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For example, if $\ce{H2}$ is converted to two $\ce{H},$ does the volume of the mixture change?

If it doesn't, does it imply that the work done is $0$ because work is $P\,\mathrm dV?$

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    $\begingroup$ Yes, that is correct. $\endgroup$ – Chet Miller Jan 26 at 13:05
  • $\begingroup$ What's correct? The volume changes or work done = 0. $\endgroup$ – HSB Jan 26 at 14:15
  • $\begingroup$ Volume doesn’t change and no work is done. $\endgroup$ – Chet Miller Jan 26 at 14:34
  • $\begingroup$ All depends on how the gas is kept, E.g. if it is kept at constant volume or constant pressure. For the former, pressure increases, but not volume, so no work is done. for constant pressure, pressure remains, but volume increases and works is done. $\endgroup$ – Poutnik Jan 29 at 9:57
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Short answer is "not necessarily". If we isolate V in the ideal gas law, we have $$V=\frac{nRT}{P}.$$

If $n$ doubles and nothing else changes, then $V$ does also double. But the doubling of $n$ could also be offset by a doubling of $P$ or a halving of $T$, either of which would result in no change in the volume.

And yes, if the volume increases, pressure-volume work has been done, and if the volume remains constant, no pressure-volume work has been done. That isn't the same as "no work has been done", as there are other types of work besides pressure-volume work.

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  • $\begingroup$ In the question i was working on, it was mentioned that the temperature was constant, and in the solution, it was somehow implied that the volume would also remain constant. $\endgroup$ – HSB Jan 26 at 14:47
  • $\begingroup$ Could you tell me why $\endgroup$ – HSB Jan 26 at 15:12
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In the dissociation of $H_2$ into $2$ $H$, one mole of something yields $ 2$ moles of something else. The number of moles increases. As $V$ is proportional to $n$, at constant pressure, the volume $V$ of the gas must increase.

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