Question

Ten moles of an ideal monoatomic gas, initially in state \(a\) at atmospheric pressure and temperature \(T_a = 27^\circ\text{C}\), is enclosed in a metal cylinder of volume \(V_0\) fitted with a frictionless piston. The gas is suddenly compressed to state \(b\) with volume \(V_0/3\). Now, keeping the piston stationary, the cylinder is submerged in a water bath of temperature \(11^\circ\text{C}\) until the gas reaches the temperature of the water bath, which is denoted as state \(c\). Finally, while still in the water bath, the piston is brought slowly to its initial position, which is denoted as state \(f\). If \(R\) is the universal gas constant, then the correct option(s) is/are:

Given: \(9^{1/3} = 2.08\)

• A. The schematic P-V diagram of the processes described above is:
  

  

• B. The change in internal energy in going from state $a$ to $b$ is $4860R$.
• C. The net change in the internal energy in the whole process is $-240R$.
• D. The pressure and temperature of the state $b$ are $2.08$ times the atmospheric pressure and $624 \text{ K}$, respectively.

Hint:

For adiabatic processes, remember that "sudden" implies no time for heat exchange. For monatomic gases, $\gamma = 1.67$, which is steeper on a P-V graph than the isothermal curve ($\gamma = 1$).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The process consists of: (1) Sudden compression (Adiabatic), (2) Constant volume cooling (Isometric), and (3) Slow expansion in a constant temperature bath (Isothermal). We need to track $P, V, T$ for each step.

Step 2: Key Formula or Approach:

• Adiabatic: $TV^{\gamma-1} = \text{const}$. For monoatomic, $\gamma = 5/3$.

• Internal energy: $\Delta U = nC_v \Delta T$. For monoatomic, $C_v = 3R/2$.

Step 3: Detailed Explanation:

• Process $a \to b$ (Adiabatic): $T_b = T_a (V_a/V_b)^{\gamma-1} = 300(3)^{2/3} = 300 \times (9^{1/3}) = 300 \times 2.08 = 624 \text{ K}$. $\Delta U_{ab} = 10 \times (1.5R) \times (624 - 300) = 15R \times 324 = 4860R$. (B) is correct. $P_b/P_a = (V_a/V_b)^\gamma = 3^{5/3} = 3 \times 3^{2/3} = 3 \times 2.08 = 6.24$. Pressure is $6.24$ times atmospheric. (D) is incorrect.
• Process $b \to c \to f$: $T_f = 11^\circ\text{C} = 284 \text{ K}$. $\Delta U_{net} = U_f - U_a = nC_v(T_f - T_a) = 15R(284 - 300) = -240R$. (C) is correct.
• P-V Diagram (A): $a \to b$ is a steep curve (adiabatic), $b \to c$ is vertical down (isometric), $c \to f$ is a shallower curve (isothermal) to $V_0$. The diagram matches the physics. (A) is correct.

Step 4: Final Answer:
The correct options are (A), (B), and (C).