Free energy Helmholtz using only the equation of state

In summary, the conversation discusses the state equation of a gas and how to demonstrate that the specific heat at constant volume, ##c_V##, only depends on temperature, as well as finding the expressions for internal energy and entropy in terms of temperature and volume. The conversation also covers how to determine ##\left(\frac{\partial S}{\partial V}\right)_T## in terms of pressure, volume, and temperature, and how to show that ##c_V## is independent of volume.
  • #1
Dario SLC

Homework Statement


This is a state ecuation of a gas:
PV=AT+B/V, where A and B there are constants.
First: Demonstrate that ##c_V## depends only of T
Second: Find U(T,V) and S(T,V)

Homework Equations


##\left(\frac{\partial U}{\partial S}\right)_V=T\text{ (1)}##
##\left(\frac{\partial U}{\partial V}\right)_S=-p\text{ (2)}##
##F=U-TS\text{ (3)}##
##\left(\frac{\partial F}{\partial V}\right)_T=-p\text{ (4)}##
##\left(\frac{\partial F}{\partial T}\right)_V=-S\text{ (5)}##

The Attempt at a Solution


For second item, I think use first (2) and then integrate respect to V using the state ecuation, I got:
##U(S,V)=-AT\ln V+B/V+u(S)##
when u(S) is a constant integration, then I use the (1) for find u(S) and the complete expression for U(S,V):
##U(S,V)=-AT\ln V+B/V+TS+U_0##
now using (4) I would like to obtain the free energy of Helmholtz:
\begin{equation}
F(T,V)=-ATlnV+B/V+f(T)
\end{equation}
when f(T) it is a new constant integration, then I use the (5) I got:
-S(T,V)=-AlnV+f'(T) then S(T,V)=AlnV-f'(T)
if F=U(S,V)-TS(T,V) and derivate respect to T, this it is the same
-AlnV=-AlnV+f'(T)
therefore f(T)=c with c constant, then
S(T,V)=AlnV-c
and
U(T,V)=-ATlnV+B/V-cT+U0

Well, I doubt that the entropy don't depends of T, I haven't see the error, it should be something like to
##S(T,V)=S_0+c_V\ln\frac{T}{T_0}+R\ln\frac{V}{V_0}## for a van der Waals gases
(Sears, Thermodynamics, chapter 9)
 
Physics news on Phys.org
  • #2
What is the equation for dU in terms of dT, Cv, P, and dV?
 
  • #3
Chestermiller said:
What is the equation for dU in terms of dT, Cv, P, and dV?
Hi, do you mean at this?
$$dU=TdS-pdV$$ or another form
$$dU=c_vdT-pdV$$
but I can't the real form for ##c_v##, except of course $$c_V=T\left(\frac{dS}{dT}\right)_V$$ but I haven't if the expression for S(T,V) found is correct.
 
  • #4
Dario SLC said:
Hi, do you mean at this?
$$dU=TdS-pdV$$ or another form
$$dU=c_vdT-pdV$$
This equation is incorrect. It should read $$dU=c_vdT+\left(T\frac{\partial S}{\partial V}-P\right)dV$$From a Maxwell relation involving dF=-SdT-PdV, what is ##\partial S/\partial V## at constant T in terms of P, V, and T?
 
Last edited:
  • #5
Chestermiller said:
This equation is incorrect. It should read $$dU=c_vdT+\left(T\frac{\partial S}{\partial V}-P\right)dV$$From a Maxwell relation involving dF=-SdT-PdV, what is ##\partial S/\partial V## at constant T in terms of P, V, and T?

Yeah! thanks a lot, reviewing a lot of writings, I see the same but I do not complete de entropy. My doubt now is the ##c_V## because except that the problem say it is constant, good! but if ##c_V=c_V(T)## there will be a integral in the expression for energy and entropy.

For the first issue, I think this:
for the first principle ##dU=TdS-pdV## and ##dU## and ##dS## there are exact differential, rewriting:
$$\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial U}{\partial V}\right)_TdV=T\left(\frac{\partial S}{\partial T}\right)_VdT+T\left(\frac{\partial S}{\partial V}\right)_TdV-pdV$$
gathering differentials for ##dT##
$$\left(\frac{\partial U}{\partial T}\right)_V=T\left(\frac{\partial S}{\partial T}\right)_V$$
when the term ##T\left(\frac{\partial S}{\partial T}\right)_V## is the specific heat at V constant $c_v$, therefore only depend of T

This is correct?
 
  • #6
Dario SLC said:
Yeah! thanks a lot, reviewing a lot of writings, I see the same but I do not complete de entropy. My doubt now is the ##c_V## because except that the problem say it is constant, good! but if ##c_V=c_V(T)## there will be a integral in the expression for energy and entropy.

