One mole of a monatomic ideal gas…

One mole of a monatomic ideal gas undergoes the processes in List-I. Match each process with the molar heat capacity C of the gas in that process…

Single correct option

One mole of a monatomic ideal gas undergoes the processes in List-I. Match each process with the molar heat capacity

C

of the gas in that process (List-II).

List-I
(P) A process in which

TV=\text{constant}

(Q) A process in which

P/V=\text{constant}

(R) A process in which

V/T=\text{constant}

(S) A process in which

TV^{2/3}=\text{constant}

List-II
(1)

0

(2)

R/2

(3)

3R/2

(4)

2R

(5)

5R/2

Show answer and solution

Answer: (C) P→2, Q→4, R→5, S→1

For one mole,

C=C_v+P\,\dfrac{dV}{dT}=\tfrac32R+P\dfrac{dV}{dT}

. (P)

TV=\text{const}

means

PV^2=\text{const}

(

T\propto PV

, so

PV\cdot V=\text{const}

): polytropic index

x=2

C=C_v+\dfrac{R}{1-x}=\tfrac32R-R=\dfrac R2

, so P→2. (Q)

P\propto V

x=-1

C=\tfrac32R+\tfrac R2=2R

, so Q→4. (R)

V\propto T

is isobaric,

C_p=\tfrac52R

, so R→5 (not

C_v

). (S)

TV^{2/3}=\text{const}

is the adiabatic of a monatomic gas (

\gamma-1=2/3

), so

C=0

, S→1. Hence P→2, Q→4, R→5, S→1; the other options swap or misassign two of the processes.

Original question written in the JEE Advanced 2025 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake