An ideal gas at pressure P_0 undergoes an iso | vedclass

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(C) Let the initial state be $A$ with $(P_0, V_0, T_0)$.In the isothermal process $A 	o B$, the temperature remains constant, so the state at $B$ is

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$2$

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(C) Let the initial state be $A$ with $(P_0, V_0, T_0)$.In the isothermal process $A 	o B$, the temperature remains constant, so the state at $B$ is $(P', 8V_0, T_0)$.In the adiabatic process $B 	o C$, the gas is compressed to volume $V_0$. The adiabatic relation is $T V^{\gamma-1} = 	ext{constant}$.For the process $B 	o C$:$T_0 (8V_0)^{\gamma-1} = T'' (V_0)^{\gamma-1}$Given $\gamma = 4/3$, so $\gamma - 1 = 1/3$.$T_0 (8V_0)^{1/3} = T'' (V_0)^{1/3}$$T_0 \cdot 8^{1/3} = T''$$T'' = 2 T_0$The average kinetic energy per molecule is given by $U_{av} = \frac{3}{2} k_B T$, which implies $U_{av} \propto T$.Therefore, the ratio of the average kinetic energy in the final state to the initial state is:$\frac{(U_{av})_f}{(U_{av})_i} = \frac{T''}{T_0} = \frac{2 T_0}{T_0} = 2$.

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