# Differential form of the first law of thermodynamics

1. Aug 28, 2011

### omberlo

Hi everyone, I'm new here.

I'm an italian student who has an exam in applied thermodynamics soon. Through the whole course as well as the Physics one I have faced a lot of equations expressed with differentials, basically all of them. I have never been taught how to use them though.

For example let's take the first law of thermodynamics
dU(S,V,N_i) = TdS - pdV + sum_{i=1}^{n} (mu_i * dN_i)

let's leave out the chemical potential part to make the equation easier:

dU(S,V) = TdS - pdV

Now my doubt is, if I wanted to calculate the actual change in the internal energy (delta U) using the above expression how would I proceed?

is the above expression supposed to be used to calculate the change in internal energy ( Delta U) of a thermodynamic process ONLY when T and P are constant throughout the whole process, as

Delta U = T * (Delta S) -p * (Delta V)

or is it supposed to be used to calculate the Delta U of a thermodynamic process even if T and P aren't constant but they must be/(are always?) a function of S and V respectively, by integrating it like this:

U_b - U_a = int from a to b(dU) = int from a to b(TdS) - int from a to b(pdV)

or how else?

Any help would be appreciated.
Thank you.

2. Aug 28, 2011

### Studiot

Good evening Omberlo, welcome to Physics Forums.

What subject are you studying thermodynamics in?

It is difficult to know where to start since your exams are imminent. However you really do need to get some basic stuff sorted.

Your equation containing entropy is not a statement of the first law. The first law is simply a statement of the law of conservation of energy. Entropy does not appear in this law - that is left to the second law.

Now some thermodynamic variables, notably P, V, T & U are called state variables because their values depend only on the system.

It is also true that state variables produce exact differentials when differentiated.

The first law states

$\Delta$U = q + w; where q is heat transferred and w is work done.

or in differential form

dU = dq + dw

however q and w are not state variables, dq and dw cannot be integrated to yield q or w . The heat exchanged or work done is a function of the environment as well as the system.

There are further equations connecting entropy with these variables.

We can only calculate along certain paths so sometimes we calculate along a path at constant volume or pressure and then a second leg at constant temperature and so on.

3. Aug 29, 2011

### omberlo

Don't worry too much about my exams being imminent, I'll probably just wait until the next exams' session to prepare for it. I decided to stop studying for this course until I completely understand equations expressed in differential form, because they are very misleading to me.

Sorry if I was inaccurate and that is not the statement for the first law (wiki calls it the state functional formulation of the first law), however I am more concerned with the meaning of that specific equation and how to use it, rather than the first law itself. I'm hoping that understanding how to use that equation will help me understand equations expressed in differential forms and their use, since the course is full of them.

Back to dU = TdS - PdV :

After reading your last sentence, I guess that equation only means that Delta U can be calculated as Delta U = T * (change in S) when T is constant and the change in volume (dV) is 0 along a path, or as Delta U = P * (change in V) when pressure is constant and the change in entropy is 0 along another path.
If a thermodynamic process is made of a leg where T is constant and a second leg where P is constant, delta U can be calculated as Delta U = T * (change in S through the first leg) + P * (change in V through the second leg), and so on if it has more than 2 legs.

Is that correct? Is that all that equation means?

4. Aug 29, 2011

### Studiot

Perhaps it would help to understand where your equation comes from.

In thermodynamics the law of conservation of energy ( the first law ) is always true.

Now let us move on to the second law, which states

dS = dq/T or dq = TdS

Note this only applies to reversible change.

Substituting this into the first law yields

dU = TdS + w

That is the positive change in internal energy (ie increase) is the sum of the heat supplied to the system and the work done on the system.

Note that usually if we supply heat to something it expands so does work on its surroundings in expansion. Work done on the system is positive and work done by the system is negative.

I asked what you are studying because there are two conventions about signs here and I am quoting the chemists convention.

Usually we are working at atmospheric pressure so the pressure is constant and the volume expands, making the negative work of expansion -PdV

so the equation becomes - at constant pressure

dU = TdS - PdV

If we were considering a steam boiler the volume would remain constant and the pressure would increase so work done = -VdP

dU = TdS - VdP

I can't do partial derivatives from this computer I'm at right now, but are we making progress?

5. Aug 30, 2011

### omberlo

Thank you Studiot, we are definitely making progress, now I understand where that equation comes from, but my doubt about HOW to use equations expressed in differential form still remains.

Even if I look at the equation from the second law, which is a bit simpler than the one I asked about, dq = TdS, I don't know if it can be used to calculate the change in heat ($\Delta Q$ or $Q_2 - Q_1$) ONLY when Temperature is constant through a process, as $\Delta Q = (constant T) * \Delta S$, or if it can be used even when Temperature isn't constant, by expressing T as a mathematical function of S and integrating it.

