JEE Advanced 2018 Paper 2 · Physics · Calorimetry and Heat Transfer · medium

One mole of a monatomic ideal gas undergoes the quasi-static processes listed in List-I. The molar heat capacity C of the gas in each process…

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One mole of a monatomic ideal gas undergoes the quasi-static processes listed in List-I. The molar heat capacity

C

of the gas in each process (defined by

\mathrm{d}Q=C\,\mathrm{d}T

) is given in List-II, where

R

is the gas constant. Match the processes with their molar heat capacities.

List-I
(P) A process in which

PV^{2}

stays constant
(Q) A process in which pressure is directly proportional to volume
(R) A process in which

PV^{3/2}

stays constant
(S) A process in which no heat is exchanged

List-II
(1)

0

(2)

R/2

(3)

2R

(4)

-R/2

(5)

7R/2

A. P→2, Q→3, R→4, S→1
B. P→2, Q→3, R→1, S→4
C. P→3, Q→2, R→4, S→1
D. P→2, Q→5, R→4, S→1

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Answer: (A) P→2, Q→3, R→4, S→1

For a polytropic process

PV^{x}=\text{const}

the first law gives

dQ=C_vdT+P\,dV

and, using

PV=RT

dW=\dfrac{R\,dT}{1-x}

, so

C=C_v+\dfrac{R}{1-x}

with

C_v=\tfrac32R

. (P)

x=2

C=\tfrac32R-R=\tfrac R2

, matching 2. (Q)

P\propto V

means

PV^{-1}=\text{const}

x=-1

C=\tfrac32R+\tfrac R2=2R

, matching 3. (R)

x=\tfrac32

C=\tfrac32R-2R=-\tfrac R2

, matching 4 (the gas gives out heat while its temperature rises or absorbs heat while it cools). (S) A process with no heat exchange is adiabatic, so

dQ=0

and

C=0

, matching 1. Check with

x=\gamma=5/3

C=\tfrac32R-\tfrac32R=0

. The value

7R/2

(for

x=\tfrac12

) is not used. So P

\to

2, Q

\to

3, R

\to

4, S

\to

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