Given, Initial volume of air at STP ${\text{V}}_1=\text{5}0\text{l}=\text{5}0\times \text{1}{0}^{–3}{\text{m}}^3$ 

Final Volume of air ${\text{V}}_{\text{2}}=\text{1}0\text{l}=\text{1}0\times \text{1}{0}^{-3}{\text{m}}^3{\text{ ; P}}_{\mathrm{atm}}=\text{1}00\text{kPa}$

Now, Process is Isothermally compressed.

∴ From 1st law of thermodynamics $\Delta \text{Q}=\Delta \text{U}+\Delta \text{W }\mathrm{....}\left(\text{i}\right),$

Now, for Isothermal process 

$\therefore \Delta \text{Q}=\Delta \text{W}$

$\Delta \text{W}=\text{5}0\times 0.00\text{1}\times \text{1}00\text{ ln}(\text{1}0/\text{5}0)\text{kJ}$

$=\text{5}\times \text{ln}\left(0.\text{2}\right)=-\text{5}\times \text{1}.\text{61kJ}$

$\Rightarrow \Delta \text{Q}=-\text{8}.0\text{5kJ}$

Negative sign shows that heat rejected from the system $\therefore \Delta \text{Q}= \text{8}.0\text{5kJ}$ flows out