Question 1

Work done on the gas during an irreversible isothermal compression is given by $W = -P_{ext} \Delta V$.

Step 1: The gas is compressed from $P_{1} = 2$ bar to $P_{2} = P$ bar against $P_{ext} = P$ bar. $W_{1} = -P(V_{2} - V_{1}) = -P \left( \frac{nRT}{P} - \frac{nRT}{2} \right) = nRT \left( \frac{P}{2} - 1 \right)$

Step 2: The gas is compressed from $P_{2} = P$ bar to $P_{3} = 8$ bar against $P_{ext} = 8$ bar. $W_{2} = -8(V_{3} - V_{2}) = -8 \left( \frac{nRT}{8} - \frac{nRT}{P} \right) = nRT \left( \frac{8}{P} - 1 \right)$

Total work done $W = W_{1} + W_{2}$: $W = nRT \left( \frac{P}{2} - 1 \right) + nRT \left( \frac{8}{P} - 1 \right) = nRT \left( \frac{P}{2} + \frac{8}{P} - 2 \right)$

Given $n = 0.5$ mol and $T = 600$ K, we have $nRT = 0.5 \times R \times 600 = 300R$. $W = 300R \left( \frac{P}{2} + \frac{8}{P} - 2 \right)$

To minimize the total work $|W|$, we must minimize the function $f(P) = \frac{P}{2} + \frac{8}{P}$. Using the AM $\ge$ GM inequality: $\frac{\frac{P}{2} + \frac{8}{P}}{2} \ge \sqrt{\frac{P}{2} \cdot \frac{8}{P}}$ $\frac{\frac{P}{2} + \frac{8}{P}}{2} \ge \sqrt{4} = 2$

So, the minimum value of $\left( \frac{P}{2} + \frac{8}{P} \right)$ is $4$. This occurs when the terms are equal: $\frac{P}{2} = \frac{8}{P} \implies P^{2} = 16 \implies P = 4 \text{ bar}$ (This is valid since it falls within the given range $2 < P < 8$).

Finally, substituting this back to find the minimum work: $|W|_{min} = 300R (4 - 2) = 600R$

Correct Option: (B)