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Thermodynamics - Q, U, W and H

  1. Mar 9, 2013 #1
    I have understood that Q=heat absorbed, ΔU=energy change, ΔH=enthalpy change, W=work done by system = Δ(PV)+Wmech (where Wmech is mechanical work, P is pressure and V is volume).

    What I don't understand is what the actual, exact relationships are between these values (which hold constant regardless of the conditions, whether we are changing pressure and volume at the same time, or whether or not there is Wmech; these complications aren't shown in any textbooks I have). First I'd like to look at these actual relationships (i.e. how to calculate each value from the others, without any approximations being made about the conditions, e.g. "pressure is constant") and then I'd like to consider what we can say under these specific conditions (e.g. "volume is constant" or "pressure is constant").

    For example, I've heard that ΔH=ΔU+PΔV. I take it this is ΔH=ΔU+P(V2-V1), which would suggest PΔV is the work done by the system in increasing volume. So this make W=Δ(PV)+Wmech the work done by the system, but how does that relate to the other values? In my expression for ΔH, what if pressure is changing? What if there is mechanical work? And what are the other similar relationships? Where does Q come into it? All of these questions!
     
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  3. Mar 9, 2013 #2

    DrDu

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    ## \Delta U=Q+W ## holds generally and also ##H=U+pV## is a general definition.
    Hence ##\Delta H= \Delta U +\Delta (pV)## generally. The last term only simplifies to ##p\Delta V## if p is constant.
     
  4. Mar 9, 2013 #3

    Philip Wood

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    H = U + pV, F = U - TS, G = U + pV - TS. These are the definitions of the commonly used 'thermodynamic potentials' for a closed system. Taking differentials and simplifying using
    dU = TdS - pdV gives
    dH = T dS + V dp, dF = -p dV - S dT, dG = V dp - S dT

    The inter-relations between U, H, F, G that you can work out by ordinary algebra (e.g. eliminating pV between H and G to obtain G = H - TS), or similarly for the differentials, are all you're going to get, without imposing special conditions, such as constant pressure (dp = 0) or constant entropy (dS = 0). It's the simple form that some of these relations assume when we do impose these conditions (e.g. dH = TdS when dp = 0) that give these potentials a lot of their importance.

    Incidentally, Q and W do not have the same status as U, H, F, G, because U, H, F, G are functions of state (for example, for a given amount of a given fluid we can express them as functions of p and V) whereas Q and W are not functions of state because they represent energy in transit to or from the system, rather than properties of the system.
     
  5. Mar 9, 2013 #4
    Thank you. Firstly, under what conditions can we rewrite Δ(pV) as Δngas*R*T?

    2) Also, what is ##W## generally? Could it be W=Δ(pV)+Wmech, i.e. the work done by the system? Or is it W=-Δ(pV)-Wmech, the work done on the system?

    3) I know ##Q## is generally the heat released. But how would you go about calculating it in general? Does it bear a relationship to ΔH? Does W bear a relationship to ΔH?

    4) I've also heard that when the pressure is constant (assuming no mechanical work), ΔH=Q, whereas when the volume is constant, ΔU=Q. When both are constant ΔH=Q=ΔU. Does this hold true even if there is mechanical work? And why are these relations the case? Perhaps if you answer my above question about Q it will start to make sense.

    Thank you. You have pre-empted my plan to move on to entropy and Gibbs' energy later. Please refer to my above reply if you can help on the mathematical definitions of Q and W.
     
  6. Mar 9, 2013 #5

    DrDu

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    1. For an ideal gas and approximately for any gas at sufficiently high temperature and low pressure.
    2. The sign of W depends on the convention chosen (whether work done on the system is positive or work done by the system), physicists and engineers often use different conventions (I count both work and heat done on the system as positive).
    3. ##Q=\Delta U-W##. In reversible processes ##\Delta H=Q## at constant p and ##\Delta U=Q## at constant V.
     
  7. Mar 9, 2013 #6
    You are confusing the form of the first law for closed systems with the form of the first law for open steady state flow systems. Go back to your textbook and study more carefully how the latter is derived from the former.
     
  8. Mar 9, 2013 #7
    1. So for any gaseous system where the gases are ideal, i.e. Δngas*R*T=Δ(pV) if the gaseous mixture is considered ideal?

    2. So you are taking W=-Δ(pV)-Wmech+Wd (let Wd be the work done on the system, Wmech the mechanical work done by the system, Δ(pV) the gas expansion-type work done by the system, so all work done by the system is negative and all done on the system positive)?