What I think is misleading me is the fact that there are differentials in those equations and I have never seen them except in the math course while learning derivatives and integrals, so whenever I see them in an equation I'm thinking that "there is a function to integrate somewhere", where in fact there is no function to integrate in these cases, and dq = T dS is just a fancy way to write $\Delta Q = (constant T) * \Delta S$..... but why even use differentials if this is the case??

I hope I've been clear on what my doubt is.

Last edited: Aug 30, 2011
6. Aug 30, 2011

### Studiot

I fully understand your question - it is quite a reasonable one.

Rest assured it will be answered as we develop the subject. I'm a bit stuck this week as I can't access my pc with Mathtype on after 7pm my time for a few days and you come online rather late, despite being ahead in time.

I should be able to expand on the exact or total differential tomorrow morning. This will take in the answer to what other equation you use to calculate the components of the first law. That is when how and why to use const T, P or V.

7. Aug 31, 2011

### omberlo

I appreciate all your help Studiot :)

8. Aug 31, 2011

### Studiot

First some stuff about exact or total differentials.

If Z is a function of two independent variables, X and Y

eg $$Z = X + Y$$ or $$Z = X{Y^m}$$

Then the exact or total differential dZ is given by

$$dZ = {\left( {\frac{{\partial Z}}{{\partial X}}} \right)_Y}dX + {\left( {\frac{{\partial Z}}{{\partial Y}}} \right)_X}dY$$

Now three variables of state are P, V and T. Only two are independent however since they are linked by an equation of state such as

$$PV = nRT$$

So we can write the internal energy as a function of any two, say V and T

$$dU = {\left( {\frac{{\partial U}}{{\partial V}}} \right)_T}dV + {\left( {\frac{{\partial U}}{{\partial T}}} \right)_V}dT$$

Or P and T

$$dU = {\left( {\frac{{\partial U}}{{\partial P}}} \right)_T}dP + {\left( {\frac{{\partial U}}{{\partial T}}} \right)_P}dT$$

And calculate the third from the equation of state.

Now we have these we look for circumstances that can reduce one of the terms to zero.

Thus we consider

Isothermal processes where dT = 0 and

$$dU = {\left( {\frac{{\partial U}}{{\partial V}}} \right)_T}dV$$

Adiabatic processes where dq = 0 thus dU = dw from the first law.

Constant volume processes where dw = 0

One final piece of the jigsaw.

As a result of the Joule and Joule Thomson experiments it was confirmed that

The internal energy depends only on the temperature, for an ideal gas.

That is

$$dT = {\left( {\frac{{\partial U}}{{\partial V}}} \right)_T} = 0$$

OK so we can use all this along with suitable gas laws to calculate stuff for ideal gasses.

Isothermal change

$$\begin{array}{l} dU = dq - PdV = {\left( {\frac{{\partial U}}{{\partial T}}} \right)_V} + {\left( {\frac{{\partial U}}{{\partial V}}} \right)_T} = 0 \\ dq = - dw = PdV \\ But\quad P = \frac{{nRT}}{V} \\ \int {dq = \int { - dw = \int {nRT\frac{{dV}}{V}} } } \\ q = - w = nRT\ln \frac{{{V_2}}}{{{V_1}}} = nRT\ln \frac{{{P_1}}}{{{P_2}}} \\ \end{array}$$

Are we still together?

Last edited: Aug 31, 2011
9. Sep 1, 2011

### omberlo

Let's see

This part is clear.

This statement is not so clear, why exactly do we want to reduce one of the terms to zero? What is the problem with having both independent variable changing?

This part is clear except for the above doubts

This part is not really intuitive to me, so I need a bit more time to think it out

10. Nov 14, 2013

### Urmi Roy

I would just like to know 2 things:
1, My book says this relation applies between any 2 equilibrium states, but it also says that it applies in an irreversible process.
In the irreversible case, Tds is greater than the corresponding reversible process between the same 2 points, and the PdV is greater than its reversible analog by the same amount, so the net sum is preserved.
I can't see how this is valid.

2. What does it mean to say T and S are canonical conjugates (similarly P and v are canonical conjugates)?

Thanks :-)

11. Nov 14, 2013

### Andrew Mason

The relationship applies between any two equilibrium states whether the process used in moving between the two states was reversible or irreversible.

Each of the terms are state variables so this is a fundamental thermodynamic relationship.

AM

12. Nov 14, 2013

### Staff: Mentor

There is an equation for the change in U along an arbitrary path involving temperature and pressure variations. It applies to a pure component without change of phase:
$$dU=C_vdT-(P\frac{\partial V}{\partial P}+T\frac{\partial V}{\partial T})dP$$

The derivation of this equation is presented in Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics, pp. 170-171.

