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 the original volume $V_0$. Finally the gas is heated at constant volume to get the original temperature.

• A. Heat absorbed in the process is $p_0{V}_{0}ln2$.
• B. Work done in the process is $p_0{V}_0(ln2-\frac{1}{2})$.
• C. Internal energy doesn't change.
• D. Heat absorbed in the process is $p_0{V}_0(ln2-\frac{1}{2})$.

Answer 1

Answer: (B), (C), (D)

Heat absorbed in the process is $p_0{V}_0(ln2-\frac{1}{2})$.

The process is cyclic so that the change in internal energy is zero. The heat supplied is therefore equal to the work done by the gas.

The work done during step ab is
$W_1=nR{T}_{0}ln2=p_0{V}_{0}ln2$

Also from the ideal gas equation,
$p_a{V}_a=p_b{V}_b$
$\Rightarrow p_b=p_a\frac{V_a}{V_b}$
$=p_0\frac{V_0}{2V_0}=\frac{p_0}{2}$

In the step bc, pressure remains constant. Hence, the work done is
$W_2=p_b(V_c-V_b)=\frac{p_0}{2}(V_0-2V_0)=-\frac{p_0{V}_0}{2}$

In the step ca, volume remains constant. Therefore work done is zero.

$\therefore$ Net work done by the gas in the process is

$W=W_1+W_2=p_0{V}_0(ln2-\frac{1}{2})$

Hence, the heat supplied to the gas is also $p_0{V}_0(ln2-\frac{1}{2})$