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

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

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

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

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

answer is A.

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Detailed Solution

${TV}^{γ - 1} = constant \Rightarrow T_{2} = {(\frac{5.6}{0.7})}^{\frac{2}{3}} (T_{1}) = 4 T_{1} Work done W = \frac{nR ∆ T}{1 - γ} = \frac{(1 / 4) R (3 T_{1})}{- (2 - 3)} = - \frac{9}{8} {RT}_{1}$

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