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A diatomic gas at pressure $P$, compressed adiabatically to half of its volume, what is the final pressure?
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Verified Answer
The correct answer is:
$(2)^{1.4} P$
For adiabatic conditions, $P V^\gamma=$ Constant
$P_1 V_1^\gamma=P_2 V_2^\gamma ; V_2=\frac{1}{2} V_1$
$P_2=P_1\left(\frac{V_1}{V_2}\right)^\gamma \quad$ [For diatomic gas, $\gamma=1.4$ ]
$P_2=P_1\left(\frac{V_1 \times 2}{V_1}\right)^{1.4}$
$P_2=P_1(2)^{1.4}=(2)^{1.4} P$
$P_1 V_1^\gamma=P_2 V_2^\gamma ; V_2=\frac{1}{2} V_1$
$P_2=P_1\left(\frac{V_1}{V_2}\right)^\gamma \quad$ [For diatomic gas, $\gamma=1.4$ ]
$P_2=P_1\left(\frac{V_1 \times 2}{V_1}\right)^{1.4}$
$P_2=P_1(2)^{1.4}=(2)^{1.4} P$
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