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. $n R T$
• B. $\frac{R T}{n}$
• C. $R T\left(1-n^{-1}\right)$
• D. $R T(n-1)$

Answer 1

Answer: (C)

$A B$ 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
$\Rightarrow Q_{A B}=C_{\nu} \Delta T=C_{V}\left(\frac{T}{n}-T\right)=C_{V} T\left(\frac{1}{n}-1\right)$
$=C_{V} T\left(\frac{1-n}{n}\right)$
$Q_{B C}=C_{P} \Delta T$ for 1 mole of the gas $=C_{P}\left(T-\frac{T}{n}\right)$
$Q_{B C}=C_{P} T\left(\frac{n-1}{n}\right)$
$Q_{\text {net }}=Q_{A B}+Q_{B C}$
$=C_{V} T\left(\frac{1-n}{n}\right)+C_{P} T\left(\frac{n-1}{n}\right)$
$=\frac{T}{n}\left(C_{V}-n C_{V}+n C_{P}-C_{P}\right)$
$=\frac{T}{n}\left\{\left(n\left(C_{P}-C_{V}\right)-\left(C_{P}-C_{V}\right)\right\}\right.$
$=\frac{T}{n}(n R-R)=\frac{T}{n}(n-1) R=R T\left(1-n^{-1}\right)$