# When is the fundamental thermodynamic relation true?

by nonequilibrium
Tags: fundamental, relation, thermodynamic
 P: 1,355 dE = TdS - PdV, or equivalently $\Delta E = \int T \mathrm d S - \int P \mathrm d V$ In general this is said to be derivable in the reversible case, however since S and V are state variables, it's also true for the irreversible case. But it can't be true for any irreversible case, since the formula only makes sense if at each moment T and P are homogeneous: only one value for each for the whole system (although it may vary in time). In most irreversible processes, T and P are definitely not homogeneous (it actually seems kind of characteristic of irreversibility that it includes inhomogeneity of the intensive parameters). So the fundamental thermodynamic relation is true for irreversible cases in case T and P stay homogeneous in their evolution in time. But how many cases are there of this? I suppose I'm looking for examples. (If one is not scared of the word quasi-static, I suppose (?) I'm looking for irreversible, quasi-static processes, although I'm not yet 100% sure of this characterisation, so comments on this are also welcome.)
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P: 6,352
 Quote by mr. vodka dE = TdS - PdV, or equivalently $\Delta E = \int T \mathrm d S - \int P \mathrm d V$ In general this is said to be derivable in the reversible case, however since S and V are state variables, it's also true for the irreversible case. But it can't be true for any irreversible case, since the formula only makes sense if at each moment T and P are homogeneous: only one value for each for the whole system (although it may vary in time). In most irreversible processes, T and P are definitely not homogeneous (it actually seems kind of characteristic of irreversibility that it includes inhomogeneity of the intensive parameters). So the fundamental thermodynamic relation is true for irreversible cases in case T and P stay homogeneous in their evolution in time. But how many cases are there of this? I suppose I'm looking for examples. (If one is not scared of the word quasi-static, I suppose (?) I'm looking for irreversible, quasi-static processes, although I'm not yet 100% sure of this characterisation, so comments on this are also welcome.)
It is true for any irreversible process between two thermodynamic states. That does not mean that it is true DURING the process. Thermodynamics deals with transitions between equilibrium states.

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 P: 1,355 I can't make much sense out of that statement. What then is the T in $\int T \mathrm d S$.
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P: 6,164

## When is the fundamental thermodynamic relation true?

 Quote by mr. vodka I can't make much sense out of that statement. What then is the T in $\int T \mathrm d S$.
Pick a quasi-static process between the initial and final states.
T is the temperature in the course of that quasi-static process.

The actual process may be something entirely different, as long as it has the same initial and final state.

Btw, an extra condition for the equation to be true is that there are no other forms of energy conversions applicable.
 P: 1,355 Okay, but then you're not using any form of "dE = TdS - PdV is true for irreversible processes", so I think you're talking about something else.
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P: 6,352
 Quote by mr. vodka Okay, but then you're not using any form of "dE = TdS - PdV is true for irreversible processes", so I think you're talking about something else.
You are correct that

$$E = \int_{(T_i,P_i,V_i)}^{(T_f,P_f,V_f)} TdS - \int_{(T_i,P_i,V_i)}^{(T_f,P_f,V_f)} PdV$$

applies whether the transition between the initial and final states is reversible or irreversible.

However, in order to calculate $\int_{(T_i,P_i,V_i)}^{(T_f,P_f,V_f)} TdS$ the you have to use a quasi-static reversible path between the initial and final states. This is because dS is defined as the reversible heat flow divided by T: dS = dQrev/T

Similarly, in order to calculate $\int_{(T_i,P_i,V_i)}^{(T_f,P_f,V_f)} PdV$you would have to use a quasi-static reversible path between the initial and final states. This is because P is the INTERNAL pressure of the system and V is the volume of the system. Internal pressure only has meaning if the system is in equilibrium.

AM
 P: 1,355 Thanks for trying to explain, but again I can't make sense out of it: Why do I need to use a quasi-static path to calculate it? If you're just saying "for practical matters", alright, but then I don't have to. In case I have to, then what does it even mean for $E= \int T \mathrm dS - \int P \mathrm d V$ to be true for an irreversible (esp. non-quasistatic) process? More simply put, how can $E= \int T \mathrm dS - \int P \mathrm d V$ be right for an irreversible process if T and P are not defined for non-quasistatic processes?
 HW Helper P: 6,164 Let's see if we can draw a comparison. Suppose you have a mass in a conservative gravitational field (like a planet around a star). If you want to know the change in gravitational energy between 2 points A and B, it does not matter how you get from A to B. The change in gravitational energy will be the same. Similarly, the thermodynamic process to get from state A to state B does not matter, be it an irreversible process, or a quasi-static one. The change in energy will be the same. However, with a quasi-static process we can use the formula you mentioned to calculate the change in energy, while in an irreversible process we would not be able to do that.
 P: 1,355 I understand (and agree with!) what you're saying, I like Serena. But Andrew Mason said in his last post that $E=\int T \mathrm d S - \int P \mathrm dV$ is also correct for irreversible processes, a claim which means that one takes the T and P of the irreversible process; however, what does that mean if T and P are not well-defined?
P: 317
 Quote by mr. vodka I understand (and agree with!) what you're saying, I like Serena. But Andrew Mason said in his last post that $E=\int T \mathrm d S - \int P \mathrm dV$ is also correct for irreversible processes, a claim which means that one takes the T and P of the irreversible process; however, what does that mean if T and P are not well-defined?
The first law of thermodynamics, in the form you have written it, is itself not meaningful during an irreversible process, because it assumes that $dE$ is the change in energy between two equilibrium states. So to even write down the integral for an irreversible process is meaningless. You can write
$$E - E_0 = \int \delta Q - \int \delta W$$
for an irreversible process, but you certainly cannot write the integral in the form you have (which is why the integral you have written down makes no sense).

