A sample of gas at temperature T is adiabatic | vedclass

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(D) For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Given $T_1 = T$,$V_1 = V$,$V_2 =

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

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(D) For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Given $T_1 = T$,$V_1 = V$,$V_2 = 2V$,and $\gamma = 3/2$.Substituting the values: $T V^{3/2-1} = T_2 (2V)^{3/2-1} \implies T V^{1/2} = T_2 (2V)^{1/2}$.Solving for $T_2$: $T_2 = T / \sqrt{2}$.The work done by the gas in an adiabatic process is $W = \frac{R(T_1 - T_2)}{\gamma - 1}$.Substituting the values: $W = \frac{R(T - T/\sqrt{2})}{3/2 - 1} = \frac{R(T - T/\sqrt{2})}{1/2} = 2R(T - T/\sqrt{2}) = 2RT(1 - 1/\sqrt{2}) = RT(2 - \sqrt{2})$.

$A$ 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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