phloem consists of sieve tubes, companion cells and phloem parenchyma and fibres. Xylem conducts water to the tip of tall trees and phloem is responsible for the translocation of food.

Question 2.
Write brief notes on:
(i) Green algae
Answer:
Green Algae: The Class (Chlorophyta: ‘GK’ choros = green: phyton = plant) has over 7,000 species. They are in several shapes and sizes. Some are unicellular and microscopic. Some are motile colonies like Volvox. Some, are multinucleated but unicellular i.e. coenocytic like cholera.

Volvox

Chlamydomonas

Chara

(ii) Brown algae
Answer:
Brown Algae: The Class (Phaeophyta: GK: pharos = brown: phyton = plant) has about 2,000 species, mostly marine. Some of, the world’s largest sea plants measuring 40-60 metres long. Brown algae occur chiefly in cooler seas. Some are filamentous. Brown algae like Laminaria are attached firmly to the rocks below by holdfasts.

Laminaria

Fucus

Dictyota

(iii) Club moss
Answer:
Club mass: It belongs to Lycopsida. In most parts of the world, Lycopodium is found. Sporangia’ are produced on mature leaves.

(iv) Horsetail
Answer:
Horsetail: Also called Sphenopsida. This group exists only Equisetuin. Because they look like the tail of a horse, so they are called horsetail. These plants are up to 1 metre in length. But some extinct species are of several metres. The root, stem and leaves are true.

(v) Sporophyll
Answer:
SporphyMs: They are special spore-bearing leaves and. produce sporangia in sori on their underside, where haploid spores are formed by meiosis. Spores germinate to form an independent, small gametophyte, the prothallus. This bears archegonia and antheridia. Male gamete from antheridia and swim in a film of water to egg cells in archegonia and fertilize them.

Question 3.
Discuss the development of seed habit.
Answer:
The seed plants have two kinds of sporangia. These sporangia are born on the sporophylls.

One type of sporangia are ovule or megasparangium. The other type of sporangia is the pollen sac or archegonium. The egg develops a pollen sac or microsporangium. The egg develops in the ovule from the megaspores. Many pollen grains are produced in the pollen sac.

The pollen grains are dispersed by the air! They reach the ovule. The male gamete and the female egg cell fuse together. The zygote is formed as a result of fertilization. Later on, the zygote forms the embryo. The seed is developed from the ovule. The development of seed habit in gymnosperm and angiosperm do not require liquid water during fertilization

Question 4.
What are the different lifestyles shown by Angiosperms?
Answer:

1. Hydrophytic plants are the plants that live in water or swampy places. Hydrophytes are categorised into, two groups:
   (a) Submerged plants like Hydrilla, Vallisneria, Utriculria and
   (b) Floating plant-like Nymphea, Wolffia and Pistia.
2. Xerophytic plants are those plants that live in the scarcity of water e.g. cactus.
3. Halophytes are a type of xerophytic plants that are present in saline conditions.
4. Insectivorous plants-A few angiosperms, though green and autotrophic trap insects to overcome the shortage of nitrogen. For example, pitcher plant, sundew, bladderwort.

Here we are providing Class 11 Biology Important Extra Questions and Answers Chapter 2 Biological Classification. Important Questions for Class 11 Biology are the best resource for students which helps in Class 11 board exams.

Class 11 Biology Chapter 2 Important Extra Questions Biological Classification

Biological Classification Important Extra Questions Very Short Answer Type

Question 1.
What are imperfect fungi?
Answer:
Asexually reproducing fungi belonging to Deuteromycites are imperfect fungi

Question 2.
How many basidiospores are formed after Karyogamy and meiosis?
Answer:
4

Question 3.
What is plasmogamy?
Answer:
Fusion of protoplasms between two motile or non-motile gametes.

Question 4.
What do you mean by mycorrhiza?
Answer:
Mycorrhiza is a symbiotic association between fungi with roots of higher plants.

Question 5.
How does the spore of slime molds disperse?
Answer:
By air currents

Question 6.
Name the protein-rich layer found in Euglenoids.
Answer:
Pellicle.

Question 7.
Give an example of red dinoflagellates?
Answer:
Gonyaulax.

Question 8.
Who is known as ‘Producers of the oceans’?
Answer:
Diatoms

Question 9.
Name the organism responsible for algal bloom?
Answer:
Blue-green algae.

Question 10.
Who has discovered viroids?
Answer:
T.O. Diener.

Biological Classification Biology Important Extra Questions Short Answer Type

Question 1.
What are the slime molds?
Answer:
1. The slime molds are both plant and animal-like.

2. They are plant-like in the production of spores during reproduction and animal-like in the mode of nutrition and rheumatic organization.

3. Their rheumatic structure consists of an acellular, multinucleate mobile mass of protoplasm which lacks a good wall called plasmodium.

4. The reproductive stage consists of sporangia and spores formed after meiosis. The spores on germination produce either flagellated swarm cells or myxamoebae.

5. These divide mitotically, then behave as gametes and fuse in pairs to form a diploid zygote. The zygote nucleus divides mitotically but the nuclear division is not followed by cell wall formation so that all nuclei lie free in the cytoplasm.

6. The enlargement of the zygote into plasmodium takes place which moves freely on the substratum and feeds on bacteria, fungal and algal spores, and also absorbs nutrients directly from the substratum. The plas¬modium then settles on dry places and develops into sporangia. Therefore asexual stage is plant-like and the sexual stage is animal-like.

Question 2.
Write a short note on kingdom Plantae.

Kingdom Plantae
Answer:
1. Kingdom Plantae includes all autotrophic plants which are photosynthetic forms called green plants starting from simple algae, bryophytes, pteridophytes to gymnosperm and angiosperms.

2. The plant cell has a cell wall mainly made of cellulose and, eukaryotic structure with prominent chloroplasts. Some plants are heterophilic like insectivorous plants which feed on insects and flies e.g. Bladderwort and Venus flytrap.

(3) The life cycle has two phases-sporophytic and gametophytic which are diploid (2n) and haploid (n) respectively. That means zygote (2n) undergoes meiosis to form haploid (n) spores these spores germinate into a gametophyte, then these gametes (male and female) fuse to form a zygote (2n) again which gives rise to the sporophyte. This phenomenon is called the alternation of generation.

Question 3.
Write a short note on Mycoplasma.
Answer:

1. Discovered by E.Nocard and E.R. Roux (1998) mycoplasma is the smallest known aerobic prokaryotes without a cell wall.
2. They were isolated from cattle suffering from bovine pleuropneumonia and hence they were designated as PPLO (pleuropneumonia-like organisms)
3. They are found in different forms as a spheroid, thin, stellate called pleomorphic.
4. They occur in soil, sewage, human, and plants.

Question 4.
Write a short note on Kingdom Animalia.
Answer:

1. Kingdom Animalia includes all animals except the protozoan. The members are multicellular eukaryotes. The cell wall is absent cells, organized into tissue. They directly or indirectly depend on plants for food.
2. They digest their food in an internal cavity and store food reserves as glycogen or fat. They are heterotrophic and the mode of nutrition is holozoic nutrition. They act as decomposers and help in the recycling of minerals.
3. Kingdom Plantae includes the multicellular, photosynthetic eukaryotic forms.
4. They have well-established mechanisms for absorption and Kingdom Animalia includes all animals except the protozoans. They are multicellular eukaryotes and are holozoic. The cells lack walls.

Question 5.
Write a short note on Lichens.
Answer:
Lichens are a symbiotic association between algae and fungi. The algae component is known as phycobiont and the fungal component as mycobiont which are autotrophic and heterotrophic respectively. Algae prepare food for fungi and fungi provides shelter and absorbs mineral nutrient and water for its partner.

On the basis of the structure of thallus, lichens can be classified into three types

1. Crustose: forms a crust on the substrate which is not easily separated from the substrate e.g. Graphis
2. Foliose: forms the leafy lobed structure attached to the substrate with the help of rhizines easily separated from substrate eg. Parmelia.
3. Fruticose: forms shrubby, cylindrical, and branched thallus. They grow erect or hang from the substrate e.g. Usnea. Lichens are the pioneer colonizers of bare rocks. They also colonize tree trunks in temperate climatic regions.

