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A perfect gas is found to obey the relation $P V^{3 / 2}=$ constant during an adiabatic process. If such a gas initially at a temperature $T$, is compressed to half of its initial volume, then its final temperature will be :
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The correct answer is:
$(2)^{1 / 2} T$
Gas obeys the equation
$P V^{3 / 2}=$ constant
Comparing with $P V^\gamma=$ constant
$\gamma=\frac{3}{2}$
For adiabatic process
$\begin{aligned} T_1 V_1^{\gamma-1} & =T_2 V_2^{\gamma-1} \\ T(V)^{3 / 2-1} & =T_2\left(\frac{V}{2}\right)^{3 / 2-1} \\ T_2 & =T(2)^{3 / 2-1} \\ & =(2)^{1 / 2} T\end{aligned}$
$P V^{3 / 2}=$ constant
Comparing with $P V^\gamma=$ constant
$\gamma=\frac{3}{2}$
For adiabatic process
$\begin{aligned} T_1 V_1^{\gamma-1} & =T_2 V_2^{\gamma-1} \\ T(V)^{3 / 2-1} & =T_2\left(\frac{V}{2}\right)^{3 / 2-1} \\ T_2 & =T(2)^{3 / 2-1} \\ & =(2)^{1 / 2} T\end{aligned}$
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