Question

A sample of an ideal gas has pressure $p_0$, volume $V_0$ and temperature $T_0$. It is isothermally expanded to twice its original volume. It is then compressed at constant pressure to have the original volume $V_0$. Finally, the gas is heated at constant volume to get the original temperature. The heat absorbed in the process is

• A. $p_0{V}_0[\text{In}2+2]$
• B. $p_0{V}_0[\text{In}2-\frac{1}{2}]$
• C. $p_0{V}_0[\text{In}2+\frac{3}{2}]$
• D. $p_0{V}_0[\text{In}2+\frac{1}{2}]$

Answer 1

The correct option is B $p_0{V}_0[\text{In}2-\frac{1}{2}]$

The process is cyclic, so the change in internal energy is zero.
Therefore, the heat supplied will be equal to the work done by the gas.

The work done during step $ab$ is

$W_1=\mu R{T}_0\text{In}\frac{2_0}{V_0}=\mu R{T}_0\text{In}2=P_0{V}_{0}In2$

From the ideal gas equation,

$p_a{V}_a=p_b{v}_b\text{or},p_b=\frac{p_a{V}_a}{V_b}=\frac{p_0{V}_0}{2V_0}=\frac{p_0}{2}$

In the step $bc$, the pressure remains constant.
Hence, the work done is

$W_2=\frac{p_0}{2}(v_0-2v_0)=-\frac{p_0{v}_0}{2}$

In the step $ca$, the volume remains constant and so the work done is zero.

The net work done by the gas in the cyclic process is

$W=W_1+W_2=p_0{V}_0[\text{In}2-\frac{1}{2}]$

Hence, the heat supplied to the gas is

$p_0{V}_0[\text{In}2-\frac{1}{2}]$

Final answer:(d)