Biological Classification Biology Important Extra Questions Long Answer Type

Question 1.
Describe the kingdom Monera.
Answer:
This kingdom comprises single-celled prokaryotic organisms like bacteria, filamentous actinomycetes, and photosynthetic blue-green algae of Cyanobacteria. The salient features are already given in Table 2.1 we will further discuss the following organisms briefly.

Bacteria: Bacteria are single-celled microscope true prokaryotic organisms which are almost omnipresent. They colonize soil, water, and air. These can survive

in extreme environmental conditions like high temperature, high salt concentration, in absence of oxygen (anaerobic) or in presence of oxygen (aerobic) in high acidic or alkaline pH, etc. (sometimes these are called Archaebacteria).

Some bacteria can be chemotrophs that derive energy from inorganic compounds in absence of oxygen e.g. methanogenic bacteria produce methane gas (CH₄) from CO₂ and H₂, some live by oxidizing hydrogen supplied eg. Thiothsix.

Some are parasites on plants and animals e.g. Xanthomonas citri and Vibrio Cholera; another form symbiotic association with plant roots e.g. Rhizobium.

The bacteria can be of various sizes and shape i.e. spherical or round, coccus (pi. cocci), rod-shaped and bacillus (pi. Bacilli), spiral-shaped spirillum (pi. Priscilla). Long and helical shaped called spirochetes. Many bacteria have one or more slender, long flagellum (pi. flagella) which helps them to move in the liquid substrate. Some bacteria form endospores under poor nutrient conditions.

Question 2.
Write the distinct characters of protozoa.
Answer:
1. Protozoa are single-celled heterotrophs or ‘first animal’. They can be free-living and parasitic members, mobile with flagellar movement, by pseudopodia, or by ciliary movements e.g. Euglena and Amoeba.

2. Cell wall is absent in some like Amoeba so they can change their shape. The Euglena is autotrophic because of the presence of chlorophyll it performs photosynthesis but in the absence of light, it becomes heterotropic and ingests other protists or food particles.

3. They reproduce asexually by binary fission but some reproduce sexually by fusion of gametes followed by meiosis. Another important member of protists is the malarial parasite, Plasmodium, causing the notorious disease malaria in man, carried by mosquitoes, it multiplies rapidly in the liver of humans and brings about the cyclic fever releasing toxins into the bloodstream of its host.

Photograph of Amoeba and Euglenas

Here we are providing Class 11 Biology Important Extra Questions and Answers Chapter 1 The Living World. Important Questions for Class 11 Biology are the best resource for students which helps in Class 11 board exams.

Class 11 Biology Chapter 1 Important Extra Questions Living World

The Living World Important Extra Questions Very Short Answer Type

Question 1.
Name two organisms that do not reproduce?
Answer:
Mules, sterile worker bees.

Question 2.
Define ‘living’?
Answer:
Organisms exhibiting distinctive characters like growth, reproduction, etc. are called living.

Question 3.
Is regeneration a characteristic of living organisms?
Answer:
Yes, because fragmented organisms regain the lost part of the body.

Question 4.
What is biodiversity? or Define Biodiversity?
Answer:
The number and variety of organisms present on earth are referred to as biodiversity.

Question 5.
Name the International Authority who gives scientific name to the plants.
Answer:
International Code for Botanical Nomenclature (ICBN)

Question 6.
Write the scientific names of the following
(i) Mango
Answer:
Mangifera indica

(ii) Human
Answer:
Homo sapiens

(iii) Cat
Answer:
Felis Domestica

(iv) Tiger
Answer:
Panthera tigris.

Question 7.
What is taxonomy?
Answer:
Taxonomy is the science of classification that is grouping them on the basis of certain similarities.

Question 8.
How does taxonomy differ from systematics?
Answer:
Systematics is the study of the diversity of plants. The study of systematics leads to their taxonomic grouping.

Question 9.
What is a species?
Answer:
A population of identical individuals which can freely interbreed to produce fertile off-springs.

Question 10.
What is a taxon?
Answer:
A level of classification is called taxon e g., species, genus, family, etc. all are taxons.

The Living World Important Extra Questions Short Answer Type

Question 1.
How are zoological parks useful to biologists.
Answer:
Zoological parks are places where animals are maintained and allowed to breed in natural habitats.
(a) It gives information about endangered animals.
(b) Helps the biologists in developing hybrids with superior quality.
(c) Support the workers of biotechnology.

Question 2.
Write the universal rules of nomenclature.
Answer:

1. Biological names are generally in Latin and written in Italics. They are Latinised or derived from Latin irrespective of their origin.
2. the First word in a biological name represents the genus while the second component denotes a specific epithet.
3. Both the words in a Biological name when written in hand are separately underlined or printed in Italics to indicate their Latin origin.
4. First-world denoting genus starts with a capital letter while the specific epithet is written starting with a small word. It can be illustrated with the example of Mangifera indica.
5. The name of the author appears after a specific epithet i.e., the end of the biological name, and is written in the abbreviated form e.g. Mangifera indica (Linn). It indicates that species was first described by Linnaeus.

Question 3.
Explain about taxonomical aids/tools?
Answer:
Identification of organisms requires intensive laboratory and field studies. The information about an organism is collected and analyzed. The collection of actual specimens of plant species is essential and is a prime source of taxonomic studies.

These are also fundamental not only to study but also to training in systematics. It is used for the classification of an organism and the information gathered is also stored along with even the specimens. In some cases, the specimen is preserved for future studies.

Biologists have established certain procedures and techniques to store and preserve the information as well as the specimens. These techniques are, in fact, aids available for the identification and classification of organisms. The knowledge of these aids is quite helpful in biological studies. Some of these are explained to help to understand the usage of these aids.

Some of the taxonomical aids are

1. Herbarium,
2. Botanical Gardens
3. Museums
4. Zoological Parks
5. Keys.

Question 4.
“Consciousness is a defining property of living organisms.” Explain.
Answer:
Flora and fauna both respond to physical-chemical or biological environmental stimuli. Awareness of their surroundings makes organisms live. Mimosa pudiea respond to touch. Photoperiodic affects flowering in plants. Thus unicellular microscopic to multicellular huge organisms show the property of consciousness.

Question 5.
Reproduction can’t be an all-inclusive defining characteristic of living organisms? Illustrate the statement.
Answer:
In nature, there are many organisms that can’t reproduce. Mules, sterile worker bees are some examples of such organisms.

But the non-living object is strictly unable to reproduce.

Viruses are placed between living and non-living. They are crystallized like non-livings but replicate when enter inside living organisms.

The Living World Important Extra Questions Long Answer Type

Question 1.
Explain two defining characteristics of living organisms.
Answer:
Growth Unicellular and multicellular organisms increase their mass and number through cell-division. Non-livings increase their size by the accumulation of matter.
(a) Cell has protoplasm which is living matter. Cell before division increases their mass through replication of genetic matter. It is absent in non-livings.

(b) Metabolic Activity: Anabolic and catabolic reaction constantly occurs in living organisms, formation and conversion of biomolecules is metabolism.

‘In Vitro, such reactions can be maintained. In non-living, there is the absence of metabolism.

Question 2.
Explain the utility of systematics for classification.
Answer:
For classification, systematic studies have to carried out.

1. First, the organisms have to be described for all their morphological and other characteristics.
2. Based on its characteristic, it is seen whether it is similar (or different) to any known group or taxa-identification is carried out.
3. Based on its similar characteristic it is then placed in known taxa or the organism is classified. Sometimes organisms are very different from the ones already described anywhere in the world, then they are placed in a new group or ‘taxa’ and named.
4. Once the organism has been placed in the right taxa-the last step is nomenclature or naming. If the organism is already known-its the correct name is determined. If an organism is not described before-it is given a new name.

Here we are providing Class 11 Physics Important Extra Questions and Answers Chapter 12 Thermodynamics. Important Questions for Class 11 Physics with Answers are the best resource for students which helps in Class 11 board exams.

Class 11 Physics Chapter 12 Important Extra Questions Thermodynamics

Thermodynamics Important Extra Questions Very Short Answer Type

Question 1.
What type of process is Carnot’s cycle?
Answer:
Cyclic process.

Question 2.
Can the Carnot engine be realized in actual practice?
Answer:
No. It is an ideal heat engine.

