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.

  • $\begingroup$ 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. $\endgroup$
    – knoftrix
    Jun 7, 2020 at 6:07
  • 1
    $\begingroup$ 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. $\endgroup$
    – Buck Thorn
    Jun 7, 2020 at 10:33
  • $\begingroup$ @BuckThorn Added references, I cannot find good one though. $\endgroup$
    – knoftrix
    Jun 7, 2020 at 15:11
  • $\begingroup$ It's clear from the official answer that there was an error in the question. It should not have said "per mole of O2". $\endgroup$
    – Buck Thorn
    Jun 8, 2020 at 14:29
  • $\begingroup$ @BuckThorn Strangely, as far as I know, no one challenged this question. $\endgroup$
    – knoftrix
    Jun 8, 2020 at 14:47


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