Maxwell relations with heat capacity

In summary: I'd rather consider S to be a function of T and V. Then I could differentiate S asdS=\left(\frac{dS}{dT}\right)_V\,dT+\left(\frac{dS}{dV}\right)_T\,dVThen I'd differentiate with respect to T at constant p:\left(\frac{dS}{dT}\right)_p=\left(\frac{dS}{dT}\right)_V+\left(\frac{dS}{dV}\right)_T\left(\frac{dV}{dT}\right)_pYou should be able to figure
  • #1
fraggedmemory
4
0

Homework Statement


Use the Maxwell relations and the Euler chain relation to express (ds/dt)p in terms of the heat capacity Cv = (du/dt)v. The expansion coefficient alpha = 1/v (dv/dt)p, and the isothermal compressibility Kt = -1/v (dV/dp)T. Hint. Assume that S= S(p,V)

Homework Equations


dQ(rev) = Tds
The maxwell relations
Euler Chain relation

The Attempt at a Solution



Alright, my attempts at this involved trying find common partial derivatives from the information already given. I couldn't find anything. But then looking at the hint I thought that there might be a way to express the change in entropy with respect to pressure and volume. I get this ds = (dU + PdV)/T assuming constant pressure. I am really not sure what I am suppose to do. I especially don't get what the expansion coefficient and thermal compressibility has to do with anything, but that might be because I can't see the big picture with this problem.

A step by step explanation would be greatly appreciated.
 
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  • #2
I'd rather consider S to be a function of T and V. Then I could differentiate S as

[tex]dS=\left(\frac{dS}{dT}\right)_V\,dT+\left(\frac{dS}{dV}\right)_T\,dV[/tex]

Then I'd differentiate with respect to T at constant p:

[tex]\left(\frac{dS}{dT}\right)_p=\left(\frac{dS}{dT}\right)_V+\left(\frac{dS}{dV}\right)_T\left(\frac{dV}{dT}\right)_p[/tex]

You should be able to figure out the rest. This is a useful trick for when you want to compare derivatives taken under different conditions.
 
  • #3
I am very thankful for your reply. However, I managed to solve the problem several hours after my post. Your method though is something that I didn't think of, so I do appreciate it.
 
  • #4
fraggedmemory said:

Homework Statement


I'm not sure about the sentence "all variables and given/known data".

fraggedmemory said:
Use the Maxwell relations and the Euler chain relation to express (ds/dt)p in terms of the heat capacity Cv = (du/dt)v. The expansion coefficient alpha = 1/v (dv/dt)p, and the isothermal compressibility Kt = -1/v (dV/dp)T. Hint. Assume that S= S(p,V)

Homework Equations


dQ(rev) = Tds
The maxwell relations
Euler Chain relation

In fact I succeed in order to obtain a relation between:
[tex] \left(\frac{\partial s}{\partial T}\right)_p[/tex]
and [tex] c_v, \alpha, k_T, T, v[/tex] just following the mapes's hint to consider S as a function of T and V. Where minuscle letter for extensive quantity means: "this quantity is molar", and all transformation are intended to involve a costant number of molecules.

In fact:

[tex] \left(\frac{\partial s}{\partial T}\right)_p = \left(\frac{\partial s}{\partial T}\right)_v + \left(\frac{\partial s}{\partial v}\right)_T \left(\frac{\partial v}{\partial T}\right)_T [/tex]

The Maxwell's relation following from [tex] d(-p dv - sdT)=0[/tex] tell us:
[tex] \left(\frac{\partial s}{\partial v}\right)_p= \left(\frac{\partial p}{\partial T}\right)_v[/tex]
Now the Euler's chain rule give us the link between the first derivative in second addend, the compressibility and the thermal expansion coefficient. In fact:

[tex] \left(\frac{\partial p}{\partial T}\right)_v \left(\frac{\partial T}{\partial v}\right)_p\left(\frac{\partial v}{\partial p}\right)_T = -1[/tex]

and:

[tex]\left(\frac{\partial T}{\partial v}\right)_p = 1/\left(\frac{\partial v}{\partial T}\right)_p [/tex]

so that:

[tex] \left(\frac{\partial p}{\partial T}\right)_v = \frac{\alpha}{k_T}[/tex]

In order to complete the derivation we need to use the given alternative definition of specific heat:

[tex] c_v = T \left(\frac{\partial s}{\partial T}\right)_v = \left(\frac{\partial u}{\partial T}\right)_v [/tex]

Which follow from:

[tex]T dS = dU + p dV [/tex]

So obtaining:

[tex] \left(\frac{\partial s}{\partial T}\right)_p = \frac{c_v}{T} + \frac{\alpha^2 v}{k_T}[/tex]

is this what was required?


fraggedmemory said:

The Attempt at a Solution



Alright, my attempts at this involved trying find common partial derivatives from the information already given. I couldn't find anything. But then looking at the hint I thought that there might be a way to express the change in entropy with respect to pressure and volume. I get this ds = (dU + PdV)/T assuming constant pressure. I am really not sure what I am suppose to do. I especially don't get what the expansion coefficient and thermal compressibility has to do with anything, but that might be because I can't see the big picture with this problem.

A step by step explanation would be greatly appreciated.
 

Related to Maxwell relations with heat capacity

1. What are Maxwell relations with heat capacity?

Maxwell relations are a set of equations in thermodynamics that relate different thermodynamic properties of a system to each other. They are derived from the fundamental laws of thermodynamics and are useful in understanding the relationships between different variables such as temperature, pressure, and volume. These relations also involve the heat capacity of a system, which is the amount of heat required to raise the temperature of the system by one degree.

2. How are Maxwell relations used in thermodynamics?

Maxwell relations are used to simplify the calculation of different thermodynamic properties of a system. They allow us to express a property in terms of other properties, making it easier to analyze and understand the behavior of a system. These relations are also helpful in predicting the behavior of a system when one of its properties is changed.

3. What is the significance of Maxwell relations in heat capacity?

Maxwell relations involving heat capacity are particularly important because they allow us to relate the heat capacity at constant pressure (Cp) to the heat capacity at constant volume (Cv). This is useful in understanding the difference in heat capacity between these two conditions and how it affects the behavior of a system.

4. How are Maxwell relations derived?

Maxwell relations are derived from the laws of thermodynamics, specifically the first and second laws. They are based on the concept of energy conservation and the idea that energy can be exchanged between a system and its surroundings in the form of heat and work. By applying these laws and manipulating the equations, we can obtain the Maxwell relations.

5. Can Maxwell relations be applied to any thermodynamic system?

Yes, Maxwell relations can be applied to any thermodynamic system as long as the system is in equilibrium. They are general equations that are valid for all systems, regardless of their specific properties or composition. However, they may be more useful for certain types of systems, such as ideal gases, where the relationships between properties are simpler.

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