A polyatomic gas is compressed to left(frac{1 | vedclass

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(D) For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = 	ext{constant}$.Let the initial pressure be

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$16$

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(D) For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = 	ext{constant}$.Let the initial pressure be $P_1 = P_0$ and initial volume be $V_1 = V$.The final volume is $V_2 = \frac{V}{8}$.The adiabatic index is given as $\gamma = \frac{4}{3}$.Using the relation $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$,we have:$P_0 V^{\gamma} = P_2 \left(\frac{V}{8}ight)^{\gamma}$.$P_2 = P_0 \left(\frac{V}{V/8}ight)^{\gamma} = P_0 (8)^{\gamma}$.Substituting $\gamma = \frac{4}{3}$:$P_2 = P_0 (8)^{4/3} = P_0 (2^3)^{4/3} = P_0 (2^4) = 16 P_0$.Thus,the new pressure is $16 P_0$.

$A$ polyatomic gas is compressed to $\left(\frac{1}{8}\right)^{\text{th}}$ of its original volume adiabatically. If its initial pressure is $P_0$,what will be its new pressure (in $P_0$)? (Given: $\gamma = \frac{C_p}{C_v} = \frac{4}{3}$)

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