Thermodynamic Cyclic Process of Ideal Gas on Volume-Temperature Diagram

To determine the correct option(s), we analyze the cyclic process $ABCA$ on the given $V-T$ diagram for $n = 1\text{ mole}$ of an ideal gas.

1. Analysis of States and Processes

• State A:
  The temperature and volume at state A are $T_A = T_0$ and $V_A = V_0$.
  From the ideal gas equation ($PV = nRT$), the pressure at A is given by:
  $P_0 = \frac{RT_0}{V_0} \implies P_0 V_0 = RT_0$

• Process AB:
  The path $AB$ is a vertical line at constant temperature $T = T_0$.
  Thus, $T_B = T_0$ and $V_B = 4V_0$. Process $AB$ is an isothermal expansion.

• Process BC:
  The path $BC$ lies on a straight line passing through the origin $(0, 0)$ and point B $(T_0, 4V_0)$.

• Process CA:
  The path $CA$ is a horizontal line at constant volume $V = V_0$. Thus, $V_C = V_0$.

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2. Verification of Options

• Option A: Internal energies at A and B are the same

  The internal energy of an ideal gas depends only on its temperature:
  $U = n C_v T$
  Since $T_A = T_B = T_0$, we have $U_A = U_B$.

  Option A is correct.

• Option B: Work done by the gas in process AB is $P_0 V_0 \ln 4$

  Since process $AB$ is isothermal at temperature $T_0$, the work done by the gas is:
  $W_{AB} = n R T_0 \ln\left(\frac{V_B}{V_A}\right)$
  Substituting $n = 1$, $V_A = V_0$, and $V_B = 4V_0$:
  $W_{AB} = 1 \cdot R T_0 \ln\left(\frac{4V_0}{V_0}\right) = RT_0 \ln 4$
  Since $RT_0 = P_0 V_0$:
  $W_{AB} = P_0 V_0 \ln 4$

  Option B is correct.

• Option D: Temperature at C is $\frac{T_0}{4}$

  Since the line $BC$ passes through the origin in the $V-T$ plane, its relation is given by $V = kT$, where $k$ is a constant slope.
  Using state B $(T_0, 4V_0)$:
  $k = \frac{V_B}{T_B} = \frac{4V_0}{T_0}$
  For state C $(T_C, V_0)$:
  $V_C = k T_C \implies V_0 = \left(\frac{4V_0}{T_0}\right) T_C \implies T_C = \frac{T_0}{4}$

  Option D is correct.

• Option C: Pressure at C is $\frac{P_0}{4}$

  For process $BC$, since $V \propto T$, the ratio $\frac{V}{T}$ is constant.
  From the ideal gas equation $P = \frac{RT}{V}$, a constant ratio of $\frac{V}{T}$ implies that pressure $P$ remains constant along line $BC$ (an isobaric process).
  Therefore, $P_C = P_B$.
  The pressure at state B is:
  $P_B = \frac{RT_B}{V_B} = \frac{RT_0}{4V_0} = \frac{P_0}{4}$
  Hence, $P_C = \frac{P_0}{4}$.

  Option C is correct.

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Conclusion

All options A, B, C, and D are correct.