Question

One mole of an ideal gas at temperature $T$ was cooled isochorically till the gas pressure fell from $P$ to $\frac{P}{n}$. Then, by an isobaric process, the gas was restored to the initial temperature. The net amount of heat absorbed by the gas in the process is

• A. $nRT$
• B. $\frac{RT}{n}$
• C. $RT\left(1-n^{-1}\right)$
• D. $RT(n-1)$

Answer 1

Answer: (C)

$AB$ is an isochoric process
$\frac{P_A}{T_A}=\frac{P_B}{T_B}$ or $\frac{P}{T}=\left(\frac{P}{n}\right)\cdot \frac{1}{T_B}$
$\Rightarrow T_B=\left(\frac{T}{n}\right)$
For 1 mole of the gas
$\Rightarrow Q_{AB}=C_{\nu }\Delta T=C_V\left(\frac{T}{n}-T\right)=C_{V}T\left(\frac{1}{n}-1\right)$
$=C_{V}T\left(\frac{1-n}{n}\right)$
$Q_{BC}=C_P\Delta T$ for 1 mole of the gas $=C_P\left(T-\frac{T}{n}\right)$
$Q_{BC}=C_{P}T\left(\frac{n-1}{n}\right)$
$Q_{\text{net }}=Q_{AB}+Q_{BC}$
$=C_{V}T\left(\frac{1-n}{n}\right)+C_{P}T\left(\frac{n-1}{n}\right)$
$=\frac{T}{n}\left(C_V-n{C}_V+n{C}_P-C_P\right)$
$=\frac{T}{n}\{\left(n\left(C_P-C_V\right)-\left(C_P-C_V\right)\right\}$
$=\frac{T}{n}(nR-R)=\frac{T}{n}(n-1)R=RT\left(1-n^{-1}\right)$