# Work done by one mole of gas during isobaric and isothermal expansion [closed]

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.

• I don't know why this question is getting closing votes. I took this question as an example to understand the concept of work in thermodynamics and to better formulate what I am asking. Jun 7, 2020 at 6:07
• Can you provide a reference (source) for the posted question? It does look like an error in the official answer. I can't see an error in your calculation. Jun 7, 2020 at 10:33
• @BuckThorn Added references, I cannot find good one though. Jun 7, 2020 at 15:11
• It's clear from the official answer that there was an error in the question. It should not have said "per mole of O2". Jun 8, 2020 at 14:29
• @BuckThorn Strangely, as far as I know, no one challenged this question. Jun 8, 2020 at 14:47