Question

One mole of an ideal monoatomic gas is taken along the path ABCA as shown in the PV diagram. The maximum temperature attained by the gas along the path BC is given by

• A. $\frac{25P_0V_0}{8R}$
• B. $\frac{3P_0V_0}{R}$
• C. $\frac{P_0V_0}{R}$
• D. $\frac{5P_0V_0}{2R}$

Answer 1

Answer: (A)

$\frac{25P_0V_0}{8R}$

Answer 2

The temperature $T$ is proportional to the product $PV$ ($PV=nRT$). For $n=1$, $T = \frac{PV}{R}$. The path BC is a line segment connecting B($V_0, 3P_0$) and C($2V_0, P_0$). The equation of this line is $P = 5P_0 - \frac{2P_0}{V_0}V$. The product $PV = V(5P_0 - \frac{2P_0}{V_0}V) = 5P_0V - \frac{2P_0}{V_0}V^2$. To find the maximum, we set the derivative with respect to $V$ to zero: $\frac{d(PV)}{dV} = 5P_0 - \frac{4P_0}{V_0}V = 0$, which gives $V = \frac{5V_0}{4}$. The corresponding pressure is $P = 5P_0 - \frac{2P_0}{V_0}(\frac{5V_0}{4}) = \frac{5P_0}{2}$. The maximum product $PV_{max} = (\frac{5P_0}{2})(\frac{5V_0}{4}) = \frac{25P_0V_0}{8}$. Thus, $T_{max} = \frac{PV_{max}}{R} = \frac{25P_0V_0}{8R}$.