An ideal gas ( 0.5 mol), initially at 2 bar pressure, is compressed...

Let the initial, intermediate, and final volumes of the gas be $V_1$, $V_2$, and $V_3$ respectively.

Using the ideal gas equation $V = \dfrac{nRT}{P}$, we have:
$V_1 = \dfrac{nRT}{2}$
$V_2 = \dfrac{nRT}{P}$
$V_3 = \dfrac{nRT}{8}$

The work done on the gas during the first step against a constant external pressure $P$ is:
$W_1 = -P_{ext,1}(V_2 - V_1) = -P \left( \dfrac{nRT}{P} - \dfrac{nRT}{2} \right) = nRT \left( \dfrac{P}{2} - 1 \right)$

The work done on the gas during the second step against a constant external pressure of $8$ bar is:
$W_2 = -P_{ext,2}(V_3 - V_2) = -8 \left( \dfrac{nRT}{8} - \dfrac{nRT}{P} \right) = nRT \left( \dfrac{8}{P} - 1 \right)$

The total work done on the gas is:
$W = W_1 + W_2 = nRT \left( \dfrac{P}{2} - 1 \right) + nRT \left( \dfrac{8}{P} - 1 \right) = nRT \left( \dfrac{P}{2} + \dfrac{8}{P} - 2 \right)$

To find the minimum value of $W$, we need to minimize the expression \left( \dfrac{P}{2} + \dfrac{8}{P} \right)$.<br><br>Using the AM-GM inequality:<br>$\dfrac{P}{2} + \dfrac{8}{P} \ge 2 \sqrt{\dfrac{P}{2} \times \dfrac{8}{P}} = 2 \sqrt{4} = 4$<br><br>The minimum value is$4$, which occurs when$\dfrac{P}{2} = \dfrac{8}{P} \Rightarrow P^2 = 16 \Rightarrow P = 4$bar. This value of$P$lies within the given range$(2 < P < 8)$.<br><br>Substituting this minimum value back into the work equation:<br>$W_\min = nRT (4 - 2) = 2nRT$<br><br>Given$n = 0.5$mol and$T = 600$K:<br>$W_\min = 2 \times 0.5 \times R \times 600 = 600R$<br><br>Answer:$600R