Question

An ideal gas having initial pressure $P$. volume $V$ and temperature $T$ is allowed to expand adiabatically until its volume becomes $27.914V$, while its temperature falls to $\frac{T}{3}$. Choose the correct option(s):

• A. Gas molecules have 6 degree of freedom.
• B. Gas molecules are diatomic.
• C. Work done by the gas is $2.02PV$.
• D. Workdone by the gas is $1.52PV$.

Answer 1

Answer: (A), (C)

The correct options are
A Gas molecules have 6 degree of freedom.
C Work done by the gas is $2.02PV$.

Initial position $\to P,V,T$
Final position $\to P_2,V_2,T_2$
Given, $V_2=27.914V,T_2=\frac{T}{3}$
For adiabatic expansion,
$T{V}^{\gamma -1}=\text{constant}$
$\Rightarrow T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$
$\Rightarrow T{V}^{\gamma -1}=\frac{T}{3}(27.914V{)}^{\gamma -1}$
$\Rightarrow (27.914{)}^{\gamma -1}=3$
$\Rightarrow \gamma =1.33$
Using $\gamma =1+\frac{2}{F}\Rightarrow 1.33=1+\frac{2}{F}$
$\Rightarrow F=6$
No. of degrees of freedom $=6$. So, gas molecules are triatomic.
From Ideal gas equation, $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
$\Rightarrow P_2=\frac{P_1{V}_1}{T_1}\times \frac{T_2}{V_2}=\frac{PV}{T}\times \frac{\frac{T}{3}}{27.914V}=\frac{P}{83.742}$
So, work done $\left(W\right)=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}=[PV-\frac{P}{83.742}\times 27.914V]\frac{1}{1.33-1}$
$=\frac{2}{3}\times \frac{PV}{0.33}=2.02PV$
So work done by gas during expansion is $2.02PV$.
Option (a) and (c) are correct.