For the first issue, I think this:
for the first principle ##dU=TdS-pdV## and ##dU## and ##dS## there are exact differential, rewriting:
$$\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial U}{\partial V}\right)_TdV=T\left(\frac{\partial S}{\partial T}\right)_VdT+T\left(\frac{\partial S}{\partial V}\right)_TdV-pdV$$
gathering differentials for ##dT##
$$\left(\frac{\partial U}{\partial T}\right)_V=T\left(\frac{\partial S}{\partial T}\right)_V$$
when the term ##T\left(\frac{\partial S}{\partial T}\right)_V## is the specific heat at V constant $c_v$, therefore only depend of T

This is correct?
Well, your math is correct. But you still haven't proven that Cv depends only on T for this particular equation of state. Do you know how to determine ##\left(\frac{\partial S}{\partial V}\right)_T## in terms of P, V, and T? Do you know how to show that
$$\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V$$
 
  • #7
Chestermiller said:
Well, your math is correct. But you still haven't proven that Cv depends only on T for this particular equation of state. Do you know how to determine ##\left(\frac{\partial S}{\partial V}\right)_T## in terms of P, V, and T? Do you know how to show that
$$\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V$$
I think I understand what you say:
I must prove that:
$$
\frac{\partial c_v}{\partial V}=0$$
how
$$\frac{\partial c_v}{\partial V}=T\left[\frac{\partial}{\partial V}\left(\frac{\partial S}{\partial T}\right)_V\right]_T$$
and how the Maxwell relation say:
$$\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V$$
then
$$\frac{\partial c_v}{\partial V}=T\left[\frac{\partial}{\partial V}\left(\frac{\partial P}{\partial T}\right)_V\right]_T$$
using the equation of state:
$$\left(\frac{\partial P}{\partial T}\right)_V=A/V$$
therefore deriving again respect to T it is null, then ##c_V## it does not depend on ##V##.
 
  • #8
Dario SLC said:
I think I understand what you say:
I must prove that:
$$
\frac{\partial c_v}{\partial V}=0$$
how
$$\frac{\partial c_v}{\partial V}=T\left[\frac{\partial}{\partial V}\left(\frac{\partial S}{\partial T}\right)_V\right]_T$$
and how the Maxwell relation say:
$$\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V$$
then
$$\frac{\partial c_v}{\partial V}=T\left[\frac{\partial}{\partial V}\left(\frac{\partial P}{\partial T}\right)_V\right]_T$$
using the equation of state:
$$\left(\frac{\partial P}{\partial T}\right)_V=A/V$$
therefore deriving again respect to T it is null, then ##c_V## it does not depend on ##V##.
Correct.
 
  • #9
Actually, you have $$dU=C_vdT-\frac{B}{V^2}dV$$. This also shows that Cv is dependent only on temperature.
 
Last edited:

What is the equation of state for calculating Free energy Helmholtz?

The equation of state used to calculate Free energy Helmholtz is the Helmholtz free energy equation, which is expressed as F = U - TS, where F is the free energy, U is the internal energy, T is the temperature, and S is the entropy.

Why is the Helmholtz free energy equation used for calculating Free energy Helmholtz?

The Helmholtz free energy equation is used because it is a thermodynamic potential that takes into account both the internal energy and the entropy of a system. This makes it a more accurate and comprehensive equation for calculating Free energy Helmholtz.

What other thermodynamic potentials can be used to calculate Free energy Helmholtz?

Besides the Helmholtz free energy, the Gibbs free energy and the Landau free energy can also be used to calculate Free energy Helmholtz. However, these equations may require additional parameters and assumptions, making the Helmholtz free energy equation a more preferred and straightforward method.

Can the Helmholtz free energy equation be used for all types of systems?

Yes, the Helmholtz free energy equation can be used for all types of systems, including gases, liquids, and solids. However, it is important to note that the equation may need to be modified or combined with other equations for more complex systems, such as mixtures or non-ideal gases.

How is Free energy Helmholtz related to other thermodynamic properties?

Free energy Helmholtz is related to other thermodynamic properties through the fundamental thermodynamic equations, such as the Maxwell equations and the Clapeyron equation. These equations allow for the calculation of other properties, such as enthalpy, entropy, and temperature, using the Free energy Helmholtz as a starting point.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
666
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
469
Replies
3
Views
913
  • Introductory Physics Homework Help
Replies
4
Views
911
  • Introductory Physics Homework Help
Replies
7
Views
910
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
664
  • Introductory Physics Homework Help
Replies
6
Views
664
  • Introductory Physics Homework Help
Replies
1
Views
446
Back
Top