Answer
\begin{equation}
2660 \mathrm{Kg}
\end{equation}
Work Step by Step
As per the information in the question, assuming methane to behave as a perfect gas under the conditions provided we can apply the perfect gas
equation i.e,
$$
p \cdot V=n \cdot R \cdot T\quad\quad\quad\quad(1)
$$
Volume of the gas $=\mathrm{V}=4 \times 10^{3} \mathrm{m}^{3}=4 \times 10^{6}$ litre
Pressure $=\mathrm{p}=1 \mathrm{atm}$
Temperature $=\mathrm{T}=20^{\circ} \mathrm{C}=293.15 \mathrm{K}$
Substituting the values in the ideal gas equation we have,
$$
\begin{aligned} p \cdot V &=n \cdot R \cdot T \\ n &=\frac{p \cdot V}{R \cdot T} \\ &=\frac{1 \times 4 \times 10^{6}}{0.0821 \times 293.15} \text { moles } \\ &=166.2 \times 10^{3} \mathrm{moles} \end{aligned}
$$
So the numbers of moles of methane required to cover the house under the
given conditions is $166.2 \times 10^{3}$ moles
Now,
Mass of methane required $=$ Moles of Methane required $\times$ Molar Mass of Methane
$$
\begin{aligned} &=166.2 \times 10^{3} \text { moles } \times 16 \frac{\mathrm{g}}{\mathrm{mole}} \\ &=2.66 \times 10^{6} \mathrm{g}=2660 \mathrm{Kg} \end{aligned}
$$
Hence 2660 $\mathrm{Kg}$ of methane gas is required to cover the volume under the
defined conditions
