The mean kinetic energy of monoatomic gas mol | vedclass

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(B) For an adiabatic process, the relation between temperature $T$ and volume $V$ is given by $T V^{\gamma-1} = 	ext{constant}$.For a monoatomic gas,

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$4$

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(B) For an adiabatic process, the relation between temperature $T$ and volume $V$ is given by $T V^{\gamma-1} = 	ext{constant}$.For a monoatomic gas, the adiabatic index $\gamma = \frac{5}{3}$.Given that the gas is compressed to $1/8$ of its initial volume, we have $V_1 = V$ and $V_2 = \frac{V}{8}$.Using the adiabatic relation: $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Substituting the values: $T_1 V^{\frac{5}{3}-1} = T_2 \left(\frac{V}{8}ight)^{\frac{5}{3}-1}$.$T_1 V^{\frac{2}{3}} = T_2 \left(\frac{V}{8}ight)^{\frac{2}{3}}$.$\frac{T_2}{T_1} = \left(\frac{V}{V/8}ight)^{\frac{2}{3}} = 8^{\frac{2}{3}} = (2^3)^{\frac{2}{3}} = 2^2 = 4$.The mean kinetic energy $\langle E angle$ of a gas molecule is given by $\langle E angle = \frac{3}{2} k_B T$, which implies $\langle E angle \propto T$.Therefore, $\frac{\langle E_2 angle}{\langle E_1 angle} = \frac{T_2}{T_1} = 4$.

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