One mole of an ideal monoatomic gas…

One mole of an ideal monoatomic gas (Cv=3/2R) at 300 K expands from volume V to 2V in four different ways. Take R=8.314 J mol^-1K^-1 and ln 2=0.693.

One or more options correct

One mole of an ideal monoatomic gas (

C_v=\tfrac32R

) at

300\ \mathrm{K}

expands from volume

V

2V

in four different ways. Take

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

and

\ln 2=0.693

. The correct statement(s) is(are)

A. In the isothermal reversible expansion, $\Delta S_{\text{system}}=+5.76\ \mathrm{J\,K^{-1}}$ and $\Delta S_{\text{universe}}=0$
B. In the free expansion into vacuum (adiabatic container), $\Delta S_{\text{universe}}=+5.76\ \mathrm{J\,K^{-1}}$
C. In the reversible adiabatic expansion, the final temperature is about $189\ \mathrm{K}$ and $\Delta S_{\text{universe}}=0$
D. In the irreversible adiabatic expansion against a constant external pressure equal to the final gas pressure, the final temperature is also about $189\ \mathrm{K}$

Show answer and solution

Answer: A, B, C

Isothermal reversible:

\Delta S_{sys}=R\ln 2=5.76\ \mathrm{J\,K^{-1}}

, the surroundings supply heat reversibly so

\Delta S_{surr}=-5.76

and

\Delta S_{univ}=0

; (A) is correct. Free expansion:

q=w=0

, the ideal gas stays at

300\ \mathrm{K}

but

S

is a state function, so

\Delta S_{sys}=R\ln 2

while the surroundings are unchanged;

\Delta S_{univ}=+5.76

and (B) is correct. Reversible adiabatic:

TV^{\gamma-1}=

constant with

\gamma=5/3

gives

T_2=300(1/2)^{2/3}=189\ \mathrm{K}

, and

q_{rev}=0

\Delta S_{univ}=0

; (C) is correct. Irreversible adiabatic against

P_{ext}=P_2=RT_2/2V

\tfrac32R(T_2-300)=-P_{ext}(2V-V)=-RT_2/2

, giving

2T_2=450

T_2=225\ \mathrm{K}

. Less work is done than in the reversible path, so the gas cools less; 225 K, not 189 K, and (D) is wrong.

Original question written in the JEE Advanced 2019 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake