Two containers $A$ and $B$ contain equal volumes of an identical gas at the same pressure and temperature.…
Two containers $A$ and $B$ contain equal volumes of an identical gas at the same pressure and temperature. The gas in container $A$ is compressed to half its original volume isothermally, while the gas in container $B$ is compressed to half its original volume adiabatically. The ratio of the final
pressure of gas in container $B$ to that of gas in container $A$ is
$(2)^{\gamma-1}$
$\left(\frac{1}{2}\right)^{\gamma-1}$
$\left(\frac{1}{1-\gamma}\right)^2$
$\left(\frac{1}{\gamma-1}\right)^2$
Solution
Given,
Initial volume of container $A, V_A=V$
Initial volume of container $B, V_B=V$
Initial pressure in $A\left(p_A\right)=$ Initial pressure in
$
B\left(p_B\right)=p
$
Let their final volumes be $V_A^{\prime}=V_B^{\prime}=\frac{V}{2}$ and their final pressure be $p_A^{\prime}$ and $p_B^{\prime}$.
As, container $A$ is performing isothermal change
$
\begin{aligned}
& \therefore & p_A V_A & =p_A^{\prime} V_A / 2 \\
& \Rightarrow & p_A^{\prime} & =2 p_A
\end{aligned}
$
and container $B$ is performing adiabatic change
$
\begin{array}{rlrl}
& \therefore & p_B V_B^\gamma & =p_B^{\prime}\left(\frac{V_B}{2}\right)^\gamma \\
\Rightarrow & p_B^{\prime} & =p_B 2^\gamma
\end{array}
$
From Eqs. (i) and (ii), we get
$
\begin{aligned}
\frac{p_B^{\prime}}{p_A^{\prime}} & =\frac{p_B 2^\gamma}{p_A 2} \\
\therefore \quad \frac{p_B^{\prime}}{p_A^{\prime}} & =(2)^{\gamma-1}
\end{aligned}
$