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

[grandpa2390]

Homework Statement

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.

Homework Equations

[/B]
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

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.

 

[Chestermiller]

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

 

[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?

 

[Chestermiller]

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?

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.

 

[grandpa2390]

Homework Statement

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.

Homework Equations

[/B]
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

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.

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.

 

[Chestermiller]

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.

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.

 

[grandpa2390]

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.

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...

 

[grandpa2390]

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.

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.

 

[grandpa2390]

Thanks by the way. $$ : ) $$

 

[Chestermiller]

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

 

[grandpa2390]

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

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

 

[Chestermiller]

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

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

 

[grandpa2390]

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

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

 

[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

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?

 

[Chestermiller]

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

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).

 

[grandpa2390]

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).

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

 

[Chestermiller]

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?

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

 

[grandpa2390]

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

Scratch this...

 

[grandpa2390]

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

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

 

[Chestermiller]

Once again I am unsure. I pretty sure for Temperature we have $$T_{int}$$ $$T_{ext}$$ and $$T_{sys}$$

I

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.

 

[grandpa2390]

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.

Yeah I am unsure I have ever heard of internal pressure or temperature or anything until now. Definitely for heat and entropy though.

 

[grandpa2390]

so if not internal pressure, what would you call: $$Π_T = \left(\frac{\partial U}{\partial V}\right)_T$$

I mean the result is the same except he multiplies it by a T during the derivation whereas you have no T

well he doesn't multiply it by T. the T is there from the internal energy formula.

 

[Chestermiller]

so if not internal pressure, what would you call: $$Π_T = \left(\frac{\partial U}{\partial V}\right)_T$$

I mean the result is the same except he multiplies it by a T during the derivation whereas you have no T

well he doesn't multiply it by T. the T is there from the internal energy formula.

I would call that the partial derivative of the internal energy with respect to volume at constant temperature (and nothing else). If you want to call it a hippopotamus you can , but that's up to you.

 

[grandpa2390]

I would call that the partial derivative of the internal energy with respect to volume at constant temperature (and nothing else). If you want to call it a hippopotamus you can , but that's up to you.

don't say you. I'm not the one choosing to call it anything. lol. I'm just a student.

I will email my professor and ask him what internal pressure is. Maybe it is a chemistry thing...

 

[grandpa2390]

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.

should it reduce to zero for an ideal gas? Is that wrong?

ok well I sent him an inquiry...

 

[Chestermiller]

should it reduce to zero for an ideal gas? Is that wrong?

That's correct. We know that, for an ideal gas, the internal energy is a function only of temperature (and not volume).

 

[grandpa2390]

That's correct. We know that, for an ideal gas, the internal energy is a function only of temperature (and not volume).

you're confusing me...
so then he is correct in calling it internal pressure. if you say it holds true... then why not call it internal pressure.

or are you saying It is correct that it is not true...?

 

[grandpa2390]

That's correct. We know that, for an ideal gas, the internal energy is a function only of temperature (and not volume).

right so at constant temperature the derivative would be 0... if we are integrating in respect to V... but what does that have to do with pressure?

I am lost. which is not necessarily a bad thing... maybe you can straighten me out.

Maybe I should wait for my professor to respond. Then I can give you what he says, so you can understand what in the world he is saying by internal pressure, and can then explain it to me :)

 

[Chestermiller]

right so at constant temperature the derivative would be 0... if we are integrating in respect to V... but what does that have to do with pressure?

I am lost. which is not necessarily a bad thing... maybe you can straighten me out.

Maybe I should wait for my professor to respond. Then I can give you what he says, so you can understand what in the world he is saying by internal pressure, and can then explain it to me :)

For an ideal gas, U is a function only of T. But, for a real gas (i.e., at pressures higher than in the ideal gas region), the internal energy U becomes a function of volume also.

 

[grandpa2390]

I would call that the partial derivative of the internal energy with respect to volume at constant temperature (and nothing else). If you want to call it a hippopotamus you can , but that's up to you.

this is driving me crazy... My professor hasn't responded yet, probably because of Spring Break!
What does this mean. it is not called anything, but what is it? It can't just be some random partial derivatives of different things. It has to mean something. it has to have some application or implication, or simplification.

 

[grandpa2390]

For an ideal gas, U is a function only of T. But, for a real gas (i.e., at pressures higher than in the ideal gas region), the internal energy U becomes a function of volume also.

My professor was of no help. He essentially said, it is not his problem if I don't understand something. :(

I want to drop the class but id get an F at this point. looks like I'll be teaching myself Thermodynamics with your help and the rest of the people here.

 

[Chestermiller]

My professor was of no help. He essentially said, it is not his problem if I don't understand something. :(

I want to drop the class but id get an F at this point. looks like I'll be teaching myself Thermodynamics with your help and the rest of the people here.

OK. I'll be glad to help. My first recommendation is to stop trying to assign physical significance to this parameter that you dreamed up. Your time is too valuable to waste obsessing over something like this. You should be focusing on solving lots of actual problems to solidify your understanding and your prowess.

 

[grandpa2390]

OK. I'll be glad to help. My first recommendation is to stop trying to assign physical significance to this parameter that you dreamed up. Your time is too valuable to waste obsessing over something like this. You should be focusing on solving lots of actual problems to solidify your understanding and your prowess.

thanks for your advice. I will focus on solving the homework problems. I just have this need to understand what can be understood. I guess I can just memorize formulas and methods to pass the tests, but I won't gain anything. but i guess that is my only option at this point. I am going to finish reading the chapter about this specific material and come back when (and it will be soon) i need help.

thankyou so much for being helpful rather than calling me an idiot and brushing me off like my professor.

 

[grandpa2390]

he basically said. as to whether the tutor has heard of the term has nothing to do with anything and is not my fault. perhaps he should google it.
did not answer my question.

but google did.
Internal pressure is a measure of how the internal energy of a system changes when it expands or contracts at constant temperature. It has the same dimensions as pressure, the SI unit of which is the pascal. Internal pressure is usually given the symbol .

that is exactly what you said. :) without the condescending attitude of course

 

[Chestermiller]

You're still obsessing. Whether or not you are able to assign a physical significance to what you are calling Pi will have no bearing on your ability to understand what is really important in thermodynamics. So let it go. There are things that you really are going to need to understand.