# In expressing Internal Energy, how do I get to the next step

1. Mar 26, 2016

### grandpa2390

1. The problem statement, all variables and given/known data
I didn't have enough space to make the topic any more precise. Here is the full question:

Express the Internal Energy with Internal Pressure.

2. Relevant equations

This is what I know:
the internal pressure is (dU/Dv)T
the fundamental equation for internal energy is dU = T*dS - P*dV

The third Maxwell relation: (dP/dT)v = (dS/dV)t

Cv = (dU/dT)v

3. The attempt at a solution

dividing by dV and imposing the constant T, then using the third Maxwell relation, we arive at
(dU/dV)t = T*(dP/dT)v - P

then I am supposed to consider the internal energy as a function of T and V.
dU(T,V) = (dU/dT)v*dT + (dU/dV)t*dV

this is where I come to my issue. Through google I found that (dU/dV)t *dV is -P

I can't find anything that gives me the relationship between T*(dP/dT)v and (dU/dT)v*dT
I know that (dU/dT)v is the Constant volume heat capacity. but I don't understand the transition... if that makes sense. I don't understand how those two terms are equivalent.

2. Mar 26, 2016

### Staff: Mentor

The expression from google is incorrect, and your expression is correct. What you have found is that :
$$dU=C_vdT+\left[T\left(\frac{\partial P}{\partial T}\right)_V-P\right]dV$$
This is correct. What does it predict for an ideal gas?

Chet

3. Mar 26, 2016

### grandpa2390

no everything is from my professor except for me finding the P on Google. the substitution of dU and then replacing it with Cv is in the notes my professor gave me.

he wrote in the notes that T*(dP/dT)_v = (dU/dT)_v dT

is the change in internal energy dU equal to the change in Pressure due to a change in Temperature?

4. Mar 26, 2016

### Staff: Mentor

This last equation is incorrect. The analysis starts out with $$dU=TdS-PdV$$You then write:
$$dS=\left(\frac{\partial S}{\partial T}\right)_VdT+\left(\frac{\partial S}{\partial V}\right)_TdV$$so
$$dU=T\left[\left(\frac{\partial S}{\partial T}\right)_VdT+\left(\frac{\partial S}{\partial V}\right)_TdV\right]-PdV$$or
$$dU=T\left(\frac{\partial S}{\partial T}\right)_VdT+\left[\left(\frac{\partial S}{\partial V}\right)_T-P\right]dV$$
So the term in brackets comes partially from the effect of volume on entropy at constant temperature. Finally, $T\left(\frac{\partial S}{\partial T}\right)_V=C_v$, so:
$$dU=C_VdT+\left[\left(\frac{\partial S}{\partial V}\right)_T-P\right]dV$$
You then substitute the Maxwell relation.

5. Mar 26, 2016

### grandpa2390

I mean I don't know. I'm not arguing with you. maybe I didn't type what I professor did clearly. I would hate to think that he is wrong or has no clue what he is doing... Because then I have no hope. :(

Let me see if I can figure out this LaTeX. Let me type his method step by step in the LaTeX and then go from there. I wish I could just upload his powerpoint slides lol.

6. Mar 26, 2016

### Staff: Mentor

In my judgment, what your professor did was correct, and the result in google is definitely incorrect. I can't say it any more clearly than that.

7. Mar 26, 2016

### grandpa2390

whew this is hard... I hate programming...

The internal Pressure $$Π_T = \left(\frac{\partial U}{\partial V}\right)_T$$

Fundamental Equation for Internal Energy $$dU = TdS - PdV$$

Divide by dV and impose constant T $$\left(\frac{\partial U}{\partial V}\right)_T = T \left(\frac{\partial S}{\partial V}\right)_T-P$$

Introduce the third Maxwell relation $$\left(\frac{\partial P}{\partial T}\right)_V = T \left(\frac{\partial S}{\partial V}\right)_T$$

The internal Pressure as a function of P, V, and T: $$Π_T = \left(\frac{\partial U}{\partial V}\right)_T = T \left(\frac{\partial P}{\partial T}\right)_V - P$$

Considering the internal Energy as a function of T and V $$dU\left({T , V}\right) = \left(\frac{\partial U}{\partial T}\right)_VdT + \left(\frac{\partial U}{\partial V}\right)_TdV$$

The First Partial Derivative is the thermodynamic observable known as the constant volume heat capacity
$$C_V = \left(\frac{\partial U}{\partial T}\right)_V$$

The second Partial Derivative is the thermodynamic observable known as the internal Pressure
$$Π_T = \left(\frac{\partial U}{\partial V}\right)_T$$

One of the main ways to express the energy is therefore:
$$dU = C_VdT + Π_TdV$$

Is this correct?

