Thermodynamics

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (b), (d) When the rod is hammered the external work is done on the rod which increases its temperature.

Heat is transferred to the gas in the small container by big reservoir at temperature T2

As the weight is added to the cylinder arrangement in the form of external pressure so it cannot reversed.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) Let us assume that T1, T2T3

According to questions there is no loss of heat in the surroundings

Heat lost by M3 = heat gained by M1+ heat gained by M2

M3s (T3-T)= M1s (T-T1)+M2s (T-T2)

T [M1+M2+M3]= M3T3+M1T1+M2T2

T = M 1 T 1 + M 2 T 2 + M 3 T 3 M 1 + M 2 + M 3

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a) Container A is isothermal and container B is adiabatic

For isothermal process P1V1=P2V2

Po (2Vo)= P2 (Vo)

P2= 2Po

for adiabatic process

P1V1y= P2V2y

Po (2Vo)y=P2 (Vo)y

P2= ( 2 V o V o )yPo= 2yPo

Hence ratio of final pressure = 2 γ P o 2 P o = 2 γ - 1

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) Work done ABCD = area of rectangle ABCDA

= AB * B C = (3Vo-Vo) * (2po-po)

= 2V0 * Po= 2poVo

And work done by the gas =- 2poVo

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c) As we know PV =constant

Hence we can say that gas is going through an isothermal process.

Clearly from the graph that between process 1 and 2 temperature is constant and the gas expands and pressure decreases. So density of 2 is less than 1 so option ii

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a) Amount of sweat evaporated /minute = s w e a t p r o d u c e d m i n u t e n u m b e r o f c a l o r i e s r e q u i r e d f o r e v a p o r a t i o n k g

 = 14.5 * 10 3 580 * 10 3 = 0.25 k g

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c) For the curve 1 volume is constant so it is isochoric process. But in curve 2 and 3 curve 2 is steeper so 2 is adiabatic and 3 is isothermal.

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

process 1 isochoric and process 2 is isothermal .

Since, work done = area under P-V curve . here area under the pV curve 1 is more . so work done is more when the gas expands in isochoric process.

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

Coefficient of β = 5

T1= 27+273=300K

Coefficient of performance β = T 2 300 - T 2

1500-5T2=T2

6T2=1500

T2= 250K

T2= 250-273=-23oC

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

Temperature of the source is 270C

T1= 27+273= 300K

T2= -3+273= 270K

Efficiency of heat engine = 1-T2/T1= 1-270/300=1/10

Efficiency of refrigerator  is 50% of a perfect engine

= 0.5 * η = 1/20

Coefficient of performance of the refrigerator β = Q 2 W = 1 - η ' η

= 1 - 1 / 20 1 / 20 = 19 / 20 1 / 20 = 19

Q2= β W =19W

= 19 * 1KW=19KW= 19kJ/s

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