Question

A cyclic process for 1 mole of an ideal gas is shown in figure in the $V-T$, diagram. The work done in $A B, B C$ and $C A$ respectively

• A. $0, R T_{2} \ln \left(\frac{V_{1}}{V_{2}}\right), R\left(T_{1}-T_{2}\right)$
• B. $R\left(T_{1}-T_{2}\right), 0, R T_{1} \ln \frac{V_{1}}{V_{2}}$
• C. $0, R T_{2} \ln \left(\frac{V_{2}}{V_{1}}\right), R\left(T_{1}-T_{2}\right)$
• D. $0, R T_{2} \ln \left(\frac{V_{2}}{V_{1}}\right), R\left(T_{2}-T_{1}\right)$

Answer 1

Answer: (C)

Process $A B$ is isochoric,
$\therefore W_{A B}=P \Delta V=0$
Process $B C$ is isothermal
$\therefore W_{B C}=R T_{2} \cdot \ln \left(\frac{V_{2}}{V_{1}}\right)$
Process $C A$ is isobaric
$\therefore W_{C A}=-P \Delta V=-R \Delta T$
$=-R\left(T_{1}-T_{2}\right)=R\left(T_{2}-T_{1}\right)$
(Negative sign is taken because of compression)