    3. We could surely rewrite this as Q=ΔU-(-Δ(pV)-Wmech+Wd), and then, if the system only has gas expansion type work, Q=ΔU+Δ(pV), which =ΔU+Δngas*R*T if the gaseous mixture is ideal?

    If ΔH=ΔU+Δ(pV) then shouldn't this conclusively mean Q=ΔH for all cases? I think, for this to be wrong, there must be some imprecision in my idea of W. What would you say is an adequate expression to sum up W (including Δ(pV), mechanical energy losses, work done on the system to give it heat, etc.)?

    Thanks for the help.
     
  9. Mar 10, 2013 #8

    Rap

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    Let's stop using Δ because it complicates things. Lets use the infinitesimal d instead. In other words, Δ is very, very small. For a simple system, the internal energy is U=TS-PV+µN for an open system, where µ is the chemical potential. Clearly dU=(SdT+TdS)-(VdP+PdV)+(Ndµ+µdN). But the fundamental law says dU=TdS-PdV+µdN. Since both are true, its clear that 0=SdT-VdP+Ndµ. Thats known as the Gibbs-Duhem relationship for a simple system.

    For a materially CLOSED system, dN is always zero.

    Work and heat are special, a small amount of work cannot be represented by dW for example. Lets represent it as [itex]\delta[/itex]W to remind us of that.

    During a transition, yes, T, P, and µ change to T+dT, P+dP and µ+dµ, but they change in such a way that dU=TdS-PdV+µdN remains true. So what T, P, and µ do you use here? Since dT, dP, and dµ are very small, you could use T+dT instead of T, for example, but then you would have dU=(T+dT)dS-PdV+µdN=dU=TdS-PdV+µdN +dTdS. If dT and dS are very very small, dTdS is very very small compared to the other terms, and you can ignore it.

    The enthalpy is defined as H=U+PV. Clearly dH=dU+(VdP+PdV). But substituting TdS-PdV+µdN for dU gives dH=TdS+VdP+µdN. You can do this sort of thing for all thermodynamic potentials.

    So you are asking is d(PV)=RT dN for an ideal gas. Not generally. d(PV)=d(NRT) and that's equal to NRdT+TRdN. If the system is closed, d(PV)=NRdT.

    No, the work done by the system is not d(PV), it is -PdV. The work done on the system is +PdV. They are equal in magnitude but opposite. Work means energy is transferred from one system to another - the work done on a system is the work done by the other system, and vice versa. If you want to figure out the work done when the change in volume is not small (i.e. from V1 to V2) then you have to integrate. The work done is the integral of PdV from V1 to V2. That's why work (and heat) are not system variables. In order to carry out that integration, you have to state clearly how P varies with V as you go from V1 to V2, and that is sometimes not easy. That's why its easier to deal with d's instead of Δ's.

    Since the previous statement was not correct, this is not either.

    [itex]\delta[/itex]Q=TdS and dH=dU+PdV+VdP. Since dU=[itex]\delta[/itex]Q-PdV+µdN, it follows that [itex]\delta[/itex]Q=dH-VdP-µdN. For a closed system, [itex]\delta[/itex]Q=dH-VdP, so [itex]\delta[/itex]Q=dH only when the pressure is constant.

    Thats why the enthalpy is so useful for situations where the system is closed and is being held at constant pressure. The Helmholtz free energy F=U-TS is likewise very useful where the system is closed and at constant temperature. The Gibbs free energy G=U+PV-TS is likewise very useful when the system is closed and held at constant temperature and pressure.
     
  10. Mar 10, 2013 #9

    DrDu

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    Think of stirring a gas or a liquid with a propeller. Even when the system is thermally insulated and ##\Delta V=0##, there will be a change U. Hence ## \Delta U## cannot equal heat when there is work other than volume work (or work related to a change of some mechanical variable characterizing equilibrium states in general).
     
  11. Mar 10, 2013 #10

    DrDu

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    The problem is that this is mainly useful for state variables. E.g. you can have a process where ##Q=-W## become arbitrary large although neither U, P, V or T change.
     
  12. Mar 10, 2013 #11
    Can I clarify before I try and absorb your posts in greater detail:

    Does a closed system simply mean no unspecified new substances can enter the system, and there are no volume or temperature changes?
     
  13. Mar 10, 2013 #12
    No. A closed system is one in which no mass can enter or leave. An isolated system is one in which no mass can enter or leave, there is no heat entering or leaving, and there is no volume change.
     
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