Answer
\begin{equation}
\begin{array}{l}{\text { For } 200 \text { t per day volume of gas is } 2.8.10^{5} \mathrm{m}^{3}} \\ {\text { For } 300 \text { t per day volume of gas is } 4.10^{5} \mathrm{m}^{3}}\end{array}
\end{equation}
Work Step by Step
In this case we have $1 \mathrm{t}=10^{3} \mathrm{kg} .$ For 200 t per day.
$$
\begin{aligned} n\left(\mathrm{SO}_{2}\right) &=\frac{200 \cdot 10^{3} \mathrm{kg}}{64 \cdot 10^{-3} \mathrm{kgmol}^{-1}} \\ &=3.1 \cdot 10^{6} \mathrm{mol} \end{aligned}
$$
$\begin{aligned} V &=\frac{n R T}{p} \\ &=\frac{\left(3.1 \cdot 10^{6} \mathrm{mol}\right) \cdot\left(0.08206 \mathrm{dm}^{3} \mathrm{atm} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right) \cdot 1073 \mathrm{K}}{1.0 \mathrm{atm}} \\ &=2.8 \cdot 10^{8} \mathrm{dm}^{3} \\ &=2.8 \cdot 10^{5} \mathrm{m}^{3} \end{aligned}$
For 300 t per day:
$$
\begin{aligned} n\left(\mathrm{SO}_{2}\right) &=\frac{300 \cdot 10^{3} \mathrm{kg}}{64 \cdot 10^{-3} \mathrm{kgmol}^{-1}} \\ &=4.7 \cdot 10^{6} \mathrm{mol} \end{aligned}
$$
$\begin{aligned} V &=\frac{n R T}{p} \\ &=\frac{\left(4.7 \cdot 10^{6} \mathrm{mol}\right) \cdot\left(0.08206 \mathrm{dm}^{3} \mathrm{atm} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right) \cdot 1073 \mathrm{K}}{1.0 \mathrm{atm}} \\ &=4 \cdot 10^{8} \mathrm{dm}^{3} \\ &=4 \cdot 10^{5} \mathrm{m}^{3} \end{aligned}$