Question 3.
A refrigerator transfers heat from a cold body to a hot body. Does this not violate the second law of thermodynamics?
Answer:
No. This is because external work is being performed.

Question 4.
What is a heat pump?
Answer:
A heat pump is a device that uses mechanical work to remove heat.

Question 5.
What forbids the complete conversion of work into heat?
Answer:
The second law of thermodynamics.

Question 6.
Does the internal energy of an ideal gas change in:
(a) an isothermal process?
Answer:
No.

(b) an adiabatic process?
Answer:
Yes.

Question 7.
What is the specific heat of a gas in an isothermal process and in an adiabatic process? Why?
Answer:
It ¡s infinite in isothermal process because ΔT = 0 (C = \(\frac{\Delta Q}{m \Delta T}\)) and zero in an adiabatic process as ΔQ = 0.

Question 8.
Can the temperature of an isolated system change?
Answer:
Yes, in an adiabatic process the temperature of an isolated system changes. It increases when the gas is compressed adiabatically.

Question 9.
Can we increase the coefficient of performance of a refrigerator by increasing the amount of working substance?
Answer:
No.

Question 10.
The door of an operating refrigerator is kept open in a closed room. Will it make the room warm or cool?
Answer:
The room will be slightly warmed.

Question 11.
How is the heat engine different from a refrigerator?
Answer:
A refrigerator is a heat engine working in the reverse direction.

Question 12.
What is the nature of the P – V diagram for isobaric and isochoric processes?
Answer:
The P – V diagram for an isobaric process is a straight line parallel to the volume axis while for an isochoric process, it is a straight line parallel to the pressure axis.

Question 13.
Mention two essential characteristics of an ideal heat engine.
Answer:

1. It should have a source of infinite thermal capacity.
2. It should have a sink of infinite thermal capacity.

Question 14.
Under that ideal condition can the efficiency of a Carnot engine be 100%?
Answer:
The efficiency of a Carnot engine can be 100% if the temperature of the sink is zero kelvin.

Question 15.
In summer, when the valve of a bicycle tube is removed, the escaping air appears cold. Why?
Answer:
This happens due to the adiabatic expansion of the air of the tube of the bicycle.

Question 16.
When the air of the atmosphere rises up, it cools. Why?
Answer:
When air rises up, it expands due to a decrease in the atmospheric pressure and thus temperature falls. So it cools.

Question 17.
Why does gas get heated on compression?
Answer:
Because work is done in compressing the gas increases the internal energy of the gas.

Question 18.
Which one among a solid, liquid, and gas of the same mass and at the same temperature has the greatest internal energy and which one has the least?
Answer:
Gas has the greatest internal energy and a solid has the least internal energy.

Question 19.
Can two isothermal curves intersect each other?
Answer:
No.

Question 20.
Does a gas work when it expands adiabatically is the source of energy?
Answer:
Yes.

Question 21.
Name the forces ¡n a system that makes the process taking place in it irreversible in nature.
Answer:
All types of dissipative forces in a system make the process taking place irreversible, e.g. force of friction is a dissipative force.

Question 22.
Is rusting of iron a reversible process? Why?
Answer:
Yes. Rusting of iron is an irreversible process as it is a chemical change.

Question 23.
State the conditions for irreversible processes.
Answer:
The following are the conditions for irreversible processes:

1. Rapidly carried out the process.
2. Dissipation of energy by any means.

Question 24.
How can we cause heat to flow from a body at a lower temperature to one at a higher temperature?
Answer:
This can be caused by doing work on an engine that works in a reverse direction.

Question 25.
Can water be made to boil without heating? How?
Answer:
Yes, by reducing the pressure on the water its boiling point can be brought to room temperature.

Question 26.
Can a room be cooled by leaving the door of an electric refrigerator opens?
Answer:
No.

Question 27.
How much is C_P greater than C_v?
Answer:
C_P is greater than C_v by R i.e. universal gas constant.

Question 28.
As C_P – C_v is constant for all gas, is C_P/C_v also constant for all gases?
Answer:
No.

Question 29.
What remains constant in an adiabatic process?
Answer:
Heat contents of the system remain constant in an adiabatic process.

Question 30.
Can heat be added to a system without changing its temperature?
Answer:
Yes.

Question 31.
Do water and ice have the same specific heat? Tell their values.
Answer:
No. For water specific heat is 1 and for ice, it is 0.5 in C.G.S. units.

Question 32.
What is the amount of work done in a cyclic process?
Answer:
The area enclosed under the P – V loop gives the amount of work done in a cyclic process.

Question 33.
What factors determine the internal energy of a real gas?
Answer:
Volume and temperature determine the internal energy of a real gas.

Question 34.
Can the temperature of gas be increased without adding heat into it? How?
Answer:
Yes. It can be done by compressing the gas adiabatically.

Question 35.
Why can’t the engine of a ship be operated using the internal energy of seawater?
Answer:
It is because a sink at low temperature is not available in the sea.

Question 36.
Is melting of ice an adiabatic or an isothermal process?
Answer:
The melting of ice is an isothermal process.

Question 37.
Can PV = RT describe an isothermal or adiabatic process?
Answer:
It can describe an isothermal process but not the adiabatic process.

Question 38.
Tell the main features of the first law of thermodynamics.
Answer:
It tells us about the conversion of mechanical energy into heat energy.

Question 39.
What are the main features of the second law of thermodynamics?
Answer:
It tells that entire heat energy can’t be converted into mechanical energy or heat can’t itself flow from lower temperature to higher temperature body.

Question 40.
What is the significance of knowing the ratio of C_P and C_v for a gas?
Answer:
It gives the atomicity of the gas as well as is an indicator of the degrees of freedom of gas molecules.

Question 41.
You enjoy shower bathing in summer but not in winter. Why?
Answer:
Water in the shower cools down due to adiabatic expansion, thus we would like to take bath in cold water in summer but not in winter.

Question 42.
Why the specific heats of argon, nitrogen, and ether are different at constant pressure?
Answer:
Argon, nitrogen, and ether have different atomicities. Hence additional heat is needed to raise the temperature by 1°C due to rotational and vibrational motion in bi, tri, and polyatomic gases, hence the difference.

Question 43.
Distinguish the formulae C_P – C_v = R, C_P – C_v = R/J and C_P – C_v = r/J.
Answer:

• C_P – C_v = R is used for a mole of the gas when C_P and C_v are measured in joule.
• C_P – C_v = R/J is used when C_P and C_v are measured in calories.
• C_P – C_v = r/J is used for a unit mass of gas, C_P and C_v are in calories.

Question 44.
Define an adiabatic process.
Answer:
It is defined as the process in which no heat enters or leaves the system.

Question 45.
State second law of thermodynamics.
Answer:
It states that heat itself can’t flow from a body at a lower temperature to a body at a higher temperature.

Question 46.
Define an isothermal process.
Answer:
It is defined as the process which takes place at a constant temperature.

Question 47.
Define an isotherm.
Answer:
It is defined as the pressure-volume curve for a fixed temperature.

Question 48.
Name the types of thermodynamic state variables.
Answer:
They are of the following two types:

1. extensive
2. intensive.

Question 49.
Define a cyclic process.
Answer:
It is defined as the process in which the system returns to its initial state.

Question 50.
(a) Is .the specific heat of water greater than that of sand?
Answer:
Yes. Actually, the specific heat of water is maximum.

(b) What is the important difference between the P – T diagram of water and that of CO2?
Answer:
The slope of the fusion curve is negative for water and positive for CO2.

Thermodynamics Important Extra Questions Short Answer Type

Question 1.
Kelvin and Clausius’s statements of the Second law of thermodynamics are equivalent. Explain?
Answer:
Suppose we have an engine that gives a continuous supply of work when it is cooled below the temperature of its surroundings.

This is a violation of Kelvin’s statement. Now if the work done by the engine is used to drive a dynamo which produces current and this current produces heat in a coil immersed in hot water, then w,e have produced a machine which causes the flow of heat from a cold body to the hot body without the help of an external agent. This is a violation of Clausius’s statement. Hence both statements are equivalent.

