What Is the Equation of State for a Gas Given Its Gibbs Function?

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SUMMARY

The Gibbs function for a gas is expressed as G = nRT ln P + A + BP + (1/2)CP^2 + (1/3)DP^3. The equation of state derived from this function is PV = nRT + PB + CP^2 + DP^3, confirming the relationship between pressure, volume, and temperature. Additionally, the discussion explores the application of Hooke's law to a spring with a temperature-dependent spring constant, leading to expressions for free energy, internal energy, and entropy as functions of temperature and displacement. The final equations incorporate the first law of thermodynamics and the initial conditions for energy states.

PREREQUISITES
  • Understanding of Gibbs free energy and its definitions
  • Familiarity with thermodynamic equations and the first law of thermodynamics
  • Knowledge of Hooke's law and its application in thermodynamics
  • Basic calculus for differentiation and integration in thermodynamic contexts
NEXT STEPS
  • Study the derivation of the Gibbs function in thermodynamics
  • Learn about the implications of temperature dependence in spring constants
  • Explore advanced applications of the first law of thermodynamics
  • Investigate the relationship between entropy and free energy in thermodynamic systems
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Students and professionals in physics and engineering, particularly those focusing on thermodynamics, material science, and mechanical systems involving gas laws and spring mechanics.

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Homework Statement


*1* The Gibbs function G(T; P) of a certain gas is:
G = nRT ln P + A + BP + (1/2)*CP^2 + (1/3)DP^3

where A, B, C and D are constants. Find the equation of state of the gas.

Homework Equations



G = nRT ln P + A + BP + (1/2)*CP^2 + (1/3)*DP^3

The Attempt at a Solution



I think I have a solution for this, but it seemed to be too easy.

The definition for Gibbs free energy is defined to be:
1) G = U - TS + PV and 2) dG = -SdT + vdP

solving 2) for dG/dP yields: dG/dP = -SdT/dP + V 3)

I thought to take 3) at constant temperature yielding: dG/dP = V
Now, taking dG/dP at constant T from the given formula yields:

dG/dP =nRT/P + B + CP + DP^2 4)

Setting 3) = 4) and solving for PV yields:

PV = nRT + PB + CP^2 + DP^3
That just seemed too easy for me >.>
The second part of this homework is:

Homework Statement


Consider a spring which follows Hookes law; namely the displacement x from equilibrium
position is proportional to the tension X when it is pulled at a constant temperature. The spring constant is temperature dependent, k = k(T). Determine the free energy F, the internal energy U, and the entropy S, as a function of T and x. Neglect the thermal expansion of spring. Use F0(T) ´ F(T; x = 0); U0(T) ´ U(T; x = 0); S0(T) ´ S(T; x = 0) where necessary.

Homework Equations



k = k(T)
F[0](T) = F(T; x = 0)
U[0](T) = U(T; x = 0)
S[0](T) =S(T; x = 0)

where the brackets denote subscripts

The Attempt at a Solution



The force of the spring can be obtained by: F = -k(T) * x
And the work: W = int(-k(T) * x) dx = -(1/2)*k(T)*x^2

and dW = k(T) *xdx

Using the first law of thermodynamics: dU = dQ - dW, so

dU = dQ + k(T)*xdx
 
Last edited:
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The elongation of the spring would add another term to your energy equation:

U=TS-PV+Fx=TS-PV+k(T)x^2

F=U-TS=-PV+k(T)x^2

S=-\left(\frac{\partial F}{\partial T}\right)_V=-\frac{\partial k(T)}{\partial T} x^2

Does this help?
 
That makes sense to me, but what about those 3 initial conditions? This just seems too easy to me >.>

F[0](T) = F(T; x = 0)
U[0](T) = U(T; x = 0)
S[0](T) =S(T; x = 0)

What do I do with those?
 
I cheated a bit by ignoring the temperature dependence of PV and by setting my energy baseline at \mu\,N. The initial conditions make these constants go away anyway. Wouldn't you end up with, for example,

S=S_0-\frac{\partial k(T)}{\partial T} x^2

I'll leave the rest for you.
 
Thanks a bunch! This helps a lot! :P
 

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