Question

One mole of an ideal gas in initial state $A$ undergoes a cyclic process $ABCA$, as shown in Fig. Its pressure at $A$ is $P_0$. Choose the correct option(s) from the following.

• A. Internal energies at $A$ and $B$ are the same
• B. Work done by the gas in process $AB$ is $P_0{V}_0\ln 4$
• C. Pressure at $C$ is $\frac{P_0}{4}$
• D. Temperature at $C$ is $\frac{T_0}{4}$

Answer 1

Answer: (A), (B)

The correct options are
A Internal energies at $A$ and $B$ are the same
B Work done by the gas in process $AB$ is $P_0{V}_0\ln 4$

Since the temperature is constant in process AB, it is isothermal process and hence no change in internal energy.

$\therefore \Delta U_{AB}=0$
Work done in process AB is
$W_{AB}=nR{T}_0\ln \frac{V_2}{V_1}$
$=P_0{V}_0\ln \left(\frac{{4V}_0}{V_0}\right)$
$=P_0{V}_0\ln \left(4\right)$

Using $\frac{V_C}{T_C}=\frac{V_B}{T_B}$ (since pressure is constant for process BC)

$\therefore$ $\frac{V_o}{T_C}=\frac{4V_o}{T_o}$ $⟹T_C=\frac{T_0}{4}$

Using $\frac{V_C{P}_C}{T_C}=\frac{V_A{P}_A}{T_A}$ where $P_A=P_0$

$\therefore$ $\frac{V_o{P}_C}{T_0/4}=\frac{V_o{P}_0}{T_o}$ $⟹P_C=\frac{P_0}{4}$