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There are two identical containers $C_1$ and $C_2$ containing to identical gases. Gas in $C_1$ is reduced to half of its original volume adiabatically, while the gas in container $C_2$ is also reduced to half of its initial volume isothermally. Find the ratio of final pressure in these containers. ( $\gamma$ be the adiabatic constant).
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Verified Answer
The correct answer is:
$2^{\gamma-1}: 1$
Since, container are identical, so initial volume and pressure will be same for identical gases.
For container $C_1$,
$$
p_1 V_1^\gamma=p_2 V_2^2
$$
(adiabitic)
$$
p_2=\left(\frac{V_1}{V_2}\right)^\gamma p_1=\left(\frac{V_1}{V_1 / 2}\right)^\gamma p_1=2^\gamma p_1
$$
For container $C_2$,
$$
\begin{aligned}
p_1 V_1 & =p_2^{\prime} V_2 \text { (isothermal) } \\
p_2^{\prime} & =\left(\frac{V_1}{V_2}\right) p_1=\left(\frac{V_1}{V_1 / 2}\right) p_1=2 p_1
\end{aligned}
$$
From Eqs. (i) and (ii), we get
$$
\frac{p_2}{p_2{ }^{\prime}}=\frac{2^\gamma p_1}{2 p_1}=2^{\gamma-1}: 1
$$
For container $C_1$,
$$
p_1 V_1^\gamma=p_2 V_2^2
$$
(adiabitic)
$$
p_2=\left(\frac{V_1}{V_2}\right)^\gamma p_1=\left(\frac{V_1}{V_1 / 2}\right)^\gamma p_1=2^\gamma p_1
$$
For container $C_2$,
$$
\begin{aligned}
p_1 V_1 & =p_2^{\prime} V_2 \text { (isothermal) } \\
p_2^{\prime} & =\left(\frac{V_1}{V_2}\right) p_1=\left(\frac{V_1}{V_1 / 2}\right) p_1=2 p_1
\end{aligned}
$$
From Eqs. (i) and (ii), we get
$$
\frac{p_2}{p_2{ }^{\prime}}=\frac{2^\gamma p_1}{2 p_1}=2^{\gamma-1}: 1
$$
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