As given that, for ideal gas equation
$$
\begin{aligned}
&p_1=1 \mathrm{~atm}, p_2=1 \mathrm{~atm}, V_1=100 \mathrm{cc}, V_2=? \\
&T_1=273+27=300 \mathrm{~K} \\
&T_2=273+327=600 \mathrm{~K}
\end{aligned}
$$
For ideal gas the relation is, $p V \propto T$
$$
\begin{aligned}
&\frac{p V}{T}=\text { constant } \\
&\quad[V=\text { Volume of gas, } T=\text { Temperature of gas }] \\
&\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2} \\
&\Rightarrow V_2=V_1\left(\frac{T_2}{T_1}\right) \times\left(\frac{P_1}{P_2}\right) \\
&\text { But, } V_2=V_1\left(\frac{600}{300}\right) \times\left[\frac{1}{1}\right] \\
&V_2=2 V_1 \\
&V_2=2 \times 100=200 \mathrm{cc}
\end{aligned}
$$
So, unit of $\left(p_1, p_2\right)$ and $\left(V_1, V_2\right)$ must same separately but unit of $T$ must be in only on Kelvin scale.