Question

For an ideal gas, the ratio of specific heat capacities is $\frac{C_P}{C_V}=\gamma$. If the gas undergoes a polytropic process according to the equation $P{V}^n=k$, find the range of $n$ for which the temperature of the gas increases when it rejects heat to the surroundings.

• A. $n>\gamma$
• B. $1<n<\gamma$
• C. $n<1$
• D. $n<0$

Answer 1

The correct option is B $1<n<\gamma$

Since it rejects heat to the surroundings on increasing temperature, specific heat capacity of the gas for the given polytropic process must be negative.
For one mole gas, work done during a polytropic process $P{V}^n=k$ is given by
$W=\int PdV=\frac{R(T_1-T_2)}{n-1}$
and, change in internal energy of the gas due to increase in temperature is given by
$\Delta U=C_V\Delta T=\frac{R(T_2-T_1)}{\gamma -1}$
From first law of thermodynamics,
$\Delta Q=\Delta U+W$
$\Rightarrow C\Delta T=\frac{R}{\gamma -1}\left(\Delta T\right)-\frac{R}{n-1}\left(\Delta T\right)$
$\Rightarrow C=\frac{R}{\gamma -1}-\frac{R}{n-1}$
$\Rightarrow C=\frac{R(n-\gamma )}{(n-1)(\gamma -1)}$
$\because \gamma$ is always > 1, $C$ can be negative if $(n-\gamma )$ is negative and $(n-1)$ is positive.

i.e $1<n<\gamma$
Thus, option (b) is the correct answer.