Q. A cyclic process for 1 mole of an ideal gas is shown in figure in the $V$-$T$, diagram. The work done in $AB,\ BC$ and $CA$ respectively
A) $0,RT_{2}\ln{\left( \frac{V_{1}}{V_{2}} \right),R(T_{1} - T_{2})}$
B) $R\left( T_{1} - T_{2} \right),0,RT_{1}\ln\frac{V_{1}}{V_{2}}$
C) $0,RT_{2}\ln{\left( \frac{V_{2}}{V_{1}} \right),R(T_{1} - T_{2})}$
D) $0,RT_{2}\ln{\left( \frac{V_{2}}{V_{1}} \right),R(T_{2} - T_{1})}$
Explanation
(c) Process $AB$ is isochoric, $\therefore W_{AB} = P\ \mathrm{\Delta}V = 0$
Process $BC$ is isothermal $\therefore W_{BC} = RT_{2}.\ln\left( \frac{V_{2}}{V_{1}} \right)$
Process $CA$ is isobaric
$\therefore W_{CA} = P\mathrm{\Delta}V = R\mathrm{\Delta}T = R(T_{1} - T_{2})$
[Negative sign is taken because of compression]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.