Question 9

1. Process $a \to b$ (Sudden Compression): This is an adiabatic process. Given: $n = 10$, $T_a = 27^\circ\text{C} = 300\text{ K}$, $V_a = V_0$, $V_b = V_0/3$. For a monoatomic gas, $\gamma = 5/3$ and $C_v = \frac{3}{2}R$. Temperature at $b$: $T_b = T_a \left(\frac{V_a}{V_b}\right)^{\gamma-1} = 300 (3)^{5/3 - 1} = 300 (3)^{2/3} = 300 (9)^{1/3}$ Given $9^{1/3} = 2.08$, so $T_b = 300 \times 2.08 = 624\text{ K}$. Pressure at $b$: $P_b = P_a \left(\frac{V_a}{V_b}\right)^\gamma = P_a (3)^{5/3} = P_a \times 3 \times 3^{2/3} = P_a \times 3 \times 2.08 = 6.24 P_a$. Since the pressure is $6.24$ times the atmospheric pressure (not $2.08$), Option (D) is incorrect.

Change in internal energy: $\Delta U_{ab} = n C_v (T_b - T_a) = 10 \times \frac{3}{2} R (624 - 300) = 15R \times 324 = 4860R$. Option (B) is correct.

2. Process $b \to c$ (Isochoric Cooling): The piston is kept stationary, so the volume is constant at $V_0/3$. The temperature reaches the bath temperature, $T_c = 11^\circ\text{C} = 284\text{ K}$.

3. Process $c \to f$ (Isothermal Expansion): The piston is brought slowly to its initial volume while submerged in the water bath. Therefore, the temperature remains constant at $T_f = T_c = 284\text{ K}$ until the volume reaches $V_f = V_0$.

4. Net Internal Energy Change: The net change in internal energy depends only on the initial and final states of the entire process. $\Delta U_{net} = n C_v (T_f - T_a) = 10 \times \frac{3}{2} R (284 - 300) = 15R (-16) = -240R$. Option (C) is correct.

5. P-V Diagram Analysis:

• $a \to b$: Adiabatic compression (steep curve to the left, volume decreases).
• $b \to c$: Isochoric cooling (vertical line downwards, pressure drops at constant volume).
• $c \to f$: Isothermal expansion (flatter curve to the right, ending at $V_0$). Additionally, at the final state $f$, $V_f = V_a$ but $T_f (284\text{ K}) < T_a (300\text{ K})$. Thus, $P_f < P_a$. The schematic correctly shows point $f$ lying below point $a$ on the same vertical volume line. Option (A) is correct.

**Correct Options: (A), (B), (C)**1. Process $a \to b$ (Sudden Compression): This is an adiabatic process. Given: $n = 10$, $T_a = 27^\circ\text{C} = 300\text{ K}$, $V_a = V_0$, $V_b = V_0/3$. For a monoatomic gas, $\gamma = 5/3$ and $C_v = \frac{3}{2}R$. Temperature at $b$: $T_b = T_a \left(\frac{V_a}{V_b}\right)^{\gamma-1} = 300 (3)^{5/3 - 1} = 300 (3)^{2/3} = 300 (9)^{1/3}$ Given $9^{1/3} = 2.08$, so $T_b = 300 \times 2.08 = 624\text{ K}$. Pressure at $b$: $P_b = P_a \left(\frac{V_a}{V_b}\right)^\gamma = P_a (3)^{5/3} = P_a \times 3 \times 3^{2/3} = P_a \times 3 \times 2.08 = 6.24 P_a$. Since the pressure is $6.24$ times the atmospheric pressure (not $2.08$), Option (D) is incorrect.

Change in internal energy: $\Delta U_{ab} = n C_v (T_b - T_a) = 10 \times \frac{3}{2} R (624 - 300) = 15R \times 324 = 4860R$. Option (B) is correct.

2. Process $b \to c$ (Isochoric Cooling): The piston is kept stationary, so the volume is constant at $V_0/3$. The temperature reaches the bath temperature, $T_c = 11^\circ\text{C} = 284\text{ K}$.

3. Process $c \to f$ (Isothermal Expansion): The piston is brought slowly to its initial volume while submerged in the water bath. Therefore, the temperature remains constant at $T_f = T_c = 284\text{ K}$ until the volume reaches $V_f = V_0$.

4. Net Internal Energy Change: The net change in internal energy depends only on the initial and final states of the entire process. $\Delta U_{net} = n C_v (T_f - T_a) = 10 \times \frac{3}{2} R (284 - 300) = 15R (-16) = -240R$. Option (C) is correct.

5. P-V Diagram Analysis:

• $a \to b$: Adiabatic compression (steep curve to the left, volume decreases).
• $b \to c$: Isochoric cooling (vertical line downwards, pressure drops at constant volume).
• $c \to f$: Isothermal expansion (flatter curve to the right, ending at $V_0$). Additionally, at the final state $f$, $V_f = V_a$ but $T_f (284\text{ K}) < T_a (300\text{ K})$. Thus, $P_f < P_a$. The schematic correctly shows point $f$ lying below point $a$ on the same vertical volume line. Option (A) is correct.

Correct Options: (A), (B), (C)