Heat Capacities Given Equation of State

Click For Summary

Homework Help Overview

The discussion revolves around deriving heat capacities from a given equation of state, specifically focusing on the relationship between pressure, volume, and temperature. The original poster presents a problem involving the equation of state and seeks to derive related equations and calculate entropy and heat capacities.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore deriving the pressure as a function of volume and temperature from the provided equation of state. There are attempts to calculate entropy based on heat capacity definitions, and questions arise regarding the dependence of entropy on volume. Some participants also discuss the implications of Maxwell's relations on the problem.

Discussion Status

The discussion is active, with participants sharing their calculations and reasoning. There is a recognition of the complexity involved in differentiating entropy with respect to temperature while holding pressure constant. Some participants suggest that constants of integration may play a role in the calculations, and there is a general exploration of how to relate different variables in the context of the problem.

Contextual Notes

Participants note the challenge of integrating constants and the implications of the given equation of state on the calculations. There is an acknowledgment of the need to maintain clarity regarding the dependencies of various thermodynamic quantities.

Yoni V
Messages
38
Reaction score
0

Homework Statement


Given the equation of state ##V(P,T)=V_1\cdot exp(\frac{T}{T_1}-\frac{P}{P_1})## where ##V_1\;,T_1\;,P_1## are constants:
a. derive an equivalent equation ##P(V,T)##;
b. given ##C_V=DT^3## where D is a const, calculate the entropy of the system ##s(V,T)## up to a const;
c. find heat capacity ##C_P##.

Homework Equations


Definitions of heat and heat capacities.
First law? Enthalpy?

The Attempt at a Solution


Starting with a, we get via a simple calculation ##P(V,T)=P_1\cdot (\frac{T}{T_1}-ln(\frac{V}{V_1}))##.
Then, using heat and heat cap. definitions we get
$$ds=dQ/T=CdT/T$$$$\Rightarrow (\frac{\partial S}{\partial T})_V = \frac{C_V}{T} = DT^2$$$$\Rightarrow s = \frac{1}{3}DT^3 + S_0$$
Of this step I'm not entirely sure. It looks okay, but s is not a function of the volume, and I think I need that dependence to work out ##C_P##. So this is where I'm stuck.

Thanks!
 
Physics news on Phys.org
Yoni V said:

Homework Statement


Given the equation of state ##V(P,T)=V_1\cdot exp(\frac{T}{T_1}-\frac{P}{P_1})## where ##V_1\;,T_1\;,P_1## are constants:
a. derive an equivalent equation ##P(V,T)##;
b. given ##C_V=DT^3## where D is a const, calculate the entropy of the system ##s(V,T)## up to a const;
c. find heat capacity ##C_P##.

Homework Equations


Definitions of heat and heat capacities.
First law? Enthalpy?

The Attempt at a Solution


Starting with a, we get via a simple calculation ##P(V,T)=P_1\cdot (\frac{T}{T_1}-ln(\frac{V}{V_1}))##.
Then, using heat and heat cap. definitions we get
$$ds=dQ/T=CdT/T$$$$\Rightarrow (\frac{\partial S}{\partial T})_V = \frac{C_V}{T} = DT^2$$$$\Rightarrow s = \frac{1}{3}DT^3 + S_0$$
Of this step I'm not entirely sure. It looks okay, but s is not a function of the volume
What is the partial derivative of entropy with respect to volume at constant temperature?
 
Hmm... from Maxwell's relations $$(\frac{\partial S}{\partial V})_T = (\frac{\partial P}{\partial T})_V$$
which equals ##\frac{P_1}{T_1}## from the given equation of state, and then $$s = \frac{P_1}{T_1}V + s_0'$$
The heat cap. is given by ##C_P = T(\frac{\partial s}{\partial T})_P##.
I could try rewriting $$(\frac{\partial s}{\partial T})_P = -\frac{(\frac{\partial P}{\partial T})_s}{(\frac{\partial P}{\partial s})_T} = -\frac{(\frac{\partial S}{\partial V})_P}{(\frac{\partial P}{\partial s})_T}$$
But I can't see how these changes in variables get me where I want to go.
Maybe I can somehow relate ##s_0## to ##s_0'##, but it doesn't seem relevant...
 
Yoni V said:
Hmm... from Maxwell's relations $$(\frac{\partial S}{\partial V})_T = (\frac{\partial P}{\partial T})_V$$
which equals ##\frac{P_1}{T_1}## from the given equation of state, and then $$s = \frac{P_1}{T_1}V + s_0'$$
What happened to the DT3/3?
 
Ok, I think I got it while writing this reply. What I started typing is:
It's part of the constant of integration ##s_0'##, or if we neglect a constant that has nothing to do with the temp. and volume, it is exactly ##s_0'##.
This I understand, but even writing it explicitly, I'm not sure on how to differentiate the entropy with respect to the temp. while holding P const, as V is free to change to keep the given pressure constant.

But..! We are given V in terms of both T and P. So plugging it to s, we finally get $$s = \frac{1}{3}DT^3 + \frac{P_1}{T_1}(
V_1⋅exp(\frac{T}{T_1}−\frac{P}{P_1}))+s_0$$ where ##s_0## is now a true constant that we can entirely ignore and differentiate happily to get $$C_P = T(\frac{\partial s}{\partial T})_P = DT^3 + \frac{P_1V_1T}{T_1^2}exp(\frac{T}{T_1}-\frac{P}{P_1})$$
Is that correct?!
And thank you very much!
 
Yoni V said:
Ok, I think I got it while writing this reply. What I started typing is:
It's part of the constant of integration ##s_0'##, or if we neglect a constant that has nothing to do with the temp. and volume, it is exactly ##s_0'##.
This I understand, but even writing it explicitly, I'm not sure on how to differentiate the entropy with respect to the temp. while holding P const, as V is free to change to keep the given pressure constant.

But..! We are given V in terms of both T and P. So plugging it to s, we finally get $$s = \frac{1}{3}DT^3 + \frac{P_1}{T_1}(
V_1⋅exp(\frac{T}{T_1}−\frac{P}{P_1}))+s_0$$ where ##s_0## is now a true constant that we can entirely ignore and differentiate happily to get $$C_P = T(\frac{\partial s}{\partial T})_P = DT^3 + \frac{P_1V_1T}{T_1^2}exp(\frac{T}{T_1}-\frac{P}{P_1})$$
Is that correct?!
And thank you very much!
Correct. Nice job.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
3
Views
3K
Replies
3
Views
1K
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K