A home owner uses $$4.00 \times 10^3$$ m$$^3$$ of methane CH$$_4$$ gas, (assume CH$$_4$$ is an ideal gas) in a year to heat his home. Under the pressure of 1.0 atm and 300 K, mass of gas used is $$x \times 10^5$$ g. The value of x is ___.
(Nearest integer)
(Given R = 0.083 L atm K$$^{-1}$$ mol$$^{-1}$$)

The volume of methane consumed in one year is given as $$4.00 \times 10^3\ \text{m}^3$$. We recall that $$1\ \text{m}^3 = 1000\ \text{L}$$, so

$$V = 4.00 \times 10^3\ \text{m}^3 \times 1000\ \frac{\text{L}}{\text{m}^3} = 4.00 \times 10^6\ \text{L}.$$

For an ideal gas the relation $$PV = nRT$$ holds, where $$P$$ is the pressure, $$V$$ is the volume, $$T$$ is the absolute temperature, $$R$$ is the universal gas constant and $$n$$ is the number of moles.

We are told that $$P = 1.0\ \text{atm}$$ and $$T = 300\ \text{K}$$, with $$R = 0.083\ \text{L atm K}^{-1}\text{mol}^{-1}$$. Substituting these values we get

$$n = \frac{PV}{RT} = \frac{(1.0\ \text{atm})(4.00 \times 10^6\ \text{L})}{(0.083\ \text{L atm K}^{-1}\text{mol}^{-1})(300\ \text{K})}.$$

First we multiply the denominator: $$0.083 \times 300 = 24.9.$$

Now we perform the division:

$$n = \frac{4.00 \times 10^6}{24.9}\ \text{mol} \approx 1.606 \times 10^5\ \text{mol}.$$

The molar mass of methane CH$$_4$$ is obtained by adding atomic masses, $$12\ (\text{for C}) + 4 \times 1\ (\text{for H}) = 16\ \text{g mol}^{-1}.$$

Hence the mass of gas used is

$$m = n \times M = (1.606 \times 10^5\ \text{mol})(16\ \text{g mol}^{-1}) = 2.5696 \times 10^6\ \text{g}.$$

We rewrite this mass in the required form $$x \times 10^5\ \text{g}$$:

$$2.5696 \times 10^6\ \text{g} = 25.696 \times 10^5\ \text{g}.$$

Taking the nearest integer, $$x \approx 26.$$

Hence, the correct answer is Option 26.