Proving Thermodynamics equations using partial derivatives

In summary: So now you can continue on to the next step?In summary, the conversation discusses proving the equation (∂V/∂T)_s/(∂V/∂T)_p = 1/1-(gamma) (gamma = Cp/Cv) using the equations (∂V/∂T)_s = -C_v (kappa)/(beta)T, C_v= - T(∂P/∂T)_v(∂V/∂T)_s, and (∂S/∂T)_p=1/T(∂U/∂T)_p. The conversation also addresses a previous question about showing (∂C_p/∂P)_T = -
  • #1
cowcow8866
11
0

Homework Statement


Prove
(∂V/∂T)_s/(∂V/∂T)_p = 1/1-(gamma) (gamma = Cp/Cv)

Homework Equations


(∂V/∂T)_s = -C_v (kappa)/(beta)T (where beta = 1/V(∂V/∂T)_p, kappa = -1/V(∂V/∂P)_T

C_v= - T(∂P/∂T)_v(∂V/∂T)_s

The Attempt at a Solution


As part(a) ask me to find C_v, I do it similar for C_p

(∂S/∂T)_p=1/T(∂U/∂T)_p
C_p=T(∂S/∂T)_p=-T(∂P/∂T)_s/(∂P/∂S)_T=-T(∂P/∂T)_s(∂V/∂T)_p
(∂V/∂T)_s/(∂V/∂T)_p = -C_v/T(∂P/∂T)_v /C_p/-T(∂P/∂T)_s=C_v(∂P/∂T)_s/ C _p(∂P/∂T)_v

Then, i cannot figure out the remaining calculation out...
 
Last edited:
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  • #2
How are you getting

[tex]\left(\frac{\partial S}{\partial T}\right)_P=\frac{1}{T}\left(\frac{\partial U}{\partial T}\right)_P\mathrm{?}[/tex]

That's not correct; starting from [itex]dU=T\,dS-P\,dV[/itex], I get

[tex]\left(\frac{\partial U}{\partial T}\right)_P=T\left(\frac{\partial S}{\partial T}\right)_P-P\left(\frac{\partial V}{\partial T}\right)_P[/tex]

[tex]\left(\frac{\partial S}{\partial T}\right)_P=\frac{1}{T}\left(\frac{\partial U}{\partial T}\right)_P+\frac{P}{T}\left(\frac{\partial V}{\partial T}\right)_P[/tex]

Also, I see a sign error on at least one of your Maxwell relations. Not sure if this resolves the problem?
 
  • #3
There is another derived equation in my book and I have been used it to prove it. However, I still have another question.
Since there is a previous question ask me to show
(∂C_p/∂P)_T = -T(∂^2V/∂T^2)_P
I use this method stated below but I get stuck.
http://i35.photobucket.com/albums/d199/cowcow8866/scan001.png

Actually, I derive this[tex]\left(\frac{\partial S}{\partial T}\right)_P=\frac{1}{T}\left(\frac{\partial U}{\partial T}\right)_P\mathrm{?}[/tex]
from another equation TdS=dU-VdP and then prove(∂C_p/∂P)_T = -T(∂^2V/∂T^2)_P successfully.
However, it is very uncommon to be used or even it maybe wrong...
 
  • #4
There is no such equation [itex]T\,dS=dU-V\,dP[/itex]. Are you thinking of [itex]T\,dS=dH-V\,dP[/itex]?
 
  • #6
cowcow8866 said:
Maybe...
Anyway, I can solve this question now.

My next question is
Prove (∂C_p/∂P)_T = -T(∂^2V/∂T^2)_P
http://i35.photobucket.com/albums/d1...66/scan001.png
I have done up to this step. How can I do the remaining part of this question??

Sorry, I know I'm not a mod or anything, heck I've been here like 3 days =]

Isn't this kinda in the wrong section? Because tbh I'm currently doing A2 physics and I'd consider that to be kind of the end of introductory physics, what you're doing here seems a bit more advanced (yes, the maths is simple, but still beyond your average physics student imo).

Also, shouldn't you create a new thread for a new question.

Again, I'm not trying to be uppity, I'm just saying.
 
Last edited by a moderator:
  • #7
cowcow8866 said:
Maybe...
Anyway, I can solve this question now.

My next question is
Prove (∂C_p/∂P)_T = -T(∂^2V/∂T^2)_P
http://i35.photobucket.com/albums/d199/cowcow8866/scan001.png
I have done up to this step. How can I do the remaining part of this question??

Your writeup contains the same mistake I pointed out in my post #2 for the previous problem. [itex](\partial U/\partial T)_P\neq C_P[/itex].
 
  • #8
I know the mistake now. Thank you.
 
  • #9
You're welcome!
 

FAQ: Proving Thermodynamics equations using partial derivatives

What is the purpose of using partial derivatives in proving thermodynamics equations?

Partial derivatives are used to determine the rate of change of a variable while holding all other variables constant. In thermodynamics, this is useful for understanding how changes in one variable, such as temperature or pressure, affect the overall system.

Can any thermodynamics equation be proven using partial derivatives?

Yes, partial derivatives can be used to prove any thermodynamics equation that involves multiple variables. By taking partial derivatives of the equation, we can show how changes in each variable contribute to the overall change in the system.

How do you use partial derivatives to prove a thermodynamics equation?

To prove a thermodynamics equation using partial derivatives, we typically start with the given equation and then take partial derivatives of both sides with respect to each variable. By manipulating these derivatives, we can show that they are equal to each other, thus proving the original equation.

Are there any limitations to using partial derivatives in proving thermodynamics equations?

While partial derivatives are a useful tool for proving thermodynamics equations, they may not always be applicable. In some cases, the relationship between variables may be too complex to be represented by a simple partial derivative. Additionally, some equations may require more advanced mathematical techniques to prove.

How do partial derivatives relate to the laws of thermodynamics?

The laws of thermodynamics are fundamental principles that govern the behavior of thermodynamic systems. Partial derivatives are often used in the mathematical representation of these laws, such as in the equations for energy conservation and entropy. By using partial derivatives, we can better understand and analyze the principles of thermodynamics.

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