Question

26. A undergoes a cyclic process ABCA, as shown in the figure. Its pressure at A is P₀. Choose the correct options(s) from the following.

• A. Internal energies at A and B are the same
• B. Work done by the gas in process AB is P₀V₀log₀4
• C. Pressure at C is $\frac{P_0}{4}$
• D. Temperature of C is $\frac{T_0}{4}$

Answer 1

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

A, B, C, D

Answer 2

From the V-T diagram:

• Point A: Volume = $V_0$, Temperature = $T_0$. Pressure = $P_0$.
• Point B: Volume = $4V_0$, Temperature = $T_0$.
• Point C: Volume = $V_0$, Temperature = $T_C$.

Process AB (Isothermal): $T_A = T_B = T_0$. Using ideal gas law $PV=nRT$: At A: $P_0 V_0 = nRT_0$. At B: $P_B (4V_0) = nRT_0$. Therefore, $P_0 V_0 = P_B (4V_0) \implies P_B = \frac{P_0}{4}$.

Process CA (Isochoric): $V_C = V_A = V_0$. At A: $P_0 V_0 = nRT_0$. At C: $P_C V_0 = nRT_C$. Dividing the two equations: $\frac{P_C}{P_0} = \frac{T_C}{T_0} \implies P_C = P_0 \frac{T_C}{T_0}$.

Process BC (Assuming it passes through origin): $V \propto T$, so $\frac{V}{T} = \text{constant}$. From point B: $\frac{4V_0}{T_0}$. From point C: $\frac{V_0}{T_C}$. Equating: $\frac{4V_0}{T_0} = \frac{V_0}{T_C} \implies T_C = \frac{T_0}{4}$.

Evaluating Options:

A) Internal energies at A and B are the same: Internal energy depends only on temperature for an ideal gas. Since $T_A = T_B = T_0$, their internal energies are the same. Correct.

B) Work done by the gas in process AB is $P_0V_0\log_e 4$: For an isothermal process, $W = nRT \ln(\frac{V_f}{V_i})$. $W_{AB} = (P_0 V_0) \ln(\frac{4V_0}{V_0}) = P_0 V_0 \ln 4$. Correct.

C) Pressure at C is $\frac{P_0}{4}$: Using $P_C = P_0 \frac{T_C}{T_0}$ and $T_C = \frac{T_0}{4}$: $P_C = P_0 \frac{T_0/4}{T_0} = \frac{P_0}{4}$. Correct.

D) Temperature of C is $\frac{T_0}{4}$: As derived from the assumption that BC passes through the origin. Correct.