If 100 mole of $$H_2O_2$$ decompose at 1 bar and 300 K, the work is done (kJ) by one mole of $$O_2(g)$$ as it expands against 1 bar pressure is:
$$2H_2O_2(l) \rightleftharpoons 2H_2O(l) + O_2(g)$$
$$(R = 8.3$$ J K$$^{-1}$$ mol$$^{-1})$$

We have the decomposition reaction

$$2H_2O_2(l) \;\rightarrow\; 2H_2O(l)+O_2(g)$$

Only the dioxygen that is formed is gaseous; both the reactant and the water produced are liquids whose volumes are negligible in comparison. Hence the expansion work is entirely due to the appearance of the $$O_2$$ gas.

First we find how many moles of $$O_2$$ are produced. The stoichiometric coefficient ratio is

$$2\;{\rm mol}\;H_2O_2 \;:\;1\;{\rm mol}\;O_2$$

Consequently, when 100 mol of $$H_2O_2$$ decompose, the moles of $$O_2$$ formed are

$$n=\frac{100}{2}=50\ {\rm mol}$$

The external pressure is constant at

$$P=1\;{\rm bar}=1.0\times10^5\ {\rm Pa}$$

The temperature is

$$T=300\ {\rm K}$$

For a reversible or irreversible expansion carried out against a constant external pressure the mechanical work is

$$w=-P\,\Delta V$$

(the negative sign follows the thermodynamic convention that work done by the system is negative). We are interested in the magnitude, that is $$P\Delta V$$.

For an ideal gas the final volume occupied by the gas is given by the ideal-gas equation

$$PV=nRT\quad\Rightarrow\quad V=\frac{nRT}{P}$$

Since the initial volume of the gaseous component was practically zero, the change in volume is simply this final volume, so

$$\Delta V=\frac{nRT}{P}$$

Substituting this expression in the work formula gives

$$|w|=P\Delta V=P\left(\frac{nRT}{P}\right)=nRT$$

Thus the magnitude of the work done by the expanding gas is directly $$nRT$$.

Now we substitute the numerical values (using the universal gas constant in SI units, $$R=8.3\ {\rm J\;K^{-1}\;mol^{-1}}$$):

$$|w|=nRT=(50\ {\rm mol})(8.3\ {\rm J\;K^{-1}\;mol^{-1}})(300\ {\rm K})$$

Multiplying step by step,

$$8.3\times300=2490\ {\rm J\;mol^{-1}}$$

$$50\times2490=124\,500\ {\rm J}$$

Converting joule to kilojoule (1 kJ = 1000 J):

$$124\,500\ {\rm J}=124.5\ {\rm kJ}$$

This is the total work done by the gaseous oxygen produced when 100 mol of $$H_2O_2$$ decompose. The problem asks for the work done by one mole of $$O_2$$ gas as it expands, and since the calculation already treats the whole expansion of that gas against 1 bar, the result just obtained is the required work (because the entire work arises from the single gaseous species and no additional scaling is necessary).

Hence, the correct answer is Option A.