5.6 litre of 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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$\frac{9}{8} R T_{1}$

$\frac{3}{2} R T_{1}$

$\frac{15}{8} R T_{1}$

$\frac{9}{2} R T_{1}$

answer is A.

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Detailed Solution

(1) Initially

$V_{1} = 5.6, T 1 = 273 K, P 1 = 1 a t m, γ = \frac{5}{3} (F o r m o n o a t o m i c g a s) T h e n u m b e r s o f m o l e s o f g a s i s n = \frac{5.6 l}{22.4 l} = \frac{1}{4} F i n a l l y (a f t e r a d i a b a t i c c o m p r e s s i o n)$

$V_{2} = 0.7 F o r a d i a b a t i c c o m p r e s s i o n T_{1} {V_{1}}^{γ - 1} = T_{2} {V_{2}}^{γ - 1} ∴ T_{2} = T_{1} {(\frac{V_{1}}{V_{2}})}^{γ - 1} = T_{1} {(\frac{5.6}{0.7})}^{\frac{5}{3} - 1} = T_{1} {(8)}^{\frac{2}{3}} = 4 T_{1} W e k n o w t h a t w o r k d o n e i n a d i a b a t i c p r o c e s s i s W = \frac{n R ∆ T}{γ - 1} = \frac{9}{8} R T_{1}$