Question

An ideal gas expands isothermally from a volume $V_1$ to $V_2$ and 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 $P_3>P_1,W<0$

slope of adiabatic process at a given state $(P,V,T)$ is more than the slope of isothermal process. The corresponding $P-V$ graph for the two processes is as shown in figure.

In the graph, $AB$ is isothermal and $BC$ is adiabatic.
$W_{AB}=$ positive (as volume is increasing)
and $W_{BC}=$ negative (as volume is decreasing) also,
$|W_{BC}|>|W_{AB}|$, as area under $P-V$ graph gives the work done.
Hence, $W_{AB}+W_{BC}=W<0$
From the graph itself, it is clear that $P_3>P_1$.

NOTE: At point $B$, magnitude of slope of adiabatic (process $BC$) is greater than the slope of isothermal (process $AB$).