The importance of the method of writing it in terms of states is to allow the above integral to be explicitly calculated in terms of different, related quantities that change in a different, related (quasi-static) situation.
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P: 6,352
 Quote by I like Serena Similarly, the thermodynamic process to get from state A to state B does not matter, be it an irreversible process, or a quasi-static one. The change in energy will be the same. However, with a quasi-static process we can use the formula you mentioned to calculate the change in energy, while in an irreversible process we would not be able to do that.
No. This is not correct. These are state parameters (T, S, P and V) and do not depend on the path. The important thing to recognize is that these state variables are DEFINED in a certain way ie. the change in entropy is DEFINED as the integral of heat flow along a reversible path between two states divided by T.

AM
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P: 6,352
 Quote by mr. vodka I understand (and agree with!) what you're saying, I like Serena. But Andrew Mason said in his last post that $E=\int T \mathrm d S - \int P \mathrm dV$ is also correct for irreversible processes, a claim which means that one takes the T and P of the irreversible process; however, what does that mean if T and P are not well-defined?
No. This does not mean that you take the T and P of the irreversible processes. It means you take the beginning and end states after any reversible or any irreversible process has occurred, and the equation will be true.

For example, a volume of gas, V is doubled but the temperature remains constant. There is a simple irreversible path between those two states: ie. an irreversible free expansion in which no heat flow occurs and no work is done. There is also a simple reversible path between those two states: an isothermal quasi-static expansion in which work is done and positive heat flow occurs. But I calculate the ∫TdS and ∫PdV the same way in each case: I calculate those integrals along any reversible path between those two states because that is how they are defined.

AM
 P: 1,355 I understand that $\Delta E$ can be calculated using $\int T \mathrm d S - \int P \mathrm d V$ across a reversible path and that it gives the same answer as for an irreversible path connecting the same initial and final states, but is that what is meant by "dE = TdS - PdV is true for irreversible processes"? If so (apparently it is), then that is an uncredibly unclear statement, since if taken literally, it should mean that the change in energy equals the temperature times the change in entropy minus etc in the irreversible process. However, apparently it only means "the left hand side of dE = TdS - PdV can be taken for an irreversible process, but the right hand side must be for a reversible path"?
 HW Helper P: 6,164 I'm just thinking of this, so I don't know if I'm right (yet). But doesn't the equation hold even for irreversible processes as well, as long as you only look at an infinitesimal mass at a time? For an infinitesimal mass T and P would be well defined (although probably hard to measure). The equation for an infinitesimal mass would be: $$\Delta e(\mathbf{r})=\int T(\mathbf{r}) ds - \int P(\mathbf{r}) dv$$ where e is the specific internal energy (internal energy per unit mass), s is the specific entropy, and v is the specific volume at some position ##\mathbf{r}##. Integrating this equation with respect to time t would give us: $$\Delta e(\mathbf{r}) = \int (T(\mathbf{r}) \dot s - P(\mathbf{r}) \dot v) dt$$ And if we integrate over the entire mass m (all positions), we would get: $$\Delta E = \int \int (T \dot s - P\dot v) dt dm$$ I think this is true. Or am I missing something?
 P: 5,462 Hello, ILS What I understand you to be saying is that you want to add up all the values for TdS and VdP on a point by point basis over the whole system? I can see how you can arrive at the TdS integral since it is the heat flow on a point by point basis. I can see how you take the pressure on a point by point basis. But what is dv, the volume change on a point by point basis?
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 Quote by Studiot But what is dv, the volume change on a point by point basis?
Hey Studiot!

On a point by point basis, the density ρ is well defined.
The specific volume v is the inverse of the density, which is also well defined.
dm / ρ = v dm is the volume of an infinitesimal mass dm.
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 Quote by mr. vodka I understand that $\Delta E$ can be calculated using $\int T \mathrm d S - \int P \mathrm d V$ across a reversible path and that it gives the same answer as for an irreversible path connecting the same initial and final states, but is that what is meant by "dE = TdS - PdV is true for irreversible processes"? If so (apparently it is), then that is an uncredibly unclear statement, since if taken literally, it should mean that the change in energy equals the temperature times the change in entropy minus etc in the irreversible process. However, apparently it only means "the left hand side of dE = TdS - PdV can be taken for an irreversible process, but the right hand side must be for a reversible path"?
The right side is defined only in terms of a reversible path between the beginning and end points. But you have to keep in mind that the path determines the end point.

If the system begins with a certain state and undergoes an irreversible process with a certain ΔV, it will have a different end point than if it undergoes a reversible process with the same ΔV. For example, an irreversible free adiabatic expansion from V to 2V (W = 0) results in ΔE = 0 but a quasi-static adiabatic expansion from V to 2V results in a decrease in temperature and a negative ΔE. To calculate TdS and PdV in the irreversible case one uses the end point resulting from the irreversible process, (ie. the value of TdS + PdV over a reversible path to the end point that results from the irreversible process) not the end point for the reversible case.

AM
 P: 5,462 Mr Vodka, please note that dE=TdS-PdV is only applicable to closed systems. To apply to open systems it must be amended to $$dE = TdS - PdV + \sum {{\mu _i}} d{n_i}$$ Where the chemical potential $\mu$i is $${\mu _i} = {\left( {\frac{{\partial E}}{{\partial {n_i}}}} \right)_{S,V,{n_{j\left( {j \ne i} \right)}}}}$$ This accounts for the energy interchanges brought about by the matter (mass) interchanges.

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