Initially

V₁ = 5.6 l, T₁ = 273 K, P₁ = 1 atm,

$$\gamma = {5 \over 3}$$ (For monoatomic gas)

The number of moles of gas is $$n = {{5.6l} \over {22.4l}} = {1 \over 4}$$

Finally (after adiabatic compression)

V₂ = 0.7 l

For adiabatic compression $${T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}$$

$$\therefore$$ $${T_2} = {T_1}{\left( {{{{V_1}} \over {{V_2}}}} \right)^{\gamma - 1}} = {T_1}{\left( {{{5.6} \over {0.7}}} \right)^{{5 \over 3} - 1}} = {T_1}{(8)^{2/3}} = 4{T_1}$$

Work done during an adiabatic process is

$$W = {{nR[{T_1} - {T_2}]} \over {(\gamma - 1)}} = {{{1 \over 4}R[{T_1} - 4{T_1}]} \over {\left[ {{5 \over 3} - 1} \right]}} = - {9 \over 8}R{T_1}$$

Negative sign shows that work is done on the gas.