Question 2.
Two identical samples of gas are expanded so that the volume is increased to twice the initial volume. However, sample number 1 is expanded isothermally while sample number 2 is expanded adiabatically. In which sample is the pressure greater? Why?
Answer:
Pressure is greater in sample number 1 as can be explained: For isothermal expansion.
P₁V₁ = P₂V₂ for no. 1 sample
Now V₂ = 2V₁
∴ P₁V₁ = P₂2V₁
or
P₂ = \(\frac{P_{1}}{2}\) …(i)

Now for adiabatic expansion (for sample 2)
P₁V₁^γ = P₂V₂^γ
or
P₂ = P₁\(\left(\frac{\mathrm{V}_{\mathrm{1}}}{\mathrm{V}_{2}}\right)^{\gamma}\) = P₁\(\left(\frac{\mathrm{V}_{1}}{2 \mathrm{~V}_{1}}\right)^{\gamma}\)

= \(\frac{P_{1}}{2^{\gamma}}\) …(ii)

∴ From (i) and (ii) we find that pressure is greater in sample 1 as γ > 1.

Question 3.
No real engine can have an efficiency greater than that of a Carnot engine working between the same two temperatures. Why?
Answer:
A Carnot engine is an ideal engine from the following points of view:

1. There is no friction between the walls of the cylinder and the piston.
2. The working substance is an ideal gas i.e. the gas molecules do not have molecular attraction and they are points in size.

However these conditions cannot be fulfilled in a real engine and hence no heat engine working between the same two temperatures can have an efficiency greater than that of a Carnot, engine.

Question 4.
Explain why two isothermal curves cannot intersect each other?
Answer:
If they intersect, then at the point of intersection, the volume and pressure of the gas will be the same at two different temperatures which is not possible.

Question 5.
What is the source of energy when gas does work when expands adiabatically?
Answer:
During adiabatic expansion, the temperature and hence the internal energy of the gas decreases. Thus work is done by the gas at the cost of its internal energy.

Question 6.
State and explain the zeroth law of thermodynamics?
Answer:
It states that if two systems A and B are in thermal equilibrium with a third system C, then A and B must.be in thermal equilibrium with each other.

Explanation: The three systems are shown in the figure. Let T1, T2, T3 be the temperatures of A, B, and C respectively.

Systems A and C, B and C will exchange heat and after a certain time, they will attain thermal equilibrium separately.
i.e. T₁ = T₃ ….(1)
and T₂ = T₃ …. (2)

Thus from (1) and (2),
T₁ = T₂
i.e. A and B are now in thermal equilibrium with each other.

Question 7.
State and explain the first law of thermodynamics. What are the sign conventions?
Answer:
It states that if an amount of heat dQ is added to a system then a part of it may increase its internal energy by an amount dU and the remaining part may be used up as the external work dW done by the system i.e. mathematically,
dQ = dU + dW
= dU + PdV

Sign conventions:

1. Work done by a system is taken as positive while the work done on the system is taken as -ve.
2. The increase in the internal energy of the system is taken as positive while the decrease in the internal energy is taken as negative.
3. Heat added (gained) by a system is taken as positive and the heat lost by the system is taken as negative.

Question 8.
Why cannot a ship use the internal energy of seawater to operate the engine?
Answer:
The heat engine can convert the internal energy of seawater if there is a sink at a temperature lower than the temperature of seawater. Since there is no such sink and hence a ship can’t use the internal energy of seawater to operate the engine.

Question 9.
A certain amount of work is done by the system in a process in which no heat is transferred to or from the system. What happens to the internal energy and the temperature of the system?
Answer:
The temperature of the system decreases as the system is doing work and no heat transfer is allowed to or from the system. As the temperature of the system decreases, the internal energy of the system also decreases.

Question 10.
If an electric fan is switched on in a closed room, will the air of the room be cooled? Why?
Answer:
No. It will not be cooled, rather it will get heated because the speed of the air molecules will increase due to the motion of the fan. We feel cooler because of the evaporation of the sweat when the fan is switched on.

Question 11.
Define coefficient of performance.
Answer:
It measures the efficiency of a refrigerator. It is defined as the ratio of the quantity of heat removed per cycle from the contents of the refrigerator to the work done by the external agency to remove it. It is denoted by β or ω or K.

Question 12.
State the principle of a refrigerator.
Answer:
A refrigerator may be regarded as Carnot’s ideal heat engine working in the reverse direction. Thus when a Carnot engine works in opposite direction as a refrigerator, it will absorb an amount of heat Q₂ from the sink (contents of the refrigerator) at lower temperature T₂. As heat is to be removed from the sink at a lower temperature, an amount of work equal to Q₁ – Q₂ is performed by the compressor of the refrigerator to remove heat from the sink and then to reject the total heat
Q₁ = (Q₂ + Q₁ – Q₂)
to the source (atmosphere) through the radiator fixed at its back. It is also called a heat pump.

Question 13.
Derive the expression for the coefficient of performance.
Answer:
We know that by def. β = \(\frac{\text { Heat removed } / \text { cycle }}{\text { Work done } / \text { cycle }}\)

Question 14.
What do you conclude about the coefficient of performance?
Answer:

1. In actual practice, β varies from 2 to 6. For an actual refrigerator, the value of β is less than that calculated from equations (1) and (2).
2. The lesser the difference between the temperatures of the cooling chamber and the atmosphere, the higher is the p.
3. P can be much higher than 100% but η of a heat engine can never exceed 100%.
4. As the refrigerator works, T₂ goes on decreasing due to the formation of too much ice. There is practically no change in T₁. This decreases the value of β. However, if the refrigerator is defrosted, T₂ shall increase, and hence the value of β. So it is necessary to defrost the refrigerator.

Question 15.
Milk is poured into a cup of tea and is mixed with a spoon. Is this an example of a reversible process? Give reason.
Answer:
No. It is an example of an irreversible process. When milk is poured into a cup of tea and mixed, some work is performed and the same gets converted into heat. It is not possible to convert the heat produced back into work which will separate the milk from the tea.

Question 16.
Explain whether the following processes are reversible?
(a) Waterfall,
Answer:
The falling of water is not a reversible process. During the waterfall, the major part of its potential energy is converted into kinetic energy of the water and on striking the ground, a part of it is converted into heat and sound. It is not possible to convert the heat and sound produced along with the K.E. of water into the potential energy so that the water will rise back to its initial height.

(b) Electrolysis.
Answer:
It is a reversible process if the electrolyte does not offer any resistance to the flow of current. If we reverse the direction of the current, the direction of motion of ions is also reversed.

Question 17.
State the conditions for reversible and irreversible processes.
Answer:
Conditions for the reversible process:

1. The process should be slow enough so that at each stage of operation the system is in:
   (a) Mechanical equilibrium with the surroundings.
   (b) Chemical equilibrium with the surroundings.
2. There should not be a loss of energy due to friction.
3. Energy should not be lost due to conduction, convection, and radiation during the process.

Question 18.
Why cooling takes place when gas suffers adiabatic expansion?
Answer:
When a gas expands under adiabatic condition, then dQ = 0 and thus dU + dW = 0. So the gas does„external work for its expansion at the cost of internal energy due to which temperature of the gas falls and hence cooling takes place.

Question 19.
Why an engine working under isothermal conditions can produce no useful work?
Answer:
We know that
η = 1 – \(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\)

For the isothermal operation of the engine,
T₁ = T₂,
So η = 0
Hence an engine working under isothermal conditions can do no useful work.

Question 20.
What do you mean by internal energy?
Answer:
When work is done on the gas, it gets compressed and its temperature rises. The work done is converted into another form of energy called internal energy of the gas and the total energy remains conserved. The internal energy is chiefly the result of kinetic energy arising from the notion of the molecules inside the gas.

Question 21.
Why C_p is greater than C_v for a gas?
Answer:
When heat is given to a gas keeping its volume constant whole of it is used to raise the kinetic energy hence, the temperature of the gas. But if the pressure is to be kept constant, then an additional amount of heat is needed to raise the temperature of the same amount of gas by the same extent as the gas has to do work against pressure in the expansion (volume is now increased). Thus according to the definition of C_p and C_v, C_p>C_v.

Question 22.
Why a room cannot be cooled by leaving. the door of a refrigerator open?
Answer:
This room cannot be cooled because the refrigerator takes out heat from stuff kept inside the refrigerator and rejects it to the surroundings i.e. room.

