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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$RT(2-\sqrt{2})$

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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 these values: $T V^{\gamma-1} = T_2 (2V)^{\gamma-1}$.$T_2 = T \left(\frac{V}{2V}\right)^{\gamma-1} = T \left(\frac{1}{2}\right)^{3/2-1} = T \left(\frac{1}{2}\right)^{1/2} = \frac{T}{\sqrt{2}}$.The work done in an adiabatic process is $W = \frac{R(T_1 - T_2)}{\gamma - 1}$.Substituting the values: $W = \frac{R(T - T/\sqrt{2})}{3/2 - 1} = \frac{R T (1 - 1/\sqrt{2})}{1/2}$.$W = 2 R T \left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) = \sqrt{2} R T (\sqrt{2}-1) = R T (2 - \sqrt{2})$.

$A$ sample of gas at temperature $T$ is adiabatically expanded to double its volume. The work done by the gas in the process is (given $\frac{C_{P}}{C_{V}}=\gamma=\frac{3}{2}$,$R=$ gas constant).

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