A sample of one mole of gas at temperature T is adiabatically compressed to half of its volume. The work done on the gas in the process is (given $γ = \frac{3}{2}$ )

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$W = R T (2 - 2 \sqrt{2})$

$W = R T (2 \sqrt{2} - 2)$

$W = R T (2 - \sqrt{2})$

$W = R T (\sqrt{2} - 2)$

answer is A.

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Detailed Solution

For one mole of ideal gas, the work done in adiabatic process
$W = \frac{R}{r - 1} [T_{1} - T_{2}]$ but $T_{1} V_{1}^{^{γ - 1}} = T_{2} V_{2}^{γ - 1}$
$T_{2} = T_{1} {(\frac{V_{1}}{V_{2}})}^{γ - 1} = {(\frac{V}{V / 2})}^{3 / 2 - 1} T = T {(2)}^{1 / 2} = \sqrt{2} T$
$W = \frac{R}{\frac{3}{2} - 1} [T - \sqrt{2} T]$
$W = 2 R T [1 - \sqrt{2}] \Rightarrow W = R T (2 - 2 \sqrt{2})$