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$$\Delta U = \Delta H - p\,\Delta V$$

Consider a gaseous system in a container with a frictionless piston, and a pressure outside the container.

  1. Is the pressure outside the container assumed to always be constant?

  2. Is the pressure inside the container assumed to always be constant? I assume no looking at the graph of Boyle's Law.

  3. For expansion of the gas, work is done by the system (i.e. the gas in the container) on the surroundings, so is $p$ the pressure of the surroundings as work is done against the pressure of the surroundings?

This doesn't seem to be the case in several examples in my lecture notes: e.g. calculate the work done against pressure of 1 bar at 298 K when 1 mol of zinc is dissolved in aq. HCl.

Here, work is being done against the pressure of the surroundings, so I'd expect $p$ = pressure of surroundings. However, the notes say 'consider $p(\Delta V) = (\Delta n)RT$' and proceeds to input the number of moles of hydrogen gas. So here $p$ is taken as pressure of the hydrogen gas produced.

  1. For compression of the gas, work is done by the surroundings on the system, so is $p$ the pressure of the gas in the container as work is done against this pressure?
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    $\begingroup$ while H = U + pV is true, which can be rearranged to give U = H - pV, it is not necessarily the case that $\Delta U = \Delta H -p \Delta V$. It is true to say that $dU = dH - d(pV)$, which you can say is $\Delta U = \Delta H - \Delta(pV)$. Take a look at relavant link to see what happens to $\Delta(pV)$ terms. I mention this because trying to explain things with an equation that is not general (which is the one you have given) can do more harm than good. $\endgroup$
    – B. Kelly
    Mar 9, 2019 at 3:08

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In the following I limit discussion to the particular expression $$ΔU = ΔH - pΔV$$ which follows (spoiler alert) from $$H\equiv U+pV$$ when the pressure $p$ of the system (the container) is identical at the beginning and end of the process.

  1. Is the pressure outside the container assumed to always be constant?

For an isobaric process such as the stated process involving zinc, by definition yes. Generally speaking the expansion work in the 1st Law is written $$dw = -p_{ext}dV$$ where $p_{ext}$ is the pressure of the surroundings against which the system does work. If $p_{ext}$ is constant work is still against or by the pressure of the surroundings and the integral can be evaluated as $$w=-p_{ext}\Delta V $$ For a reversible process the pressure of the system and surroundings are equal and we evaluate $$w=-\int_{Vi}^{Vf} pdV$$ where $p$ is the pressure of the system.

  1. Is the pressure inside the container assumed to always be constant? I assume no looking at the graph of Boyle's Law.

In the case of a reversible process against constant external pressure, yes.

BTW Boyle's law ($p \propto V^{-1}$) does not consider changes in the number of particles during a reaction.

  1. For expansion of the gas, work is done by the system (i.e. the gas in the container) on the surroundings, so is p the pressure of the surroundings as work is done against the pressure of the surroundings?

The pressure of the system (container) and surroundings are assumed to be equal (balanced) for a reversible process. However, if the pressure of the system is greater than that of the surroundings (the latter assumed constant) then the work is still $$w=-p_{ext}\Delta V$$

This doesn't seem to be the case in several examples in my lecture notes: e.g. calculate the work done against pressure of 1 bar at 298 K when 1 mol of zinc is dissolved in aq. HCl

$pV$ work involves a change in volume. Here the volume change is due to formation of gas which causes the pressure in the container to increase by an infinitesimal amount $\mathrm{d}p$. The gas pushes against the piston, increasing the volume and returning the system to mechanical equilibrium when the internal pressure equals the external pressure $p$.

  1. For compression of the gas, work is done by the surroundings on the system, so is p the pressure of the gas in the container as work is done against this pressure?

Assuming no chemical processes, compression requires the surrounding pressure to be higher than that of the system. It suffices that the difference be infinitesimal in magnitude, resulting in a reversible process. During such a process, the pressure $p$ of the system may vary during the entire process, remaining always equal (to within an infinitesimal amount) to the applied pressure.

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  • $\begingroup$ $\Delta U = \Delta H - p\,\Delta V$ does not require that the pressure be constant. Because U and H are state functions, this expression is independent of path, and thus only requires that the initial pressure be equal to the final pressure. $\endgroup$
    – theorist
    Dec 1, 2019 at 5:30
  • $\begingroup$ @theorist I was presumably assuming a reversible isothermal process but see that that was not justified. I'll update the answer. $\endgroup$
    – Buck Thorn ♦
    Dec 1, 2019 at 9:31

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