A sample of gas at temperature T is adiabatic | vedclass

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(D) For an adiabatic process, the relationship between temperature and volume is $TV^{\gamma-1} = 	ext{constant}$.Given initial temperature $T_1 = T$

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$W = RT[2 - \sqrt{2}]$

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(D) For an adiabatic process, the relationship between temperature and volume is $TV^{\gamma-1} = 	ext{constant}$.Given initial temperature $T_1 = T$ and initial volume $V_1 = V$.Final volume $V_2 = 2V$ and $\gamma = 3/2$.Thus, $\gamma - 1 = 3/2 - 1 = 1/2$.Using $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$, we get $T(V)^{1/2} = T_2(2V)^{1/2}$.$T_2 = T 	imes (V/2V)^{1/2} = T / \sqrt{2}$.The work done in an adiabatic process is given by $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.Assuming $n = 1$ mole, $W = \frac{R(T - T/\sqrt{2})}{3/2 - 1} = \frac{R(T - T/\sqrt{2})}{1/2} = 2R(T - T/\sqrt{2}) = 2RT(1 - 1/\sqrt{2})$.$W = 2RT - \frac{2RT}{\sqrt{2}} = 2RT - \sqrt{2}RT = RT(2 - \sqrt{2})$.

$A$ sample of gas at temperature $T$ is adiabatically expanded to double its volume. The work done by the gas in the process is (given, $\gamma = 3/2$):

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