Two moles of an ideal monatomic gas undergo a cyclic process ABCA as shown in the figure. The ratio of temperature at B and A can be represented as $\frac{3^{x}}{2^{y}}$ . Find (x + y).

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Solution

Equation of line AB,
$P - P_{0} = tan 60^{o} (V - V_{0}) \Rightarrow P - P_{0} = \sqrt{3} (V - V_{0}) \cdot \cdot \cdot (1)$
Equation of line BC,
$P - P_{0} = \frac{- 1}{\sqrt{3}} (V - 6 V_{0}) \cdot \cdot \cdot (2)$
Subtracting $(2)$ from $(1)$
$0 = \sqrt{3} (V - V_{0}) + \frac{1}{\sqrt{3}} (V - 6 V_{0})$

$\Rightarrow (\sqrt{3} + \frac{1}{\sqrt{3}}) V - (\sqrt{3} + \frac{6}{\sqrt{3}}) V_{0} = 0$

$\Rightarrow V = \frac{9 V_{0} / \sqrt{3}}{4 / \sqrt{3}} = \frac{9 V_{0}}{4}$
$∴ Temparature of B, T_{B} = \frac{P_{B} V_{B}}{2 R} = \frac{3 P_{0} (\frac{9 V_{0}}{4})}{2 R}$

$Temparature of A , T_{A} = \frac{P_{A} V_{A}}{2 R} = \frac{P_{0} V_{0}}{2 R}$
$∴ \frac{T_{B}}{T_{A}} = \frac{\frac{27 P_{0} V_{0}}{8 R}}{\frac{P_{0} V_{0}}{2 R}} = \frac{27}{4} = \frac{3^{3}}{2^{2}}$
$Hence, \frac{3^{x}}{2^{y}} = \frac{3^{3}}{2^{2}}$
Therefore, $x + y = 3 + 2 = 5$

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