A sample of gas at temperature T is adiabatic | vedclass

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(D) For an adiabatic process,the work done by the gas is given by $W = \frac{nR(T_i - T_f)}{\gamma - 1}$.Given $n = 1$,$T_i = T$,and $V_f = 2V_i$.The

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

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(D) For an adiabatic process,the work done by the gas is given by $W = \frac{nR(T_i - T_f)}{\gamma - 1}$.Given $n = 1$,$T_i = T$,and $V_f = 2V_i$.The adiabatic relation is $T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$.Substituting the values: $T V^{\gamma-1} = T_f (2V)^{\gamma-1}$.$T_f = T \left(\frac{V}{2V}\right)^{\gamma-1} = T \left(\frac{1}{2}\right)^{3/2-1} = T \left(\frac{1}{2}\right)^{1/2} = \frac{T}{\sqrt{2}}$.Now,substitute $T_f$ into the work formula:$W = \frac{R(T - T/\sqrt{2})}{3/2 - 1} = \frac{R T (1 - 1/\sqrt{2})}{1/2} = 2RT \left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) = RT \left(\frac{2\sqrt{2}-2}{\sqrt{2}}\right) = RT(2 - \sqrt{2})$.

$A$ sample of gas at temperature $T$ is adiabatically expanded to double its volume. The adiabatic constant for the gas is $\gamma = 3/2$. The work done by the gas in the process is: $(\mu = 1 \text{ mole})$

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