$5.6 l i t r e$ of helium gas at STP is adiabatically compressed to $0.7 l i t r e$ . Taking the initial temperature to be $T_{1}$ , the work done in the process is

\frac{9}{8} R T_{1}

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

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

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

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Solution

The correct option is A $\frac{9}{8} R T_{1}$
Number of moles of $H e = 5.6 / 22.4 = 1 / 4$
Now $T (5.6)^{γ - 1} = T_{2} (0.7)^{γ - 1}$
$T_{1} = T_{2} {(\frac{1}{8})}^{2 / 3} \Rightarrow 4 T_{1} = T_{2}$
Work done $= \frac{n R [T_{2} - T_{1}]}{γ - 1} = \frac{\frac{1}{4} R [3 T_{1}]}{\frac{2}{3}} = \frac{9}{8} R T_{1}$ .

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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

The correct option is A 98RT1Number of moles of He=5.6/22.4=1/4Now T(5.6)γ−1=T2(0.7)γ−1T1=T2(18)2/3⇒4T1=T2Work done =nR[T2−T1]γ−1=14R[3T1]23=98RT1.

$5.6 l i t r e$ of helium gas at STP is adiabatically compressed to $0.7 l i t r e$ . Taking the initial temperature to be $T_{1}$ , the work done in the process is