Question

An ideal gas expands isothermally from volume $V_1$ to $V_2$ and is then compressed to original volume $V_1$ adiabatically. Initial pressure is $P_1$ and final pressure is $P_3$. The total work done is $W$. Then

• A. $P_3>P_1$, $W>0$
• B. $P_3<P_1$, $W<0$
• C. $P_3>P_1$, $W<0$
• D. $P_3=P_1$, $W=0$

Answer 1

The correct option is C $P_3>P_1$, $W<0$

Let the thermodynamic coordinates of various states be:
initial state $(P_1,V_1,T_1)$
state after first process $(P_2,V_2,T_2)$
final state $(P_3,V_3,T_3)$
First process being isothermal, $P_2=\frac{P_1{V}_1}{V_2}$
and $T_2=T_1$
Next one being adiabatic, $P_2{V}_2^{\gamma }=P_3{V}_3^{\gamma }$
Here $V_3=V_1$
So $P_3=\frac{P_2{V}_2^{\gamma }}{V_1^{\gamma }}$
Replacing $P_2$ from above,$P_3=P_1\frac{V_2^{\gamma -1}}{V_1^{\gamma -1}}$
We know that $\gamma >1\Rightarrow \gamma -1>0$
given $V_2>V_1\Rightarrow (\frac{V_2}{V_1}{)}^{\gamma -1}>1$
Hence $P_3>P_1$
Work done in first process=$P_1{V}_1\ln \frac{V_2}{V_1}=\frac{P_1{V}_1}{\gamma -1}\ln (\frac{V_2}{V_1}{)}^{\gamma -1}<P_1{V}_1\frac{(\frac{V_2}{V_1}{)}^{\gamma -1}-1}{\gamma -1}$
and in second process work= $\frac{P_2{V}_2-P_3{V}_1}{\gamma -1}=P_1{V}_1\frac{1-(\frac{V_2}{V_1}{)}^{\gamma -1}}{\gamma -1}$

Hence net work=$W<0$