If $100\ \mathrm{mol}$ of $\ce{H2O2}$ decomposes at $1\ \mathrm{bar}$ and $300\ \mathrm K$, the work done ($\mathrm{kJ}$) by one mole of $\ce{O2(g)}$ as it expands against $1\ \mathrm{bar}$ pressure is:
$$\ce{2H2O2(l) <=>2H2O(l) +O2(g)}$$
My attempt to solve the problem:
Since $2\ \mathrm{mol}$ of $\ce{H2O2}$ decomposes to give $1\ \mathrm{mol}$ of $\ce{O2}$, so $100\ \mathrm{mol}$ of $\ce{H2O2}$ will give $50\ \mathrm{mol}$ of $\ce{O2}$.
Now, The work done by $50\ \mathrm{mol}$ of $\ce{O2}$ as it expands is,
$$w_{\ce{O2}}=-w_\text{surr}=\int{p_\text{ext}\,\mathrm dV}=p_\text{ext}\,\Delta V=(\Delta n_\text{gas})RT=50\times8.3\times300\times10^{-3}\ \mathrm{kJ}=124.5\ \mathrm{kJ}$$
$124.5\ \mathrm{kJ}$ is in fact the answer given but we are asked to find work done by one mole of $\ce{O2}$.
Shouldn't the answer be $w={124.5\over50}\ \mathrm{kJ\ mol^{-1}}=2.49\ \mathrm{kJ\ mol^{-1}}$?
If $50\ \mathrm{moles}$ of gas expanded to a specific volume, say $V$, then $1\ \mathrm{mole}$ of gas should expand to ${V\over50}$. Then according to me work done by $1\ \mathrm{mole}$ of gas should be ${1\over50}$ of work done by $50\ \mathrm{moles}$ of gas. Does work add up like extensive properties of system?
Reference: This question appeared in JEE Main entrance examination held in India in 2016, the official answers and question papers are not publicly available, all I can find is this pdf https://www.resonance.ac.in/answer-key-solutions/JEE-Main/2016/papers/jee-main-2016-online-CBT-paper-Dt-10-04-2016.pdf#page=19 (Question no. 20 page 19) and the answers (also pdf) https://www.resonance.ac.in/answer-key-solutions/JEE-Main/2016/answer-key/jee-main-2016-online-CBT-solution-Chemistry-10-04-2016.pdf#page=7 (Question 20 Page 7)
$2.49\ \mathrm{kJ\ mol^{-1}}$ as calculated by me doesn't seem to be in the options at all.
