Question
A certain gas at atmospheric pressure is compressed adiabatically so that its volume becomes half of its original volume. Calculate the resulting pressure in $Nm^{-2}$. Take $\gamma=1.4$ for air.
✓
Answer
Let the original volume, $V_1 = V $
$\therefore$ Final valume, $\text{V}_2=\frac{\text{V}}{2}$ Initial pressure, $P_1 = 0.76$ metre of Hg column. Let $P_2$ be the fanal pressure after compression. As the change is adiabatic, $\therefore\text{P}_1\text{V}_1^\gamma=\text{P}_2\text{V}_2^\gamma$
$\text{P}_2=\text{P}_1\Big(\frac{\text{V}_1}{\text{V}_2}\Big)^\gamma=\text{P}_1\bigg(\frac{\text{V}}{\frac{\text{V}}2{}}\bigg)^{1.4}$
$\text{P}_2=0.76\times(2)^{1.4}$
$\text{P}_2=2.00$ metre of Hg column, As $\text{P}=\text{h}\rho\text{g}$
$\therefore\text{P}_22.00\times(13.6\times10^3)\times9.8\text{ Nm}^{-2}$
$\text{P}_2=2.672\times10^5\text{Nm}^{-2}$
$\therefore$ Final valume, $\text{V}_2=\frac{\text{V}}{2}$ Initial pressure, $P_1 = 0.76$ metre of Hg column. Let $P_2$ be the fanal pressure after compression. As the change is adiabatic, $\therefore\text{P}_1\text{V}_1^\gamma=\text{P}_2\text{V}_2^\gamma$
$\text{P}_2=\text{P}_1\Big(\frac{\text{V}_1}{\text{V}_2}\Big)^\gamma=\text{P}_1\bigg(\frac{\text{V}}{\frac{\text{V}}2{}}\bigg)^{1.4}$
$\text{P}_2=0.76\times(2)^{1.4}$
$\text{P}_2=2.00$ metre of Hg column, As $\text{P}=\text{h}\rho\text{g}$
$\therefore\text{P}_22.00\times(13.6\times10^3)\times9.8\text{ Nm}^{-2}$
$\text{P}_2=2.672\times10^5\text{Nm}^{-2}$
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