A monoatomic ideal gas,initially at temperatu | vedclass

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Question

(A) For an adiabatic process,the relation between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Therefore,the ratio of

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$\left[\frac{L_1}{L_2}\right]^{2/3}$

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(A) For an adiabatic process,the relation between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Therefore,the ratio of temperatures is $\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}ight)^{\gamma-1}$.For a monoatomic ideal gas,the adiabatic exponent is $\gamma = \frac{5}{3}$.Thus,$\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$.Since the gas is in a cylinder of constant cross-sectional area $A$,the volume is $V = A 	imes L$. Therefore,$V_1 = A L_1$ and $V_2 = A L_2$.Substituting these into the temperature ratio equation:$\frac{T_2}{T_1} = \left(\frac{A L_1}{A L_2}ight)^{2/3} = \left(\frac{L_1}{L_2}ight)^{2/3}$.

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