A total of $$48 \text{ J}$$ heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by $$2°C$$. The work done by the gas is: Given, $$R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$$.

We need to find the work done by one mole of helium gas when 48 J of heat is supplied and the temperature increases by 2°C.

We know that according to the First Law of Thermodynamics $$Q = \Delta U + W$$, where $$Q$$ is heat supplied, $$\Delta U$$ is the change in internal energy, and $$W$$ is the work done by the gas. For an ideal gas, the change in internal energy depends only on temperature change, giving $$\Delta U = n C_v \Delta T$$. Since helium is a monoatomic ideal gas with three translational degrees of freedom, its molar heat capacity at constant volume is $$C_v = \frac{3}{2}R$$.

Substituting $$n = 1$$ mol, $$\Delta T = 2$$ K (because 2°C corresponds to 2 K), and $$R = 8.3 \, \text{J K}^{-1}\text{mol}^{-1}$$ into the expression for the change in internal energy yields $$\Delta U = n C_v \Delta T = 1 \times \frac{3}{2} \times 8.3 \times 2$$.

This gives $$\Delta U = 1 \times 1.5 \times 8.3 \times 2 = 1 \times 12.45 \times 2 = 24.9 \, \text{J}$$.

Applying the First Law then results in the work done by the gas as $$W = Q - \Delta U = 48 - 24.9 = 23.1 \, \text{J}$$.

The correct answer is Option (4): 23.1 J.