Question

The temperature $T$ of one mole of an ideal gas varies with its volume $V$ as $T=-\alpha V^3+\beta V^2$, where $\alpha$ and $\beta$ are positive constants. The maximum pressure of the gas during this process is

• A. $\frac{\alpha \beta }{2R}$
• B. $\frac{{\beta }^{2}R}{4\alpha }$
• C. $\frac{(\alpha +\beta )R}{2{\beta }^2}$
• D. $\frac{{\alpha }^{2}R}{2\beta }$

Answer 1

The correct option is B $\frac{{\beta }^{2}R}{4\alpha }$

Given,
$T=-\alpha V^3+\beta V^2.....\left(1\right)$
Ideal gas equation for 1 mole of an ideal gas is given by
$PV=RT.....\left(2\right)$
Using $\left(2\right)$ in $\left(1\right)$, we can write that
$P=(-\alpha V^2+\beta V)R......\left(3\right)$
On differentiating both sides with respect to $V$,
$\frac{dP}{dV}=(-2\alpha V+\beta )R.....\left(4\right)$

To find the maxima, $\frac{dP}{dV}=0$ and $\frac{d^{2}P}{d{V}^2}<0$
From $\left(4\right)$, we can say that,
$0=(-2\alpha V+\beta )R$
$\Rightarrow V=\frac{\beta }{2\alpha }....\left(5\right)$
and $\frac{d^{2}P}{d{V}^2}=-2\alpha R<0$
Using $\left(5\right)$ in $\left(3\right)$, we get
$P=(-\alpha \frac{{\beta }^2}{4{\alpha }^2}+\frac{{\beta }^2}{2\alpha })R$
$\Rightarrow P=\frac{{\beta }^{2}R}{4\alpha }$
Thus, option (b) is the correct answer.