What is the work done in the formation of 1 mole gaseous methane...

The balanced chemical equation for the formation of $1$ mole of gaseous methane is:

$\text{C(s)} + 2\text{H}_2\text{(g)} \rightarrow \text{CH}_4\text{(g)}$

The change in the number of moles of gaseous substances ($\Delta n_g$) is:

$\Delta n_g = n_{g,\text{products}} - n_{g,\text{reactants}}$

$\Delta n_g = 1 - 2 = -1$

The work done ($W$) during a chemical reaction at constant temperature and pressure is given by:

$W = -\Delta n_g R T$

Given:
$R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$
T = 27^{}^{\circ}\text{C} = 27 + 273 = 300 \text{ K}

Substituting the values:

$W = -(-1) \times 8.314 \times 300$

$W = +2494.2 \text{ J}$

$W = +2.4942 \text{ kJ}$

Answer: $+2.4942$ kJ