This brings up a next question. when it says first and second partial derivative. what is the original function it is referring to. What are we taking the partial derivative of to get internal pressure and heat capacity. I probably know this already, but I forgot.

edit: well it appears like it is the first and second partial derivative of that dU function...

8. Mar 26, 2016

### grandpa2390

So P is not equal to $$\left(\frac{\partial U}{\partial V}\right)_TdV$$

because that is what was subbed in for P from the previous step...

so where did that piece come from? how did we get from the previous equation with the P in it to the dU function.

9. Mar 26, 2016

### grandpa2390

Thanks by the way. $$: )$$

10. Mar 26, 2016

### Staff: Mentor

Pressure is minus the partial of U with respect to V at constant S, not at constant T. Where did you get the idea that it was at constant T?

Chet

11. Mar 26, 2016

### grandpa2390

I'm just trying to figure out what replaced what and why... I need to be able to do this :(

12. Mar 27, 2016

### Staff: Mentor

I have never heard the term "internal pressure" used before.

The derivation I presented in post #4 is correct, and the final result I presented in post #2 is correct. Both of these are consistent with what your professor presented. I'm having trouble understanding what your issue is.

Chet

13. Mar 27, 2016

### grandpa2390

Internal Pressure is what he has written... I could take a screenshot of his powerpoint for you...

my issue is that I don't know how he went from
The internal Pressure as a function of P, V, and T: $$Π_T = \left(\frac{\partial U}{\partial V}\right)_T = T \left(\frac{\partial P}{\partial T}\right)_V - P$$

to

Considering the internal Energy as a function of T and V $$dU\left({T , V}\right) = \left(\frac{\partial U}{\partial T}\right)_VdT + \left(\frac{\partial U}{\partial V}\right)_TdV$$

unless, as it just occurred to me, that the internal Pressure formula and internal energy are indeed not connected in a linear way. That he just pulled the the internal energy function out of thin air...
and then plugged in the internal pressure and Cv into it...

at which point your solution makes sense

14. Mar 27, 2016

### grandpa2390

My professor replaced the $$\left(\frac{\partial U}{\partial V}\right)_TdV$$ with $$Π_T$$
where you replaced it with $$T \left(\frac{\partial P}{\partial T}\right)_V - P$$

well not exactly, but it looks like you just didn't use the maxwell relation...
and I am not sure what happened to your T?

15. Mar 27, 2016

### Staff: Mentor

The internal energy U is a state function, and, as such, its value (per mole) is determined by specifying any to intensive properties. In the original equation, it was expressed as U = U(S,V). But it can also be expressed as U = U(T,V). So, using the chain rule for partial differentiation, you have:
$$dU\left({T , V}\right) = \left(\frac{\partial U}{\partial T}\right)_VdT + \left(\frac{\partial U}{\partial V}\right)_TdV$$. He is just showing how to express the partial derivatives in this equation in terms of Cv, P, V, and T (starting from the original equation).

16. Mar 27, 2016

### grandpa2390

ok
oh and I believe the internal pressure is pressure that is internal as opposed to external pressure or system pressure

17. Mar 27, 2016

### Staff: Mentor

If that's what he's calling the "internal pressure" (for whatever reason), then all the derivations make sense mathematically.

18. Mar 27, 2016

### grandpa2390

Scratch this...

19. Mar 27, 2016

### grandpa2390

or maybe I am confused....
$$S_{sys}$$ $$S_{surr}$$ $$S_{univ}$$ and that for q also.
I may have just invented the internal and external and system for temp and pressure

20. Mar 27, 2016

### Staff: Mentor

Like I said, I have never heard the term "internal pressure" before, and it has no meaning to me. It certainly isn't anything like the actual gas pressure P. For an ideal gas, his relationship for the "internal pressure" reduces to zero.