Question 23.
Explain how heat can be added to a system without changing its temperature?
Answer:
if heat is added to a system to do only the external work, then there will be.no a rise in temperature.

Question 24.
Why a stove-pin does not burn one’s fingers even when its other end is red hot?
Answer:
This is because of the fact that the pin, has a very small area of cross-section, thus the amount of heat conducted which is directly proportional to the area of the cross-section is very small.

Question 25.
Although mechanical energy can be fully converted into heat energy, yet the reverse is not true? Why?
Answer:
This is because a part of heat energy is always retained by the system in the form of its internal energy.

Question 26.
How can a refrigerator be used as a heat pump to heat a room in winter?
Answer:
Install the refrigerator in a window such that its back is inside the room. The refrigerator will now reject the heat inside the room which will get warmed up.

Question 27.
What is the difference between the internal energy of an idea and a real gas?
Answer:
The internal energy of an ideal gas depends upon K.E only while for a real gas, it depends upon both K.E and P.E

Question 28.
Why can’t a Carnot engine be ever designed?
Answer:
It is because the conditions imposed on the source, sink, insulating stand, cylinder-piston, and the working substance can’t be achieved fully.

Question 29.
Air escaping from a cycle tube becomes cool on removing the value. Why?
Answer:
The air in a cycle tube is present at a present greater than the atmospheric pressure. When the valve is removed, the air expands suddenly i.e. adiabatic expansion takes place and the temperature decreases, thus the tube becomes cool.

Question 30.
Explain the need fofthesecbndiaWbf thermodynamics.
Answer:
Its need is as follows:

1. The heat flows from a body at a higher temperature to a body at a lower temperature, but why not from a body at a lower temperature to a higher temperature.
2. Heat engine converts heat into Work, but Why is the efficiency of an engine less than one?
3. When we rotate a paddle in a beaker containing water, we know that the Work was done is converted into heat but when we put a paddle in hot water, no mechanical work is done. Why?

The above questions lead us to the need for the second law of thermodynamics.

Thermodynamics Important Extra Questions Long Answer Type

Question 1.
Discuss the Carnot cycle and give essential features of a Carnot engine.
Answer:
Carnot cycle: Heat engines essentially have

1. a source of heat,
2. a W0fkthg substance
3. a sink (at a temperature lower than that source) and
4. mechanical parts.

Carnot designed an idea engine that operated in the reversible cycle. The cycle consisted of two isotherms and two adiabatic. The heat was taken in or rejected during isothermal expansion or contraction. The Carnot cycle thus consists of four steps (see fig.) Carnot took a perfect gas as the working substance enclosed in a cylinder with perfectly insulting walls fitted with an insulating piston but the bases of the cylinder were conducting

(1) In the first step of the cycle let P₁, V₁, by the pressure of the gas. It is placed, in contact with the. source of heat at temperature T1 i.e the cylinder is out on the source. As the gas expands isothermally it absorbs some amount of heat to keep the temperature constant (curve AB)

The heat absorbed from the source Q₁ is equal to the work done W, in expanding the gas volume from V₁ to V₂ at temperature T₁ so that


In = Area ABMKA ….(1)

(2) The cylinder is put on insulating and gas is allowed to expand from V₂ to V₃ adiabatically. Its temperature falls from T₁ to T₂ and pressure becomes P₃ and P₂. The work done W is then

(3) In this part of the cycle the cylinder is put with its conducting base in contact with a sink as temperature T2 and gas is compressed isothermally. It rejects Q₂ heat at constant temperature T₂, the work done on the gas is [pressure volume change to (P₄, V₄) from (P₃, V₃)].

(4) In the last step of the cycle, the cylinder’s base is again put on the insulating stand, and the gas is compressed adiabatically so that the system returns back to its original state at A i.e. from (P_4, V₄) to (P₁, V₁) at temperature T₁ via curve DA. Now the work done on the gas is

= Area DLKAD …. (4)

From equation (2) and (4), it is clear that W₄ = W₂
If W = net work done by the engine in one cycle, then
W = W₁ + W₂ + (- W₃) + (- W₄)
= W₁ – W₃ = Area ABCDA = Q₁ – Q₂ …..(5)

The efficiency of the Carnot engine (η): It is defined as the ratio of work done by the engine to the energy supplied to the engine in a cycle.
i.e η = \(\frac{\mathrm{W}}{\mathrm{Q}_{1}}=\frac{\mathrm{Q}_{1}-\mathrm{Q}_{2}}{\mathrm{Q}_{1}}\)
= 1 – \(\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}}\)

Using equations (1) and (3)
\(\frac{Q_{1}}{Q_{2}}=\frac{R T_{1} l n \frac{V_{2}}{V_{1}}}{R T_{2} l n \frac{V_{3}}{V_{4}}}\) …(7)

Since B and C lie on the same adiabatic so
T₁V₂^γ-1; = T₂V₃^γ-1
or
\(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{V}_{3}}{\mathrm{~V}_{2}}\right)^{\gamma-1}\) ….(8)

Also D and A lie on the same adiabatic so
T₁V₁^γ-1 = T₂V₄^γ-1
or
\(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{V}_{4}}{\mathrm{~V}_{1}}\right)^{\gamma-1}\) ….(9)

∴ from (8) and (9), we get
\(\left(\frac{V_{3}}{V_{2}}\right)^{\gamma}\) = \(\left(\frac{V_{4}}{V_{1}}\right)^{\gamma}\)
ln \(\frac{\mathrm{V}_{3}}{\mathrm{~V}_{\mathrm{4}}}\) = ln \(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{\mathrm{1}}}\) …(10)

∴ from (7) and (10), we get

1. The interesting aspect of η of Carnot engine is that it is independent of the nature of the working substance. But Carnot used an ideal gas operation which is not strictly followed by real gases or fuel
2. Theoretically, η can be 100%.
3. The efficiency of Carnot’s ideal engine depends only on the temperature of the scarce and the sink.
4. The efficiency of any reversible engine working between the same two temperatures is the same.

Question 2.
Derive the expression for the work done during:
(a) Isothermal process
(b) Adiabatic process
Answer:
Consider one mole of a perfect gas contained in a cylinder having conducting walls and fitted with a movable piston.

Let P, V be the pressure and volume of the gas corresponding to this state.
Let dx = distance by which piston moves outward at constant pressure P so that its volume increases by dV.
Let a = area of cross-section of the piston.

(a) If dW = work done in moving the piston by dx, then .
dW = force on piston × dx
= P a dx
= PdV …(i)
Where dV = a dx = volume
Let the system goes from initial state A(P₁, V₁) to final state B(P₂, V₂)

If W = total work done from A to B, then

= RT (log_eV₂ – logeV₁)
= RT log₂ \(\frac{V_{2}}{V_{1}}\)
= 2.303 RT log₁₀ \(\frac{V_{2}}{V_{1}}\)

(b) From equation (ii) of case (a), we get

We know that an adiabatic process is represented mathematically by the equation:
PV^γ = constant = K
or
P = \(\frac{\mathrm{K}}{\mathrm{V}^{\gamma}}\) …(iii)

∴ from (ii) and (iii), we get

Numerical Problems:

Question 1.
A gas is suddenly compressed to 1/3 of its original volume. Calculate the rise in temperature, the original temperature being 300K and γ = 1.5.
Answer:
Let V₁ = Initial volume
V₂ = Final volume = \(\frac{\mathrm{V}_{1}}{3}\)
or
\(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\) = 3
T₁ = 300K
T₂ – T₁ = ?
γ = 1.5

We know that for an adiabatic change,

Question 2.
The efficiency of an ideal engine is \(\frac{1}{8}\). By lowering the temperature of the sink by 100 K, it increases to \(\frac{1}4}\). Find the initial and final temperature of the sink.
Answer:
Here, η₁ = \(\frac{1}{8}\) in 1st case
η₂ = \(\frac{1}{4}\) in 2nd case

Let T₁ = Temperature of source
T₂ = Initial Temperature of sink
and
T’₂ = final Temperature of sink
= T₂ – 100

∴ Using the relation

T₁ = 800 K …(3)

from (1) and (3), we get
T₂ = \(\frac{7}{8}\) × 800 = 700 K
∴ T’₂ = 600 K

Question 3.
A perfect Qarjiotreiigifae utilizes an ideal gas. The source temperature is 500K and since the temperature is 375 K. If the engine takes 600 Kcal per cycle from the source, compute:
(a) the efficiency of The engine.
Answer:
Here, T₁ = 50.0 K
T₂ = 375 k
Q₁ = Heat absorbed per cycle
= 600 K cal

∴ (a) Using t|ig relation,
η = 1 – \(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{\mathrm{1}}}\), we get
η = \(\frac{T_{1}-T_{2}}{T_{1}}=\frac{500-375}{500}\)
= \(\frac{125}{500}\) = 0.25

η% = 0.25 × 100 = 25%

(b) work done per cycle,
Answer:
Let W = work done per cycle

∴ Using relation
η = \(\frac{\mathrm{W}}{\mathrm{Q}_{1}}\), we get

W = ηQ1
= 0.25 × 600 K cal
= 150 K cal
= 150 × 10³ × 4.2 J
= 6.3 × 10⁵ J.

