$5.6$ liter of helium gas at $STP$ is adiabatically compressed to $0.7$ liter. Taking the initial temperature to be $T _1$, the work done in the process is
IIT 2011, Diffcult
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$V_1=5.6 l, T_1=273 K, P_1=1 a t m$
$\gamma=\frac{5}{3}(F \text { or monoa } \rightarrow \text { micgas })$
The number of moles of gas is $n=\frac{5.6 l}{22.4 l}=\frac{1}{4}$
Finally (after adiabatic compression)
$V_2=0.7 l$
For adiabatic compression $T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}$
$\therefore T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}=T_1\left(\frac{5.6}{0.7}\right)^{\frac{5}{3}-1}=T_1(8)^{2 / 3}$
$=4 T_1$
We know that work done in adiabatic process is
$W=\frac{n R \Delta T}{\gamma-1}=\frac{9}{8} R T_1$
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