Question

An ideal gas (\(0.5\) mol), initially at \(2\) bar pressure, is compressed at a constant temperature of \(600\ \text{K}\) in two steps: first, against a constant external pressure of \(P\) bar \((2 < P < 8)\), and then against constant external pressure of \(8\) bar. At each step, the compression is stopped only when the pressure of the gas becomes equal to the external pressure. The total work done on the gas in these steps is \(W\). Considering all possible values of \(P\ (2 < P < 8)\) and taking the gas constant as \(R\) (in \(\text{J K}^{-1}\text{mol}^{-1}\)), the minimum value of \(|W|\) (in J) is equal to:

• A. \(207R\)
• B. \(600R\)
• C. \(630R\)
• D. \(900R\)

Hint:

For irreversible compression against constant external pressure, \[ W = -P_{\text{ext}}(V_f - V_i) \] In multi-step isothermal processes, first use \[ PV = nRT \] to determine intermediate volumes, then calculate work separately for each step. For minimization problems involving expressions of the form \[ aP + \frac{b}{P}, \] the minimum occurs when \[ P = \sqrt{\frac{b}{a}} \] which follows from differentiation or the AM-GM inequality.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Work done in an irreversible compression (against constant external pressure) is given by: \[ W = -P_{\text{ext}}(V_{\text{final}} - V_{\text{initial}}) \] Since work is done *on* the gas, \(W\) will be positive. We are compressing 0.5 mol of an ideal gas from 2 bar to 8 bar in two discrete stages. We want to find the intermediate pressure \(P\) that minimizes the total work.
Step 2: Key Formula or Approach:
Initial state: \(P_1 = 2\) bar, \(V_1 = \frac{nRT}{P_1} = \frac{0.5RT}{2} = \frac{RT}{4}\).
Intermediate state: \(P_{\text{int}} = P\) bar, \(V_{\text{int}} = \frac{0.5RT}{P}\).
Final state: \(P_2 = 8\) bar, \(V_2 = \frac{0.5RT}{8} = \frac{RT}{16}\).
Total work \(W = W_1 + W_2 = P(V_1 - V_{\text{int}}) + 8(V_{\text{int}} - V_2)\).
Step 3: Detailed Explanation:
Substitute the volumes into the work expression: \[ W = P\left(\frac{RT}{4} - \frac{RT}{2P}\right) + 8\left(\frac{RT}{2P} - \frac{RT}{16}\right) \] \[ W = RT \left[ \frac{P}{4} - \frac{1}{2} + \frac{4}{P} - \frac{1}{2} \right] \] \[ W = RT \left[ \frac{P}{4} + \frac{4}{P} - 1 \right] \] To find the minimum \(W\), differentiate with respect to \(P\) and set to zero: \[ \frac{dW}{dP} = RT \left[ \frac{1}{4} - \frac{4}{P^2} \right] = 0 \] \[ \frac{1}{4} = \frac{4}{P^2} \implies P^2 = 16 \implies P = 4 \text{ bar} \] Note that \(P=4\) lies in the given range \(2<P<8\). Now calculate the minimum work at \(P=4\) and \(T=600\) K: \[ W_{\text{min}} = R(600) \left[ \frac{4}{4} + \frac{4}{4} - 1 \right] \] \[ W_{\text{min}} = 600R [1 + 1 - 1] = 600R \]
Step 4: Final Answer:
The intermediate pressure that minimizes the work is the geometric mean of the initial and final pressures. The minimum work done is 600R.