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In which of the following processes is energy transferred into the substance by work ($w > 0$)?

a) Expansion of a gas against the surroundings

b) Expansion of a gas into a vacuum

c) Vaporization of one mole of water at 50˚C in an open container

d) Melting of 100 g of ice on a laboratory benchtop

e) Combustion of methane in a constant volume container.

The correct answer is given as (d), but I needed to find a physicist to explain it to me. This problem can being thought of entirely in terms of thermodynamic work.

So, we know that

$w = P \Delta V$

From this, we see that no work is done in (b) and (e). Also, since the gas expanded against its surroundings, it did work on it's surroundings, which isn't what the question asked, thus (a) is not the correct answer.

Now, although (c) appears to be a heat transfer, and it is, there is a little more going on. The heat that was given to the water is converted into work as the water vaporizes, this causes a volume expansion (the gas rising). Intuitively, one can think of the surroundings as a tube with a movable piston. When vaporizing, the water vapor will push against the piston, increasing the pressure and thus the piston will move, causing the water to do work.

The last choice is (d), the correct answer, but there is also an intuitive way to think about this. As the ice is heated, it melts (this process is also, as we correctly deduced, is a heat transfer), however, this heat also causes the surrounding atmosphere to do work on the ice.

Again, think of a tube with a movable piston with a cube of ice that is slowly melting in the presence of normal air. As the ice melts, the volume will decrease, and thus the nitrogen and oxygen molecules in the air have more space to move about, decreasing the pressure, which causes the piston to move down, thus doing work upon the ice.

Hopefully this makes sense to everyone who has been either watching or contributing to this discussion.

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  • $\begingroup$ Do you know, is the question asking about mechanical work (any force acting over a distance), or thermodynamic work? $\endgroup$ – hBy2Py Aug 26 '15 at 18:09
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    $\begingroup$ Unfortunately, I don't know what the authors of this question had in mind. This doozy came up as I was working through an old Chemistry GRE exam. It wouldn't have occurred to me to consider this problem through mechanical work. As a side note, only 18% of the people who answered this question got it right, which essentially means that people just guessed the answer. $\endgroup$ – coloratura Aug 26 '15 at 21:46
  • $\begingroup$ Yeah, Chem GRE almost certainly means thermodynamic work. Physics GRE might(?) have implied mechanical work. $\endgroup$ – hBy2Py Aug 26 '15 at 23:18
  • $\begingroup$ GRE also often means the actual answer is terrifically obscure, and the goal of the question is eliminating answers that CAN'T be correct, until you're left with only one that you may not really understand but has to be it. This lends strong support to Curt F.'s and Karl Ratzsch's argument for (d). $\endgroup$ – hBy2Py Sep 6 '15 at 11:37
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This is a bad question. But the answer is probably supposed to be (d) because:

  • Water is more dense than ice.
  • Thus when ice melts, the total volume of "substance" goes down.
  • This means that atmospheric pressure did PV work on the melting ice.

Of course, heat transfer would also be required to melt the ice but the question isn't asking about that.

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TL;DR: If the question is asking about thermodynamic work, then none of the answers is strictly correct.

As you note, only (a) comes close to being correct: $Pv$ work is done by the gas on the surroundings as it expands. In this case, however, energy is transferred out of the substance as it expands against its surroundings, so it is not a correct answer per the question wording.

As well, you're right that in (b), since no force is required to push the gas against vacuum (there is no resisting pressure afforded by the vacuum at the gas boundary), the $Pv$ work is zero.

I also concur that (c) and (d) are examples of heat transfer, and thus are not work interactions. There's no electricity or shaft work involved, and no system boundary against which volume expansion/compression (i.e., $Pv$ work)

Agree also, the fixed volume of (e) means that no work interaction is possible.

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    $\begingroup$ I agree it's a bad question but since ice occupies a greater volume than water, when it melts, the system shrinks. That means that the atmosphere did some PV work on the system. See my answer below. $\endgroup$ – Curt F. Aug 26 '15 at 22:35
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In (c+d) you clearly need to transfer energy into the water. But that'd be a heat flux, normally abbreviated $Q$, and not work, $W$, as was pointed out in the disk below.

In d) the surroundings of the ice block must be at least a tiny bit above $0\:\mathrm{^\circ C}$, and in c) the container gets colder.

In case d) the surrounding atmosphere does a tiny bit of work on the melting ice block, because its volume shrinks slightly (water has a larger density). It sort of gets "compressed". Not sure if the teacher thought about that when he set the question.

Aside: Ice won't melt unless you transfer a substancial amount of energy into it. Melting ice at zero degrees Celsius takes up as much heat as heating the molten water to around 75 degrees Celsius afterwards. $300\:\mathrm{kJ/kg}$ heat of melting divided by $4\:\mathrm{kJ/kg\,K}$ heat capacity $\approx 75\:\mathrm{K})$

(The exact numbers are a bit different, also the heat capacity is not constant.)

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    $\begingroup$ In answer (c), though, heat is also being transferred to the system, and is causing vaporization so wouldn't that also be a case where work is being transferred into a substance? $\endgroup$ – coloratura Aug 26 '15 at 15:46
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    $\begingroup$ Disagree. The thermodynamic definition of work only includes certain specific modes of energy transfer. Heat interactions are, by definition, not work. $\endgroup$ – hBy2Py Aug 26 '15 at 15:53
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    $\begingroup$ When the water evaporates, it doesn't stay an isolated system against the atmosphere -- the $\ce{H2O}$ molecules diffuse into the air. This isn't expansion work against the atmosphere. $\endgroup$ – hBy2Py Aug 26 '15 at 15:56
  • $\begingroup$ @Brian If i heat something from the outside, by whatever means, i definitely do work. Heat transfer in a system is, you're right, not work. $\endgroup$ – Karl Aug 26 '15 at 16:08
  • $\begingroup$ Not in formal thermodynamic terms. "Thermodynamic work does not account for any energy transferred between systems as heat." en.wikipedia.org/wiki/Work_(thermodynamics) $\endgroup$ – hBy2Py Aug 26 '15 at 17:32

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