Physics NCERT Exemplar Solutions Class 11th Chapter Twelve

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P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (c) Q1= W+Q2

W=Q1-Q2>0

Q1>Q2>0

We can also write Q21

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Payal Gupta

Contributor-Level 10

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(a) the given process is a cyclic process i.e returns to the original state 1

Hence change in internal energy dU =0

dQ= dU+dW=0+dW

hence total heat supplied is converted to work done by the gas which is not possible by second law of thermodynamics.

(c) When the gas expands adiabatically from 2 to 3 . it is not possible to return to the same state without being heat supplied hence 3 to 1 cannot be adiabatic.

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Payal Gupta

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(b), (c) Change in internal energy for process A to B

dU=nCvdT=nCv (dT)=nCv (TB-TA)

work done from A to B  = area under the PV curve which is maximum for path I

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Payal Gupta

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(a), (d) For isothermal dT= 0 so T=constant

For an ideal gas dU = change in internal energy = nCvdT=0

From first law of thermodynamics dQ= dU+dW

dQ= dW

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Payal Gupta

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(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

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(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

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(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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