A sample of 1 mole of gas at temperature T is | vedclass

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(A) For an adiabatic process,the relation between temperature and volume is given by $TV^{\gamma-1} = 	ext{constant}$.Given $n = 1$,initial temperatu

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

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(A) For an adiabatic process,the relation between temperature and volume is given by $TV^{\gamma-1} = 	ext{constant}$.Given $n = 1$,initial temperature $= T$,initial volume $= V$,final volume $= 2V$,and $\gamma = \frac{3}{2}$.Applying the relation: $T(V)^{\frac{3}{2}-1} = T_f(2V)^{\frac{3}{2}-1}$.$T(V)^{\frac{1}{2}} = T_f(2)^{\frac{1}{2}}(V)^{\frac{1}{2}}$.$T = T_f \sqrt{2} \Rightarrow T_f = \frac{T}{\sqrt{2}}$.The work done in an adiabatic process is given by $W = \frac{nR(T_i - T_f)}{\gamma - 1}$.Substituting the values: $W = \frac{1 \cdot R(T - \frac{T}{\sqrt{2}})}{\frac{3}{2} - 1}$.$W = \frac{R T (1 - \frac{1}{\sqrt{2}})}{\frac{1}{2}} = 2RT \left(1 - \frac{1}{\sqrt{2}}ight)$.$W = RT(2 - \sqrt{2})$.

$A$ sample of $1$ mole of gas at temperature $T$ is adiabatically expanded to double its volume. If the adiabatic constant for the gas is $\gamma = \frac{3}{2}$,then the work done by the gas in the process is:

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