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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 $T_1$, the magnitude work done in the process is
$\frac{9}{8}\,R{T_1}$
$\frac{3}{2}\,R{T_1}$
$\frac{15}{8}\,R{T_1}$
$\frac{9}{2}\,R{T_1}$
Solution
Initially
$V_{1}=5.6 l, T_{1}=273 K, P_{1}=1 a t m$
$\gamma=\frac{5}{3}(F \text { or monoa } \rightarrow \text { micgas })$
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
$W=\frac{n R \Delta T}{\gamma-1}=\frac{9}{8} R T_{1}$