13. Nov 15, 2013

### Urmi Roy

Thanks a lot, Andrew Mason and Chestermiller. Just one last question: why is this called the 'canonical' relation? I mean it doesn't look all that special...its just the first law, right?

14. Nov 18, 2013

### Andrew Mason

Not quite. The first law is: Q = ΔU + W where W is the work done by the system. TdS = dU + PdV relates the state variables of the system after a change in state. It is equivalent to the first law only if TdS = δQ and PdV = δW (ie where the process involved in the change of state was a reversible process).

Sometimes academics want to make things mysterious so they invent fancy words. I am not sure how the term "canonical" entered the physics vocabulary, but it is not a particularly illuminating term. It is mainly used in statistical mechanics, as in canonical and microcanonical ensembles, which is just a fancy term for generalized macro and micro systems in which the states are determined by the statistical behaviour of its components (such as the randomly moving molecules of a gas).

So don't worry about terminology. Just think of Q = ΔU + W as the first law of thermodynamics and TdS = dU + PdV as a fundamental relationship between thermodynamic states. Both relate to thermodynamic changes. (The first is about about energy flow. The latter is about the changes in thermodynamic states).

AM

15. Nov 19, 2013

### DrDu

Your second answer is correct. To calculate the change in U between two equilibrium states you calculate the integral
$U_b-U_a=\int [T(\lambda) \frac{dS}{d\lambda}-p(\lambda) \frac{dV}{d\lambda}]d\lambda$
along any path $\lambda$ through equilibrium states joining a and b.
it is irrelevant whether the actual thermodynamic process leading from a to b is reversible or not.

16. Nov 26, 2013

### Urmi Roy

Andrew Mason and Dr Du,
dU=TdS-PdV gives the change in internal energy even if it is an irreversible process (in which case TdS is larger by the same amount from the reversible case as the PdV)....but this equation applies only when we know the path through equilibrium states...but is it not true that in irreversible processes, we don't really have equilibrium states? Seems paradoxical to me :-/

Also, from a discussion I had with my Professor, it seems that the dU=TdS-PdV is significant because it is an 'exact' differential and one can define other quantities such as Gibb's free energy in terms of its conjugate variables from it...so in thermodynamics, are 'exact' differentials rare or something?

17. Nov 27, 2013

### DrDu

You can't even say so. In irreversible states neither T nor P must be homogeneous over the system, if they are defined at all.

18. Nov 27, 2013

### Jano L.

The equation may apply to irreversible process, if in the process system proceeds through equilibrium states. The quantities $T,S,P$ need to have meaning - as soon as they do, they are always related through the above relation.

It depends on what process we consider. Explosion is irreversible process in which gas goes through non-equilibrium states and the above equation is not valid. Slow expansion of gas in cylinder against piston which experiences friction is also irreversible due to friction, but since the process is slow, the gas goes through equilibrium states and the above equation is valid. In such case, TdS is greater than accepted heat and the performed volume work on the atmosphere is lower than pdV.

19. Nov 27, 2013

### Andrew Mason

Thermodynamics is the study of changes in systems and surroundings between different equilibrium states. The laws of thermodynamics apply whether the process involved in changing from one equilibrium state to another was reversible or irreversible. Thermodynamics does not equip us to analyse the process itself unless the system and surroundings are in thermodynamic equilibrium during the process. But once the system and surroundings reach equilibrium, the relationship between the initial and final equilibrium states obey the laws of thermodynamics.

The fundamental relation dU = TdS - PdV implies that $\Delta U = \int_A^B TdS - \int_A^B PdV$ where A and B are equilibrium states and the integration is over a reversible path between A and B. This is the case whether or not the actual process that occurred in passing from state A to state B was reversible.

AM

20. Dec 14, 2013

### Urmi Roy

What you mentioned in the quotes is like gold for me, since we recently did a problem on the equivalent of the canonical relation for a paramagnetic substance. It helps to know that du=Tds-Pdv is just to relate the states of a system and is not identical to the first law.

What I'm finding hard to picture now is as to whether this (du=Tds-Pdv ) is valid despite phase changes that occur? Also, why is it equally valid for different models of substance, like ideal gas, and van der waals?

Last edited: Dec 14, 2013
21. Dec 15, 2013

### Andrew Mason

It is valid for different gases and phase changes because all the variables are state variables. You can measure T, P and V. So you just have to know the way U and S behave for the gas as functions of T, P and V. These will be different for different kinds of substances.

AM

22. Dec 15, 2013

### Staff: Mentor

This equation is valid even for state changes, as long as you have a closed system. Also, why do you think it would not be valid when dealing with more complicated P-V-T behavior of the material? This generality of the equation is part of its beauty and utility.

23. Dec 15, 2013

### Urmi Roy

Thanks a lot Andrew Mason and Chestermiller.
Chestermiller-Yeah...I'm just not comfortable with it yet...still struggling to understand it!