Heat Capacities Given Equation of State

AI Thread Summary
The discussion revolves around deriving the pressure equation from a given equation of state and calculating the entropy and heat capacities of a system. The derived equation for pressure is P(V,T)=P_1·(T/T_1 - ln(V/V_1)). The entropy is expressed as s = (1/3)DT^3 + (P_1/T_1)V·exp((T/T_1)-(P/P_1)) + s_0, where s_0 is a constant. The heat capacity at constant pressure, C_P, is calculated as C_P = DT^3 + (P_1V_1T/T_1^2)·exp((T/T_1)-(P/P_1)). The calculations and reasoning presented in the thread confirm the correctness of these derivations.
Yoni V
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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!
 
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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.
 
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