Thermodynamic Process of Ideal Monoatomic Gas under Compression and Thermal Equilibrium

To determine the correct options, let us analyze each process step-by-step.

1. Initial State ($\mathbf{a}$)

• Number of moles: $n = 10$
• Monoatomic gas: $C_v = \frac{3}{2}R$, $\gamma = \frac{5}{3}$
• Initial volume: $V_a = V_0$
• Initial temperature: $T_a = 27^\circ\text{C} = 300\text{ K}$
• Initial pressure: $P_a = P_0$ (atmospheric pressure)

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2. Process $\mathbf{a} \to \mathbf{b}$ (Sudden Compression)

A sudden compression is a rapid thermodynamic process where heat exchange with the surroundings is negligible, making it an adiabatic process ($\Delta Q = 0$).

• Final volume: $V_b = \frac{V_0}{3}$
• Using the adiabatic relation $T V^{\gamma-1} = \text{constant}$: $T_b V_b^{\gamma-1} = T_a V_a^{\gamma-1}$ $T_b = T_a \left(\frac{V_a}{V_b}\right)^{\gamma-1} = 300 \times \left(\frac{V_0}{V_0/3}\right)^{\frac{5}{3}-1} = 300 \times 3^{2/3}\text{ K}$

Given $9^{1/3} = \left(3^2\right)^{1/3} = 3^{2/3} = 2.08$: $T_b = 300 \times 2.08 = 624\text{ K}$

• Using the adiabatic relation $P V^\gamma = \text{constant}$: $P_b = P_a \left(\frac{V_a}{V_b}\right)^\gamma = P_0 \times 3^{5/3} = P_0 \times 3 \times 3^{2/3} = P_0 \times 3 \times 2.08 = 6.24 P_0$

• The change in internal energy ($\Delta U_{a \to b}$): $\Delta U_{a \to b} = n C_v (T_b - T_a) = 10 \times \left(\frac{3}{2} R\right) \times (624 - 300) = 15 R \times 324 = 4860 R$

• Analysis of Option B: $\Delta U_{a \to b} = 4860 R$. (Correct)

• Analysis of Option D: The temperature at state $\mathbf{b}$ is $624\text{ K}$, but the pressure is $6.24 P_0$ (not $2.08 P_0$). (Incorrect)

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3. Process $\mathbf{b} \to \mathbf{c}$ (Isochoric Cooling)

• The piston is held stationary, so $V_c = V_b = \frac{V_0}{3}$ (Isochoric process).
• The cylinder is submerged in a water bath at $T_{\text{bath}} = 11^\circ\text{C} = 284\text{ K}$ until equilibrium is reached: $T_c = 284\text{ K}$
• On the $P-V$ diagram, this is represented by a vertical line going downward from $\mathbf{b}$ to $\mathbf{c}$ as pressure drops at constant volume.

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4. Process $\mathbf{c} \to \mathbf{f}$ (Slow Isothermal Expansion)

• The piston is brought slowly back to its initial position $V_f = V_0$ while inside the water bath.
• Since the process is slow and in thermal contact with the bath, it is an isothermal process at $T_f = T_c = 284\text{ K}$.

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5. Net Change in Internal Energy ($\Delta U_{\text{net}}$)

Since internal energy $U$ is a state function depending only on temperature: $\Delta U_{\text{net}} = U_f - U_a = n C_v (T_f - T_a)$ $\Delta U_{\text{net}} = 10 \times \left(\frac{3}{2} R\right) \times (284 - 300) = 15 R \times (-16) = -240 R$

• Analysis of Option C: Net change in internal energy is $-240 R$. (Correct)

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6. P-V Diagram (Option A)

• $\mathbf{a} \to \mathbf{b}$: Adiabatic curve compression from $V_0$ to $V_0/3$ (pressure increases).
• $\mathbf{b} \to \mathbf{c}$: Isochoric vertical line downwards from $P_b$ to $P_c$ at $V = V_0/3$.
• $\mathbf{c} \to \mathbf{f}$: Isothermal curve expansion from $V_0/3$ to $V_0$ (pressure decreases).

This perfectly matches the schematic shown in option A.

• Analysis of Option A: (Correct)

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Conclusion

The correct options are A, B, and C.