(c) heat rejected to the sink per cycle.
Answer:
Let Q₂ = heat rejected to the sink

∴ Using the relation
W = Q₁ – Q₂, we get
Q₂ = Q₁ – W = 600 – 150 = 450 K cal

Question 4.
A refrigerator has, to transfer an average of 263 J of heat per second from temperature – 10°C to 25°C. Calculate the average power consumed assuming ideal reversible cycle and no other l0sses.
Answer:
Here, T₁ = 25 + 273 = 298 K
T₂ = – 10 + 273 = 263 K
Q₂ = 263 Js^-1

we know that

= 298 Js^-1

∴ Average power consumed = Q₁ – Q₂
= (298 – 263) Js^-1
= 35 W

Question 5.
Assuming the domestic refrigerator as a reversible engine working between the melting point of ice and the room temperature of 27°C, calculate the energy in Joule that must be supplied to freeze one kg of water. Given the temperature of water 0°C, L = 80 cal g^-1.
Answer:
Here, T₁ = 27 + 273 = 300 K
T₂ = 0 + 273 = 273 K
m = 1 kg = 1000 g
L = 80 cal g^-1

Heat to be removed, Q₂ = mL
= 1000 × 80 cal
= 8 × 10⁴ cal

Using the relation
\(\frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\), we get
Q₁ = \(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\) × Q2 = \(\frac{300}{273}\) × 8 × 10⁴
= 87912.1 cal

Energy required to be supplied,
W = Q₁ – Q₂
= (87912.1 – 80000) cal
= 7912.1 cal
= 7912.1 × 4.2 J
= 33230.8 J.

Question 6.
What is the coefficient of performance (β) of a Carnot refrigerator working between 30°C and 0°C?
Answer:
Here, T₂ = 0°C = 273 K
T₁ = 30°C = 273 + 30 = 303 K
β = ?
Using the relation,
β = \(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}-\mathrm{T}_{2}}\), we get
β = \(\frac{273}{303-273}=\frac{273}{30}\) = 9.1

Question 7.
A certain volume of dry air at NTP is allowed to expand 4 times its original volume under
(a) isothermal conditions
(b) adiabatic conditions.
Calculate the final pressure and temperature in each case γ = 1.4.
Answer:
Here, let V₁ = V
∴ V₂  = 4V
P₁ = 76 cm of Hg
P₂ = ?
γ = 1.4
T₁ = 273 K
T₂ = ?

(a) For isothermal expansion,
P₁V₁ = P₂V₂
or
P₂ = P1\(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\)
= \(\frac{76}{4}\) = 19 cm of Hg.

Since the process is isothermal, therefore the final temperature will be the same, as the initial temperature.
T₁ = 273 K

(b) Adiabatic expansion Using the relation
P₁V₁^γ = P₂V₂^γ

Question 8.
A liter of hydrogen at 27°G and 106 dyne cm^-2 pressure expands isothermally until its volume is doubled and then adiabatically until redoubledi’Find the final temperature, pressure and work done in each case, γ = 1.4
Answer:
Here, P₁ = 106 dyne cm^-2
V₁ = 103 cm3 = 1 litre
T₁ = 27 °C = 300 K.
V₂ = 2V₁
= 2000 cm³ for isothermal expansion
T₂ =?
P₂ =?

For adiabatic expansion
V₁ = 2000 cm³
V’₁ = 2V’₁ = 4000 cm³

(a) For isothermal expansion: Using the relation
P₂V₂ = P₁V₁,we get
P₂ = \(\frac{P_{1} V_{1}}{V_{2}}=\frac{10^{6} \times 10^{3}}{2 \times 10^{3}}\)
= 5 × 10⁵ dyne cm^-2

T₂ = T₁ for isothermal expansion = 300 K = 27°C

Work done during isothermal expansion is given by
W = 2.303 RT log₁₀\(\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)\)
= 2.303 P₁V₁ log₁₀\(\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)\)
= 2.303 × 10⁶ × 10³ log₁₀2
= 2.303 × 10⁹ × 0.3010
= 6.93 × 10⁸ erg

(b) For adiabatic expansion:

= – 45.6 °C

Work dorie during adiabatic expansion is given by

Question 9.
A 50g lead bullet (specific heat 0.02) is initially at 30°C. It is fired vertically upward with a speed of 840 ms^-1. On returning to the starting level, it strikes a cake of ice at 0°C. How much ice is melted? Assume that all energy is spent in melting only. Latent heat of ice = 80 cal g^-1.
Answer:
When bullet falls to the same level its velocity is same i.e. 840 ms^-1.
K.E. of bullet on returning to the same level
= \(\frac{1}{2}\)mv² = \(\frac{1}{2}\) × \(\frac{500}{1000}\) × (840)²
or
W = 17640 J
This K.E. will be converted into heat energy on striking the cake of ice.
∴ Heat produced Q is given by
Q = \(\frac{\mathrm{W}}{\mathrm{J}}=\frac{17640}{4.2}\) = 4200 cal

Heat given up by the bullet ih cooling to 0°C from 30°C is given by
Q₂ = mcΔθ = 50 × 0.02 × 30
= 30 cal

∴ Total heat given to the ice = 4200 + 30 = 4230 cal
Let m = mass ice melted
∴ 4230 = mL
or
m = \(\frac{4230 \mathrm{cal}}{\mathrm{L}}=\frac{4230}{80 \mathrm{cal} \mathrm{g}^{1}}\)
or
m = 52.875 g.

Question 10.
Two moles of He gas (γ = \(\frac{5}{3}\)) J are initially at temperature 27°C and occupy a volume of 20 liters. The gas is first expanded at constant pressure until the volume is doubled. Then it undergoes an adiabatic change until the temperature returns to its initial value.
(a) Sketch the process on a P – V diagram.
(b) What are the final volume and pressure of the gas?
(c) What is the work done by the gas?
Answer:
Here, T₁ = 27 + 273 = 300 K
V₁ = 20 × 10^-3 m^-3
R = 8.3 J mol^-1 K^-1
n = 2

Using P₁V₁ = nRT₁, we get
P₁ = \(\frac{\mathrm{nRT}_{1}}{\mathrm{~V}_{1}}=\frac{2 \times 8.3 \times 300}{2 \times 10^{-2}}\)
= 24.9 × 10⁴ = 2.49 × 10⁵ Nm^-2

Now when the gas is expanded at constant pressure, then let V₂ and T₂ be the new values of volume and temperature.
∴ According to Charle’s law
\(\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}}\)
Here, V₁ = 2V₁

∴ T₂ = \(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\) × T₁
= 2 × 300
= 600 K

∴ V₂ = 2 × 2 × 10^-2 = 4 × 10^-2 m^-3

Finally gas expands adiabatically.
Let P₃, V₃, T₃ be the new values of pressure, volume and temperature.
Here, T₃ = T₁ = 300 K


The process is shown in the figure:

Total work done (W) is given by
W = work done (W₁) during expansion at constant pressure + work done (W₂) during adiabatic expansion
or
W = W₁ + W₂

Question 11.
The motor in a refrigerator has a power output of 250 watts. The freezing compartment is at 270 K and outside air at 300 K. Assuming ideat efficiency, what is the amount of heat that can be extracted from the freezing compartment in 10 minutes? What is the shortest time in which 10 kg of water at 273 K can be converted into ¡ce? J = 4.2 × 10^-3 J K ca^-1.
Answer:
We know that β = \frac{\mathrm{Q}_{2}}{\mathrm{~W}}=\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}-\mathrm{Q}_{2}}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}-\mathrm{T}_{2}}\(\)
Here, T₁ = 300 K
T₂ = 270 K
W = 250 W = 250 J s^-1
Q =?
t =?

