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. $RT(1-\frac{1}{n}$)
• D. $RT(n-1)$

Answer 1

Answer: (C)

$RT(1-\frac{1}{n}$)

The temperature remains unchanged therefore
$U_f=U_i$
Also, $\Delta Q=\Delta W$.

In the first step which is isochoric, $\Delta W=0$.

In second step, pressure = $\frac{P}{n}$. Volume is increased from $V$ to $nV$.
$\therefore$$W=\frac{P}{n}(nV-V)$
= $PV\left(\frac{n-1}{n}\right)$