1. Write the reaction and find moles of $O_2$ formed

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

From stoichiometry:

• $2$ mol $H_2O_2$ produce $1$ mol $O_2$

So, if $100$ mol $H_2O_2$ decompose, $n(O_2)=\frac{100}{2}=50\text{ mol}$

2. Work done during expansion against constant external pressure

For expansion at constant external pressure, $w = -P_{ext}\Delta V$

Magnitude of work done by the gas is: $|w| = P_{ext}\Delta V$

Since liquids have negligible volume change contribution, the expansion is due to formation of $O_2(g)$.

At $1$ bar and $300$ K, using ideal gas equation: $V = \frac{nRT}{P}$

For $50$ mol $O_2$, $\Delta V = \frac{nRT}{P} = \frac{50\times 8.3\times 300}{10^5} \text{ m}^3$

But it is easier in energy form: $P\Delta V = nRT$

Hence, $|w| = nRT = 50\times 8.3\times 300 \text{ J}$ $= 124500 \text{ J} = 124.5 \text{ kJ}$

This is the work done by 50 mol of $O_2$.

3. Work done by one mole of $O_2$

Therefore, work done by 1 mole of $O_2$ is: $\frac{124.5}{50}=2.49\text{ kJ}$

4. Compare with options

The computed value for one mole of $O_2$ is: $2.49\text{ kJ}$

This does not match any option.

However, the option $124.50\text{ kJ}$ corresponds to the work done by all 50 mol of $O_2$ formed.

The stored answer $249.00$ kJ is also inconsistent with the statement "by one mole of $O_2$" and with the calculation.

It seems the question/options may contain an error.

5. Final conclusion

• If the question truly asks work done by one mole of $O_2$, then the answer is: $2.49\text{ kJ}$
• If it intended total work done by the $O_2$ formed from 100 mol $H_2O_2$, then the answer would be: $124.50\text{ kJ}$ which is option B.