∴ Q₂ = Wβ = W\(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}-\mathrm{T}_{2}}\)
= 250 × \(\frac{270}{300-270}\)
= 250 × \(\frac{270}{30}\)
= 2250 J s^-1

(i) Let Q = heat extracted from the freezing compartment in 10 minutes = ?
∴ Q = Q₂ × 10 min = 2250 × 10 × 60
= 1350000 J
= \(\frac{135 \times 10^{4}}{4.2 \times 10^{3}}\) K cal.
= 321.4 K cal.

(ii) Heat required to convert 1 Kg of water at 273 K into ice,
Q’ = m × L = 1 × 80 K cal = 80 × 4.2 × 10³ J

Let Q’ be extracted in a time t.
∴ rate of extraction of heat from freezing compartment
= \(\frac{80 \times 4.2 \times 10^{3}}{\mathrm{t}}\) J s^-1

This rate must be equal to Q₂
i.e. 2250 = \(\frac{80 \times 4.2 \times 10^{3}}{\mathrm{t}}\)
= 149.33 s.

Question 12.
In a refrigerator, heat from inside at 277 K is transferred to a room at 300 K. How many joules of heat will be delivered to the room for each joule of electric energy consumed ideally
Answer:
Here, T₂ = 277 K
T₁ = 300 K
Q₁ =?
W = 1 J

Using the relation
β = \(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}-\mathrm{T}_{2}}\), we get
β = \(\frac{277}{300-277}=\frac{277}{23}\) = 12.04

Also we know that
β = \(\frac{\mathrm{Q}_{2}}{\mathrm{~W}}=\frac{\mathrm{Q}_{2}}{\mathrm{l}}\)
or
β = Q₂
∴ Q₂ = 12.04 J
Now W = Q₁ – Q₂
or
Q₁ = W + Q₂
or
Total heat delivered to the room,
Q₁ = 1 + 12.04 = 13.04 J.

Question 13.
A Carnot-type engine is designed to operate between 480 and 300 K. Assuming that the engine actually produces 1.2 K J of mechanical energy per kilo cal of heat absorbed, compare the actual efficiency with the theoretical maximum efficiency.
Answer:
Here, T₁ = 480 K
T₂ = 300 K
W = 1.2 K J
Q₁ = heat absorbed =1 K cal = 4.2 K J

If ηa be its actual efficiency, then
η_a % = \(\frac{\text { Energy output }}{\text { Energy input }}\) × 100
= \(\frac{\mathrm{W}}{\mathrm{Q}_{1}}\) × 100
= \(\frac{1.2}{4.2}\) × 100 = 28.57%

Also, let ηm be its maximum theoretical efficiency,

i.e. η_a is nearly \(\frac{3}{4}\) th of η_m.

Question 14.
20 m³ of a monoatomic gas at 12°C and pressure 100 K Pa is suddenly compressed to 0.5 m³. What are its new temperature and pressure? y for monoatomic gas = 1.67.
Answer:
Here, P₁ = 100K Pa
V₁ = 20m3
t₁ = 12°C

∴ T₁ = 273 + 12 = 285 K
γ = 1.67

V₂ = 0.5 m³
= \(\frac{1}{2}\) m³
P₂ =?
T₂ =?

(a) Using the relation,
T₁V₁^γ-1 = T₂V₂^γ-1, we get
T₂ = T₁\(\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma-1}\) = 285\(\left(\frac{20}{1 / 2}\right)^{167-1}\)

= 285 × (40)0.67
= 285 × 11.84
= 3374 K.

(b) Again using the relation
P₁V₁^γ = P₂V₂^γ, we get
P₂ = P₁\(\left(\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}\right)^{\gamma}\)
= 100 × \(\left(\frac{20}{1 / 2}\right)^{1.67}\)
= 100 × (40)^1.67
= 100 × 474
= 474 × 10² K Pa.

Question 15.
1 m³ of Helium originally at OC and 1 atmospheric pressure is cooled at constant pressure until the volume ¡s 0.75 m³. Calculate the heat removed. Sp. the heat of He = 3.1 KJ/kg K.
Answer:
Here, P₁ = 1 atm = 1.013 × 10⁵ Pa
V₁ = 1 m³
T₁ = 273 + 0 = 273 K
P₂ = 1 atm
V₂ = 0.75 m³ = \(\frac{3}{4}\) m³

Let T₂ = final temperature
C_v = 3.1 KJ/kg K
∴ ΔV = (\(\frac{3}{4}\) – 1) m³

Using equation

According to first law of thermodynamics,
ΔQ = ΔU + ΔW

For an idea gas, ΔU = μ Cv ΔT
ΔQ = μ Cv ΔT + PΔV

Also at S.T.P., one mole of a gas occupies a volume
= 22400 cc
or
1 K_mol = 22.4 m³

∴ 1 m³ volume contain = \(\frac{1}{22.4}\) K mol
∴ ΔQ = \(\frac{1}{22.4}\) × 3.1 (205 – 273) + \(\frac{1.013 \times 10^{5}(0.75-1)}{4184}\)
= – 9.11 – 6.05 × 10^-5 × 10⁵
= – 9.11 – 6.05
= – 15.2 K cal
Here – ve sign shows that heat is removed.

Question 16.
Determine the P – V relation for a monoatomic ideal gas undergoing an idiabatic process.
Answer:
For adiabatic process, 1st law of thermodynamics
dQ = dU + dW becomes
dU = – dW (∵ dQ = 0) …. (1)

The U for a monoatomic ideal gas is given by
U = \(\frac{3}{2}\) NkT
∴ dU = \(\frac{3}{2}\) NkdT …. (2)

Also dW = PdV ….(3)
Using (1), (2) and (3), we get


integrating equation (4), we get

log T + \(\frac{2}{3}\) logV = constant
or
log (TV^2/3) = constant
or
TV^2/3 = constant

In terms of P and V, PV^5/3 = constant (∵ PV = RT)

Question 17.
The specific heat capacity of water is 4200 J/kg K. What is the change In the internal energy of 5 kg of water when it is heated from 20°G to 80°C?
Answer:
Here, C = 4200 J/kg K
m = 5 kg
T₁ = 20°C + 273 = 293 K
T₂ = 80 + 273 = 353 K
ΔT = T₂ – T₁
= 353 – 293 = 60 K
ΔU = ?

If ΔQ be the heat supplied to water for its heating, then using
ΔQ = mCΔT, we get
ΔQ = 5 × 4200 × 60
= 1260 KJ

Let us ignore the slight expansion in the volume of water,
∴ ΔW = PdV = 0
∴ ΔQ = ΔU + ΔW, gives
ΔU = ΔQ = 1260 KJ.

Question 18.
The specific heat of argon at constant pressure is 0.125 cal/gm and at constant volume is 0.075 cal g^-1. Calculate the density of argon at N.T.P. J = 4.18 × 10⁷ ergs ca^-1 and normal pressure = 1.01 × 106 dynes cm^-2.
Answer:
Here, C_p = 0.125 cal g^-1
C_v = 0.075 cal g^-1
J = 4.18 × 10⁷ ergs cal^-1
p = 1.01 × 10⁶ dynes cm²
C_p – C_v = 0.125 – 0.075
= 0.050 cal g^-1

Let ρ = density of argon at N.T.P. = ?
Using the relation,

Question 19.
One gram of water at 373 K is converted into steam at the same temperature. Volume of 1 c.c. of water becomes 1671 cm³ on boiling. Calculate the change in internal energy of the system if heat of vaporization is 540 cal g^-1. Atmospheric pressure is 1.013 × 10⁵ Nm^-2
Answer:
Here, m = mass of water = 1 g
V₁ = initial volume = 1 cm³
V₂ = final volume = 1671 cm³
P = atmospheric pressure = 1.013 × 10⁵ Nm^-2
L = heat of vaporization = 540 cal g^-1

∴ dV = change in volume
= V₂ – V₁
= 1671 – 1 = 1670 cm³
= 1670 × 10⁶ m³

∴ dW = PdV
= 1.013 × 10⁵ × 1670 × 10^-6 J
= 169.17 J
= \(\frac{169.17}{4.2}\) cal
= 40.3 cal
and dQ = mL = 1 × 540 cal

Let dU = change in internal energy of the system = ?
∴ According to first law of thermodynamics,
dQ = dU + dW
or
dU = dQ – dW
or
dU = mL – PdV
= 1 × 540 – 40.3
= 499.7 cal.

