Question

$P-T$ diagram of one $\text{mole}$ of an ideal monatomic gas is shown. Process $AB$ and $CD$ are adiabatic. Work done in the complete cycle is

• A. $2.5RT$
• B. $-2RT$
• C. $1.5RT$
• D. $-3.5RT$

Answer 1

The correct option is A $2.5RT$


From the figure, $AB$ and $CD$ are adiabatic $\therefore Q_{AB}=Q_{CD}=0$
$BC$ and $AD$ are isobaric,
$\therefore Q_{BC}=n{C}_P\Delta T=1\times \frac{5R}{2}\times (4T-T)=7.5RT$
$Q_{AD}=n{C}_P\Delta T=1\times \frac{5R}{2}\times (3T-5T)=-5RT$
Hence, $Q_{net}=Q_{BC}+Q_{AD}=2.5RT$
From the figure, it is evident that it is a cyclic process
In a cyclic process, $Q_{net}=W_{net}$
$\Rightarrow W_{net}=2.5RT$