One mole of an ideal monatomic gas…

One mole of an ideal monatomic gas (CV=3/2R, R=8.314 J K^-1 mol^-1) at 300 K and volume 2 L is expanded reversibly to 16 L by two different paths…

One or more options correct

One mole of an ideal monatomic gas (

C_V=\tfrac32R

R=8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}

) at

300\ \mathrm{K}

and volume

2\ \mathrm{L}

is expanded reversibly to

16\ \mathrm{L}

by two different paths: Path I is isothermal and Path II is adiabatic. Which of the following statement(s) is(are) correct? (Use

\ln 8=2.08

A. The final temperature in Path II is $75\ \mathrm{K}$
B. The entropy change of the system in Path II is $R\ln 8$
C. The magnitude of work done by the gas in Path I is about $5.19\ \mathrm{kJ}$
D. The magnitude of work done by the gas in Path II is greater than that in Path I

Show answer and solution

Answer: A, C

Path II:

TV^{\gamma-1}

= constant with

\gamma=5/3

gives

T_2=300\,(2/16)^{2/3}=300/4=75

K (checked numerically), so A is correct. A reversible adiabatic process has

q_{rev}=0

, hence

\Delta S_{sys}=0

, not

R\ln 8

; B is incorrect. Path I:

|w|=RT\ln(V_2/V_1)=8.314\times300\times2.08\approx5.19

kJ; C is correct. Path II:

|w|=|\Delta U|=\tfrac32R(300-75)=2.81

kJ, which is less than

5.19

kJ, because in the adiabatic path the gas cools and its pressure falls faster; D is incorrect. Correct options: A and C.

Original question written in the JEE Advanced 2024 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