Question 20.
A perfect Carnot engine utilizes an ideal gas. The temperature of the source is 500 K and that of the sink is 375 K. If the k engine takes 600 Kcal per cycle from the source, then calculate:
Answer:
Here, T₁ = temp, of source = 500 K
T₂ = temp, of sink = 375 K
Q₁ = heat absorbed from the source per cycle
= 600 Kcal

(i) the efficiency of the engine.
Answer:
Let η = thermal efficiency of the Carnot engine,
then η = \(\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{500-375}{500}\)
= \(\frac{125}{500}\) = 0.25

∴ η = 0.25 × 100 = 25%

(ii) work done per cycle in Joule.
Answer:
Let W be the work done/cycle, then
η = \(\frac{\text { Work done per cycle }}{\text { Heat absorbed per cycle }}=\frac{\mathrm{W}}{\mathrm{Q}_{1}}\)
or
\(\frac{25}{100}=\frac{\mathrm{W}}{600}\)
or
W = 25 × 6 = 150 K cal
= 150 × 10³ cal
= 150 × 10³ × 4.2 J
= 6.3 × 10⁵ J.

(iii) heat rejected to the sink per cycle.
Answer:
Let Q₂ = heat rejected to the sink per cycle, then
Q₁ = W + Q₂
or
Q₂ = Q₁ – W
= 600 – 150
= 450 K cal.

Question 21.
A diatomic gas is heated at constant pressure. What fraction of the heat energy is used to increase the internal energy?
Answer:
Let ΔQ = heat given to the gas
ΔU = increase in the internal energy
ΔW = work done from first law of thermodynamics,

∴ from first law of thermodynamics,
ΔQ = ΔU + ΔW

For diatomic gas,
ΔQ = C_p ΔT
and ΔU = C_v ΔT

∴ \(\frac{\Delta U}{\Delta Q}=\frac{C_{V}}{C_{P}}=\frac{\frac{5}{2} R}{\frac{7}{2} R}=\frac{5}{7}\)

Question 22.
A gas is enclosed in a vessel of volume 1000 cc at a pressure of 72.6 cm of Hg. It is being evacuated with the help of a piston pump, which expels 10% gas in each stroke. What will be the pressure after the second stroke?
Answer:
Here, V₀ = initial volume = 1000 cm³
P₀ = initial pressure = 72.6 cm of Hg
\(\frac{\Delta \mathrm{V}}{\mathrm{V}_{0}}\) = rate of expelling the gas per stroke
= 10% = \(\frac{10}{100}=\frac{1}{10}\)

∴ After first stroke,
P₁(V₀ + ΔV) = P₀V₀

∴ After second stroke,
P₂(V₀ + ΔV) = P₁V₀
or
P₂ = P₀ \(\left(\frac{\mathrm{V}_{0}}{\mathrm{~V}_{0}+\Delta \mathrm{V}}\right)^{2}\)
= P₀\(\left(\frac{1}{1+\frac{\Delta V}{V_{0}}}\right)^{2}\)

Where P₁ and P₂ are the pressures after first and second strokes.
∴ P₂ = 72.6\(\left(\frac{1}{1+\frac{1}{10}}\right)^{2}\)
= 72.6 × \(\left(\frac{10}{11}\right)^{2}\)
= \(\frac{726}{10} \times \frac{100}{121}\)
= 60 cm of Hg.

Question 23.
A gas expands with temperature according to the relation V = kT^2/3. What is the work done when the temperature changes by 30°C?
Answer:
Here, V = kT^2/3

= R × \(\frac{2}{3}\)(T₂ – T₁)
= \(\frac{2}{3}\)R × 30 = 20 R.

Question 24.
A Carnot engine has the same efficiency when working:
(a) between 100 K and 500 K and
(b) between 180 K and T K.
Calculate the temperature T of the source in case (b).
Answer:
Here,
(a) T₁ = temperature of source = 500 K
T₂ = temperature of sink = 100 K

(b) T’₁ = temperature of source = T K = ?
T’₂ = temperature of sink = 180 K

Let η and η’ be the efficiency in case (a) and (b) respectively.
As η = η’ (given)

Question 25.
An ideal gas in a cylinder is compressed adiabatically to one-third of its original volume. During the process. 45 J of work is done on the gas by the compressing agent. By how much did the internal energy of the gas change in the process? How much heat flowed into the gas.
Answer:
Here, ΔQ = 0, ΔW = – 45 J, ΔU = ?
According to first law of thermodynamics,
ΔQ = ΔU + ΔW, we get
ΔU = 0 – ΔW
= – (- 45) = 45 J .
As the process is adiabatic, so the heat flow is zero.

Value-Based Type:

Question 1.
Amita who is a student of class XI asked a question to her friend Smita. Smita let me answer: Can you design a heat engine of 100% efficiency? Explain your answer. She was totally confused and did not able to answer it. Then, Amita explained it as under:
(i) Which value is displayed by Amita?
Answer:
The values displayed by Amita are:
(a) Helping nature
(b) Keen to share, her ideas
(c) Interested in teaching
(d) Co-operative.

(ii) How did she explain it?
Answer:
The efficiency of a heat engine is given by
η = 1 – \(\frac{T_{2}}{T_{1}}\)

The efficiency will be 100% or 1 if T₂ = OK
Since these conditions cannot be attained practically.

So, a heat engine can not have 100% efficiency:

Question 2.
Sweta, a student of class XI asked her physics teacher “Can a room be cooled by opening the door of a refrigerator in a closed room?
Her teacher answered as under:
No, when a refrigerator is working in a closed room with its door closed, it is rejecting heat from inside to the air in the room. So the temperature of the room increases gradually.
When the door of the refrigerator is kept open, heat rejected by the refrigerator to the room will be more than the heat taken by the refrigerator from the room (by an amount equal to work done by the compressor). Therefore, the temperature of the room will increase at a slower rate compared to the first case.
(i) What values are displayed by Sweta?
Answer:
The values shown by Sweta are: Sharp mind, observing nature, awareness, scientific attitude.

(ii) Is there any case when the opening of the door of a refrigerator will cool the room gradually?
Answer:
Yes, if a heat-rejecting portion of the refrigerator is outside the closed room, then the opening of the refrigerator’s door will cool the room gradually.

(iii) A refrigerator is to remove heat from the eatable kept inside at 10°C. Calculate the coefficient of performance, if the room temperature is 36°C.
Answer:
Here, T₁ = 36°C = (36 + 273) K = 309 K
T₂ = 10°C = (10 + 273)K = 283K

Coefficient of performance = \(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}-\mathrm{T}_{2}}=\frac{283}{309-283}\)

= \(\frac{283}{26}\) = 10.9

Question 3.
Shikhar told his friend that the transfer of heat can not take place from a colder body to a holler body. His friend Raman was surprised as there were no such explanations in the first law of thermodynamics about the direction in which change can occur. Finally, they went to ask their physics teacher the reason behind it.
(i) What values are shown by them?
Answer:
The values shown by them are: Curiosity, Group Discussion, and keenness to know the scientific reason.

(ii) How did the teacher explain it?
Answer:
The teacher explained that this is the limitation of the first law of thermodynamics. However, heat transfer occurs from the body, at higher temperatures to lower temperatures. According to the second law of thermodynamics, ds ≥ 0 i.e system cannot move towards the direction of decreasing entropy/probability. Hence, heat cannot flow by itself from a colder body to a hotter body.