A monoatomic ideal gas,initially at temperatu | vedclass

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(D) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $T V^{\gamma - 1} = 	ext{constant}$.Thus,$T_1 V_1^{\

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$(\frac{L_2}{L_1})^{2/3}$

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(D) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $T V^{\gamma - 1} = 	ext{constant}$.Thus,$T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$,which implies $\frac{T_1}{T_2} = (\frac{V_2}{V_1})^{\gamma - 1}$.Since the gas is monoatomic,the adiabatic index $\gamma = 5/3$. Therefore,$\gamma - 1 = 5/3 - 1 = 2/3$.The volume of the gas in a cylinder is $V = A 	imes L$,where $A$ is the cross-sectional area and $L$ is the length of the gas column.Substituting $V_1 = A L_1$ and $V_2 = A L_2$,we get $\frac{V_2}{V_1} = \frac{A L_2}{A L_1} = \frac{L_2}{L_1}$.Substituting these into the temperature ratio equation: $\frac{T_1}{T_2} = (\frac{L_2}{L_1})^{2/3}$.

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