5.6 liter of helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be T1, the work done in the process is

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5.6 liter of helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be T₁, the work done in the process is

$\frac{9}{8} {RT}_{1}$

$\frac{15}{8} {RT}_{1}$

$\frac{3}{2} {RT}_{1}$

$\frac{9}{2} {RT}_{1}$

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$T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$

$T_{2} = T_{1} {(\frac{V_{1}}{V_{2}})}^{γ - 1}$

$= T_{1} {(\frac{5 .7}{0 .7})}^{\frac{5}{3} - 1} = T_{1} {(8)}^{2 / 3} = 4 T_{1}$

no. of mole = ¼

$W = \frac{nR (T_{1} - T_{2})}{γ - 1} = \frac{\frac{1}{4} \times R (4 T_{1} - T_{1})}{2 / 3} = \frac{9}{8} {RT}_{1}$

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