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^{\ga

Vedclass · Accepted Answer

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

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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,$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.For a monoatomic gas,the adiabatic index $\gamma = 5/3$,so $\gamma - 1 = 2/3$.The volume of the gas in a cylinder of cross-sectional area $A$ is $V = A 	imes L$,where $L$ is the length of the gas column.Substituting $V_1 = A L_1$ and $V_2 = A L_2$ into the adiabatic equation:$T_1 (A L_1)^{2/3} = T_2 (A L_2)^{2/3}$.Dividing both sides by $T_2 (A L_1)^{2/3}$,we get:$\frac{T_1}{T_2} = \frac{(A L_2)^{2/3}}{(A L_1)^{2/3}} = (\frac{L_2}{L_1})^{2/3}$.

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