Thermodynamics - Maxwell Relations

In summary, the conversation discusses the use of Maxwell's relations and state variables in solving a thermodynamics problem. The approach involves expressing dT as a function of both P and v, reducing the derivatives, and using the coefficients of dv and dP in terms of alpha and kappa_T. However, there is a discrepancy with the given answer, which may be due to a mistake in the sign of the first term on the right. The conversation also suggests expressing ln(P_b/P_a) in terms of v_a and v_b.
  • #1
Lucas Mayr
18
0

Homework Statement



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2. The attempt at a solution
I've tried using the relation Cp = T(dS/dT), isolating "T" for T = Cv2(dT/dS) and using the maxwell relations to reduce the derivatives, reaching, T = Cv2/D (dV/dS), but i don't think this is the right way to do solve this problem, i couldn't find a similar example on the chapter either.
 
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  • #2
##T## is a state variable that can be thought of as a function of ##v## and ##P##: ##\;T(v,P)##.

Consider ##dT## which will be something times ##dv## plus something times ##dP##. Can you express the coefficients of ##dv## and ##dP## in terms of ##\alpha## and ##\kappa_T##?
 
  • #3
Ok, so I've tried looking at dT as a function of both P and v and reached dT = (∂T/∂P)v dP + (∂T/∂v)P dv
And after reducing the derivatives, dT = KT/α dP + 1/vα dv , and using the problem's KT and α.
dT = 1/D dv + Ev2/D dP
dT = 1/D dv + EPava2/(PbD) dP
Tb = Ta + (vb - va)/D + EPava2 Ln(Pb/Pa)/D

which is close but still different from the answer given on the question and i can't find a reason why, what did i miss?
 
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  • #4
Lucas Mayr said:
Ok, so I've tried looking at dT as a function of both P and v and reached dT = (∂T/∂P)v dP + (∂T/∂v)P dv
And after reducing the derivatives, dT = KT/α dP + 1/vα dv ,...
Check the sign of the first term on the right. Otherwise, that looks good.

Tb = Ta + (vb - va)/D + EPava2 Ln(Pb/Pa)/D
Can you express ##\ln(P_b/P_a)## in terms of ##v_a## and ##v_b##?
 
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Related to Thermodynamics - Maxwell Relations

1. What are Maxwell Relations in thermodynamics?

Maxwell relations are a set of equations that relate the four thermodynamic potentials: internal energy, enthalpy, Helmholtz free energy, and Gibbs free energy. They are derived from the four fundamental thermodynamic laws and can be used to calculate changes in one potential based on changes in another.

2. How many Maxwell Relations are there?

There are four Maxwell Relations, each one corresponding to one of the four thermodynamic potentials. They are also known as the "Maxwell Equations."

3. What is the significance of Maxwell Relations in thermodynamics?

Maxwell Relations are significant because they allow us to relate changes in one thermodynamic potential to another, making it easier to calculate and understand thermodynamic processes. They also provide a useful tool for checking the consistency of experimental data.

4. How are Maxwell Relations derived?

Maxwell Relations are derived from the four fundamental thermodynamic laws: the first law (conservation of energy), the second law (entropy), the third law (absolute zero), and the fourth law (thermal equilibrium). By combining these laws and using mathematical manipulations, we can obtain the four Maxwell Relations.

5. What is an example of using Maxwell Relations in thermodynamics?

One example is using the Maxwell Relation between internal energy and entropy to calculate the change in internal energy when the temperature and volume of a system are held constant. This can be useful in determining the efficiency of a heat engine.

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