Question

An ideal gas follows a path given by $P=-aV+b$ where a and b are positive constants. At $V=V_0$ the ideal gas attains the maximum temperature. If a real gas is taken with the same a and b as respective Van der Waal's constants then select the correct statement(s):

• A. Inversion temperature of the real gas is $R{V}_0$
• B. If the initial volume is $\frac{b}{a}$ temperature will always increase with the expansion of the ideal gas
• C. If the initial volume is $\frac{b}{a}$ the temperature will first increase then decrease with the compression of the ideal gas
• D. Inversion temperature of the real gas is $\frac{1}{R{V}_0}$

Answer 1

Answer: (C), (D)

The correct options are
C If the initial volume is $\frac{b}{a}$ the temperature will first increase then decrease with the compression of the ideal gas
D Inversion temperature of the real gas is $\frac{1}{R{V}_0}$

In the given process the $P-V$ equation will be a straight line with a negative slope and a positive intercept, i.e.,
$P=-aV+b$
Here, $a$ and $b$ are positive constants.
$P=-aV+b$
$\Rightarrow PV=-a{V}^2+bV$
$\Rightarrow nRT=-a{V}^2+bV[\because PV=nRT\left(\text{From ideal gas equation}\right)]$
$T=\frac{1}{nR}(-a{V}^2+bV)$
$\frac{dT}{dV}=0$
$\Rightarrow \frac{1}{nR}(-2aV+b)=0$
$\Rightarrow -2aV+b=0$
$\Rightarrow V=\frac{b}{2a}=V_0$
Now,
$\frac{d^{2}T}{d{V}^2}=-2a=-ve$
$\therefore T$ has some maximum value.
Inversion temperature is $\frac{2a}{Rb}$
$\therefore T_i=\frac{1}{R{V}_0}$