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

Concept: For an isothermal process involving an ideal gas, \[ PV = nRT \] Since temperature remains constant throughout the compression, the quantity \(nRT\) remains constant. In an irreversible compression against a constant external pressure \(P_{\text{ext}}\), the work done by the gas is: \[ W = -P_{\text{ext}}(V_f - V_i) \] For compression, \[ V_f < V_i \] therefore the work done on the gas becomes positive in magnitude. The problem involves a two-step irreversible isothermal compression process. The external pressure changes in two stages, and we must determine the minimum possible magnitude of the total work.

Step 1: Determining the initial state of the gas. The gas initially has: \[ n = 0.5\ \text{mol} \] \[ P_1 = 2\ \text{bar} \] \[ T = 600\ \text{K} \] Using the ideal gas equation, \[ P_1V_1 = nRT \] Hence, \[ V_1 = \frac{nRT}{P_1} \] Substituting values, \[ V_1 = \frac{0.5 \times R \times 600}{2} \] \[ V_1 = 150R \] Thus, \[ V_1 = 150R \]

Step 2: Analyzing the first compression step. The gas is compressed against a constant external pressure of \(P\) bar, where \[ 2 < P < 8 \] Compression stops when the pressure of the gas becomes equal to the external pressure \(P\). Since the process is isothermal, \[ PV = nRT \] At the end of the first step, \[ PV_2 = nRT \] Therefore, \[ V_2 = \frac{nRT}{P} \] Substituting \(nRT = 0.5 \times 600R = 300R\), \[ V_2 = \frac{300R}{P} \]

Step 3: Calculating work done in the first step. For irreversible compression against constant external pressure \(P\), \[ W_1 = -P(V_2 - V_1) \] Substituting values, \[ W_1 = -P\left(\frac{300R}{P} - 150R\right) \] Simplifying carefully, \[ W_1 = -\left(300R - 150PR\right) \] \[ W_1 = 150PR - 300R \] Thus, \[ W_1 = R(150P - 300) \]

Step 4: Analyzing the second compression step. Now the gas is further compressed against a constant external pressure of \(8\) bar. At the end of this step, the gas pressure becomes \(8\) bar. Using the isothermal relation again, \[ P_3V_3 = nRT \] where \[ P_3 = 8\ \text{bar} \] Hence, \[ V_3 = \frac{300R}{8} \] \[ V_3 = 37.5R \]

Step 5: Calculating work done in the second step. The external pressure during the second step is \(8\) bar. Hence, \[ W_2 = -8(V_3 - V_2) \] Substituting the expressions for \(V_3\) and \(V_2\), \[ W_2 = -8\left(37.5R - \frac{300R}{P}\right) \] Expanding, \[ W_2 = -300R + \frac{2400R}{P} \] Thus, \[ W_2 = R\left(\frac{2400}{P} - 300\right) \]

Step 6: Finding the total work done. Total work done on the gas is: \[ W = W_1 + W_2 \] Substituting the obtained expressions, \[ W = R(150P - 300) + R\left(\frac{2400}{P} - 300\right) \] \[ W = R\left(150P + \frac{2400}{P} - 600\right) \] Therefore, \[ W = R\left(150P + \frac{2400}{P} - 600\right) \] We must find the minimum possible value of \(|W|\).

Step 7: Minimizing the work expression. Define \[ f(P) = 150P + \frac{2400}{P} - 600 \] To minimize \(W\), differentiate with respect to \(P\): \[ \frac{df}{dP} = 150 - \frac{2400}{P^2} \] For minimum value, \[ \frac{df}{dP} = 0 \] Hence, \[ 150 - \frac{2400}{P^2} = 0 \] \[ 150 = \frac{2400}{P^2} \] \[ P^2 = \frac{2400}{150} \] \[ P^2 = 16 \] \[ P = 4 \] Since \(2 < P < 8\), this value is valid.

Step 8: Calculating the minimum work. Substitute \(P = 4\) into the work expression: \[ W_{\min} = R\left(150(4) + \frac{2400}{4} - 600\right) \] \[ W_{\min} = R(600 + 600 - 600) \] \[ W_{\min} = 600R \] Therefore, \[ |W|_{\min} = 600R \] Final Answer: \[ \boxed{600R} \] Hence, the correct option is: \[ \boxed{\text{(B) } 600R} \]