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

Step 1: Identify the coordinates of the states from the $V$-$T$ diagram.

For $n = 1$ mole of an ideal gas, the coordinates of the states are given by:

$$A = (T_0, V_0)$$ $$B = (T_0, 4V_0)$$ $$C = (T_C, V_0)$$

The pressure at state $A$ is:

$$P_A = P_0$$

Step 2: Analyze the process $AB$.

Since the temperature remains constant at $T = T_0$, the process is isothermal.

$$U_A = U_B$$

The work done during this isothermal expansion is:

$$W_{AB} = n R T_0 \ln\left(\frac{V_B}{V_A}\right)$$

Using the ideal gas equation at state $A$, $P_0 V_0 = n R T_0$:

$$W_{AB} = P_0 V_0 \ln\left(\frac{4V_0}{V_0}\right)$$ $$W_{AB} = P_0 V_0 \ln 4$$

Step 3: Analyze the process $BC$.

The path $BC$ is a straight line passing through the origin on the $V$-$T$ diagram, which represents an isobaric process.

$$P_C = P_B$$

Using the ideal gas equation at state $B$:

$$P_B = \frac{n R T_0}{4V_0}$$ $$P_B = \frac{P_0}{4}$$

Thus, the pressure at state $C$ is:

$$P_C = \frac{P_0}{4}$$

Step 4: Determine the temperature at state $C$.

Using the relation for the isobaric process $BC$:

$$\frac{V_C}{T_C} = \frac{V_B}{T_B}$$ $$\frac{V_0}{T_C} = \frac{4V_0}{T_0}$$ $$T_C = \frac{T_0}{4}$$

Answer: The correct options are:

• Internal energies at $A$ and $B$ are equal ($U_A = U_B$)
• Work done by the gas in process $AB$ is $P_0 V_0 \ln 4$
• Pressure at $C$ is $\dfrac{P_0}{4}$
• Temperature at $C$ is $\dfrac{T_0}{4}$