Solve this:

Dear Student,

Initial volume of the gas is not given in the question so we assume it to be 60 cm³ .

20 % of 60 = $\frac{60}{100}\times 20 =12 c{m}^3$
$$\begin{gathered} {\text{Thus final volume of the gas = 12 + 60 = 72 cm}}^3 \\ \mathrm{Now}, \mathrm{Initial} \mathrm{Temperature} = 273 + 27 = 300 \mathrm{K} \\ \mathrm{Applying} \mathrm{Charle}\text{'}\mathrm{s} \mathrm{Law} , \\ \frac{{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{V}}_2}{{\mathrm{T}}_2} \\ {\mathrm{T}}_2 =\frac{{\mathrm{V}}_2{\mathrm{T}}_1}{{\mathrm{V}}_1} =\frac{ 72 \times 300}{60}=360 \mathrm{K} \\ \mathrm{Regards} \end{gathered}$$