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(A) For an adiabatic process,the relation between pressure and volume is given by $P_1 V_1^\gamma = P_2 V_2^\gamma$. Given that the final volume $V_2

Vedclass · Accepted Answer

$32$

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(A) For an adiabatic process,the relation between pressure and volume is given by $P_1 V_1^\gamma = P_2 V_2^\gamma$. Given that the final volume $V_2 = V_1 / 8$,so the ratio $V_1 / V_2 = 8$. The adiabatic index for a monoatomic gas is $\gamma = 5/3$. Substituting these values into the equation: $\frac{P_2}{P_1} = \left( \frac{V_1}{V_2} \right)^\gamma = (8)^{5/3}$. Since $8 = 2^3$,we have $(2^3)^{5/3} = 2^5 = 32$. Therefore,the ratio of the final pressure to the initial pressure is $32$.

$A$ monoatomic gas is suddenly compressed to $(1/8)^{\text{th}}$ of its initial volume adiabatically. The ratio of the final pressure to initial pressure of the gas is (given $\gamma = 5/3$).

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