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Gibbs free energy and the first law of thermodynamics

  1. Jul 17, 2013 #1
    Hi thanks for reading. I have 2 questions regarding the topics mentioned that I'm having trouble with. Hope you guys can help me out here :)

    1) The first law is ΔU=Q+W so we can rearrange this to get ΔH=ΔU+PΔV so would this indicate that ΔH only measures the heat given out at constant pressure? Because ΔU=ΔH-PΔV so they would cancel out right? But I was thinking even if pressure isn't constant, won't the heat evolved be the same since they would still cancel out?

    2) Then when combining ΔH and ΔS, my teacher briefly mentioned that ΔH is doing non expansive work which makes sense if what I said in the first paragraph above was correct that enthalpy change only tells us how much heat is evolved or absorbed. But in the example where enthalpy change was negative and entropy change was negative also but the overall Gibbs free energy was still negative. So, he said some energy from the enthalpy change is used for the non spontaneous entropy change for the expansive work. What does that mean?

    I understand mathematically that ΔG=ΔH-TΔS so one factor might outweigh the other in terms of magnitude. However, I don't quite understand the theory on why the terms expansive work were used in explaining these concept and how I should explain it if it was the other way round where ΔS is positive while ΔH is positive also. I can explain mathematically that if they are subtracted to give a negative answer, then there is free energy. But I don't understand the reasons on why expansive work is done and the general theory behind them.

    Sorry for the long post. Hope you guy can help me out here :) thanks
  2. jcsd
  3. Jul 17, 2013 #2
    Part of the reason for your confusion, and least in item (1), is that you used the wrong relationship for enthalpy change. You wrote, ΔH=ΔU+PΔV, but the correct relationship is ΔH=ΔU+Δ(PV). Then, if the pressure isn't constant, the items you are referring to do not cancel out.

    I really didn't understand what you were saying in item (2), but, aside from that, here again you used the wrong relationship. The correct relationship should be ΔG=ΔH-Δ(TS).
  4. Jul 17, 2013 #3
    Hi thanks for the help :)

    For question 1: but even if its Δ(PV) won't they still cancel out? Something like this link: http://www.chemicalforums.com/index.php?topic=24911.0 so even if pressure isn't constant won't the pressure times volume component cancel out since change in internal energy encompasses the pressure times volume change already? But I also stumbled upon this link http://www.chemicalforums.com/index.php?topic=23028.0 which also explains why they don't cancel out at constant pressure. However, I don't quite get why they shouldn't cancel. Because if the pressure isn't constant then now my ΔH wouldn't be just the heat produced only. So how can this be?

    For question 2: sorry for being vague I'll try to explain my confusion here. The mathematical formula is ΔG=ΔH-Δ(TS) but since we assume temperature remains constant we can change it to ΔG=ΔH-TΔS? I don't quite understand this assumption too because in a chemical reaction won't temperature definitely change?

    So I understand mathematically that if both my enthalpy change and entropy change is negative, ΔG can either be positive or negative depending on the magnitude of ΔH, T and ΔS. however, I don't really understand the theory on why we can put these two different variables together when we combine them together.

    Our teacher gave us a reaction where both ΔH and ΔS was negative however, the overall ΔG was negative. Then he mentioned that because ΔS was negative, it would mean that component isn't spontaneous so some energy from ΔH was used to do the expansive work from ΔS. and that the remaining energy is the Free Energy remaining to do non expansive work. So I didn't get what he meant by the energy from ΔH was used do expansive work actually. Why would the heat from the ΔH be used in ΔS be for expansion? And why can the Free Energy be termed as the energy to do non expansive work? I thought that that Free Energy would go into heating up the surroundings?

    Thanks again!
    Last edited: Jul 17, 2013
  5. Jul 17, 2013 #4
    For a closed system, the first law is expressed as

    ΔU=Q - W

    where W is the work done by the system on the surroundings, such that:
    So, [tex]ΔH=Q-\int{P_{Surr}dV}+Δ(PV)[/tex]
    where final term refers to the conditions of the system only in the initial and final equilibrium states. Even if a change is carried out reversibly so that, between the initial and final states, PSurr ≈ P, the above equation reduces to:
    where we have integrated by parts.
    Note that, unless P is constant, Q is not equal to ΔH.
  6. Jul 18, 2013 #5
    Ohh I get it now.

    Regarding the second question, what does it mean by ΔG is used for non expansive work? And if both ΔH and ΔS are positive what does it mean by energy from ΔH is used for expansive work of ΔS?

    I was also doing a tutorial question where they ask about a perfume diffusing and they wanted the ΔG, ΔH and ΔS. I guess that the first was negative as it occurs spontaneously, second negative as energy is absorbed to break bonds and positive for the last one as it becomes more disorderly. However, I don't quite understand the term "free energy" according to various sources its the energy to do non expansive work. What does that mean? So in the perfume case won't there be expansion? So why would ΔG be negative?

    Thanks so much :)
    Last edited: Jul 18, 2013
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