Question

One mole of ideal monotonic gas is taken through process AB given by $p=\alpha -\beta v^2$ where $\alpha$ and $\beta$ are +ve constants as shown in the P-V diagram. Which of the following are the correct statements regarding the process

• A. Temp is maximum at $V=\sqrt{\frac{\alpha }{3\beta }}$
• B. Temp is minimum at $V=\sqrt{\frac{\alpha }{3\beta }}$
• C. Rate of increse of temp of gas with volume is maximum at A
• D. Rate of increase of temp of gas with volume in maximum at $V=\sqrt{\frac{\alpha }{3\beta }}$

Answer 1

Answer: (A), (C)

Temp is maximum at $V=\sqrt{\frac{\alpha }{3\beta }}$
Rate of increse of temp of gas with volume is maximum at A

Given: $p=\alpha -\beta v^2$

For one mole of an ideal gas, $pv=RT$

$RT=(\alpha -\beta v^2)v⟹T=\frac{\alpha v-\beta v^3}{R}$

For extrimas, $\frac{dT}{dv}=0$

$\frac{\alpha -3\beta v^2}{R}=0⟹v=\sqrt{\frac{\alpha }{3\beta }}$

Also, $\frac{d^{2}T}{d{v}^2}=\frac{-6\beta v}{R}<0$

Thus at $v=\sqrt{\frac{\alpha }{3\beta }}$ Temp is maximum.

Also as $v_A<v_B⟹\frac{dT}{dv}$ is max at A