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

Solution :
Initially
$${V_1} = 5.6\,\ell ,{T_1} = 273\,K,{P_1} = 1\,{\text{atm,}}$$
$$\gamma = \frac{5}{3}$$   (For monoatomic gas)
The number of moles of gas is $$n = \frac{{5.6\,\ell }}{{22.4\,\ell }} = \frac{1}{4}$$
Finally (after adiabatic compression)
$${V_2} = 0.7\,\ell $$
For adiabatic compression $${T_1}{V_1}^{\gamma - 1} = {T_2}{V_2}^{\gamma - 1}$$
$$\eqalign{ & \therefore \,\,{T_2} = {T_1}{\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma - 1}} \cr & = {T_1}{\left( {\frac{{5.6}}{{0.7}}} \right)^{\frac{5}{3} - 1}} \cr & = {T_1}{\left( 8 \right)^{\frac{2}{3}}} \cr & = 4{T_1} \cr} $$
We know that work done in adiabatic process is
$$W = \frac{{nR\Delta T}}{{\gamma - 1}} = \frac{9}{8}\,R{T_1}$$