Question

The initial pressure & volume of a given mass of gas having $\gamma =\frac{C_P}{C_V}$ are $P_0$ and $V_0$. The gas is seperated from the surroundings by a diathermic wall. It is suddenly compressed to $\frac{V_0}{6}$ and then slowly compressed to $\frac{V_0}{36}$. If the final pressure of the gas is $n^{\gamma +1}$, value of $n$ is

Answer 1

The entire process can be represented as : $(P_0,V_0)\underset{\text{Step}-I}{\underset{\text{Compressed}}{\overset{\text{suddenly}}{\to }}}(P_1,\frac{V_0}{6})\underset{\text{Step}-II}{\underset{\text{Compressed}}{\overset{\text{slowly}}{\to }}}(P_2,\frac{V_0}{36})$

A diathermic wall means the gas is capable of exchanging heat with the surroundings.
During step - I, when the gas is suddenly compressed, the process is adibatic as there is no chance for the heat to escape from the system.
Thus, using the equation of state of an adiabatic process in terms of $P$ and $V$, we get
$P_0{V}_0^{\gamma }=P_1{V}_1^{\gamma }$
From the data given in the question,
$P_0{V}_0^{\gamma }=P_1{\left(\frac{V_0}{6}\right)}^{\gamma }$
$\Rightarrow P_1=\left(6^{\gamma }\right)P_0.....\left(1\right)$

During step - II, when the gas is slowly compressed, the gas can exchange heat with the surroundings through the diathermic wall, thus the process will be isothermal in nature.
Using equation of state of an isothermal process,
$P_1{V}_1=P_2{V}_2$,
We get, $P_2\times \left(\frac{V_0}{36}\right)=P_1\times \left(\frac{V_0}{6}\right)$
$\Rightarrow P_2=6P_1$
But from $\left(1\right)$, we know that $P_1=\left(6^{\gamma }\right)P_0$
$\therefore P_2=6\times \left(6^{\gamma }\right)P_0$
$\Rightarrow P_2=\left(6^{\gamma +1}\right)P_0$
Thus, $n=6$