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A given mass of a gas is compressed isothermally until its pressure is doubled. It is then allowed to expand adiabatically until its original volume is restored and its pressure is then found to be 0.75 of its initial pressure. The ratio of the specific heats of the gas is approximately :
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The correct answer is:
1.41
isothermal process, temperature of the gas remains constant,so the gas obeys Boyle's law. That is,
$P \propto \frac{1}{V}$
$\Rightarrow \quad \frac{P_2}{P_1}=\frac{V_1}{V_2}$
$\Rightarrow \quad \frac{2 P}{P}=\frac{V_1}{V_2}$
$\therefore \quad \frac{V_1}{V_2}=2$ $\ldots(1)$
Now, the gas is expanded adiabatically, so $P V^\gamma=$ constant
$\frac{P_1}{P_2}=\left(\frac{V_2}{V_1}\right)^\gamma$
$\Rightarrow \quad \frac{2 P}{0.75 P}=\left(\frac{2}{1}\right)^\gamma$
(since volume is restored)
$\Rightarrow \quad \log \left(\frac{8}{3}\right)=\gamma \log 2$
$\Rightarrow \quad \log 8-\log 3=\gamma \log 2$
$\therefore \quad \gamma=1.41$
$P \propto \frac{1}{V}$
$\Rightarrow \quad \frac{P_2}{P_1}=\frac{V_1}{V_2}$
$\Rightarrow \quad \frac{2 P}{P}=\frac{V_1}{V_2}$
$\therefore \quad \frac{V_1}{V_2}=2$ $\ldots(1)$
Now, the gas is expanded adiabatically, so $P V^\gamma=$ constant
$\frac{P_1}{P_2}=\left(\frac{V_2}{V_1}\right)^\gamma$
$\Rightarrow \quad \frac{2 P}{0.75 P}=\left(\frac{2}{1}\right)^\gamma$
(since volume is restored)
$\Rightarrow \quad \log \left(\frac{8}{3}\right)=\gamma \log 2$
$\Rightarrow \quad \log 8-\log 3=\gamma \log 2$
$\therefore \quad \gamma=1.41$
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