# Maxwell relations with heat capacity.

• fraggedmemory
In summary, using the Euler chain relation and the Maxwell relations, we can express the change in entropy with respect to temperature at constant pressure in terms of the heat capacity and the change in pressure.
fraggedmemory
Maxwell relations with heat capacity. Solved.

1. 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)

2. Homework Equations
dQ(rev) = Tds
The maxwell relations
Euler Chain relation

3. 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.

For anyone who cares about the answer... ds = (ds/dp)T dp + (ds/dT)p dT. Then you use the euler chain relations on both.
Then you use maxwell's relations for the denominator of both. Then you can simplify the partial derivative. After that, you euler's chain relation again... At this point it is easy to see what else you have to do. It would have been impossible for me to solve this problem had I not finally figured out that the hint was really really REALLY important, and I just euler chaining everything I saw.

Last edited:

du = Tds - PdV

We can rearrange this to solve for ds:

ds = (du + PdV)/T

Next, we can use the Maxwell relations to express the partial derivative (ds/dp)T in terms of other partial derivatives:

(ds/dp)T = (dV/dT)p

Similarly, we can use the Maxwell relations to express the partial derivative (ds/dT)p in terms of other partial derivatives:

(ds/dT)p = -(dP/dT)V

Now, let's substitute these expressions into our previous equation for ds:

ds = [(du + PdV)/T][(dV/dT)p] - [(du + PdV)/T][(dP/dT)V]dP

Next, we can use the Euler chain relation again to rewrite du and PdV in terms of other partial derivatives:

du = Cv dT + T(dP/dT)V dV

PdV = -Kt dT + T(dV/dT)p dP

Substituting these expressions into our equation for ds, we get:

ds = [Cv dT + T(dP/dT)V dV + PdV]/T [(dV/dT)p] - [Cv dT + T(dP/dT)V dV + PdV]/T [(dP/dT)V]dP

Simplifying this expression, we get:

ds = Cv dT + [T(dP/dT)V (dV/dT)p - T(dV/dT)p (dP/dT)V]dP

Finally, we can use the Maxwell relations again to simplify the expression in brackets:

(dP/dT)V (dV/dT)p - (dV/dT)p (dP/dT)V = -1

Therefore, our final expression for ds is:

ds = Cv dT - dP

This shows that the change in entropy (ds) with respect to temperature (dT) at constant pressure is equal to the heat capacity (Cv) at constant volume times dT, minus the change in pressure (dP).

This solution may seem complicated, but it is important to remember to use the Euler chain relation and the Maxwell relations to simplify the partial derivatives and express them in terms of other partial derivatives. This will help you see the relationships between different therm

I appreciate your effort in trying to solve this problem. It is clear that you have a good understanding of the basic principles involved. Let me provide a step-by-step explanation to help you understand the solution better.

Firstly, we need to use the Maxwell relations to express the change in entropy (ds) in terms of the heat capacity (Cv). From the first Maxwell relation, we know that (ds/dp)T = (dV/dT)p. Using this, we can rewrite the expression for ds as ds = (dV/dT)p dp + (ds/dT)p dT.

Now, we can use the Euler chain relation, which states that for any function f(x,y), we have the following relation: (df/dx)y = (df/dy)x (dy/dx). In our case, the function is entropy (s) and the variables are pressure (p) and temperature (T). So, we can rewrite (ds/dp)T as (ds/dT)p (dT/dp).

Next, we can substitute the values of (ds/dp)T and (ds/dT)p in our expression for ds, and we get ds = (dV/dT)p dp + (ds/dT)p (dT/dp) dp.

Now, we can use the second Maxwell relation, which states that (dV/dT)p = -(dS/dP)T. This allows us to simplify our expression for ds to ds = -(dS/dP)T dp + (ds/dT)p (dT/dp) dp.

Finally, we can use the Euler chain relation again to rewrite (dS/dP)T as (dS/dT)p (dT/dP). This gives us our final expression for ds as ds = -(dS/dT)p (dT/dP) dp + (ds/dT)p (dT/dp) dp.

Now, we can see that we have two terms with (ds/dT)p and (dT/dp) in our expression. We can combine them to get our final result: ds = [(ds/dT)p + (dT/dp)(ds/dT)p] dp.

We can rewrite this as ds = (ds/dT)p [1 + (dT/dp)] dp. Now, we can use the definition of heat capacity (Cv = (dU/dT)v) to replace (ds/dT

## 1. What are Maxwell relations with heat capacity?

Maxwell relations with heat capacity are a set of equations derived from the four fundamental thermodynamic potentials (internal energy, enthalpy, Helmholtz free energy, and Gibbs free energy) that describe the relationship between various thermodynamic properties, such as temperature, pressure, and volume.

## 2. Why are Maxwell relations important in thermodynamics?

Maxwell relations are important in thermodynamics because they allow us to determine the values of one thermodynamic property based on the values of other properties. This is especially useful in situations where it is difficult or impossible to directly measure a particular property.

## 3. How do Maxwell relations relate to heat capacity?

Maxwell relations relate to heat capacity by showing the mathematical relationships between heat capacity and other thermodynamic properties, such as temperature and volume. These relations can be used to calculate the heat capacity of a system using other known properties.

## 4. Can Maxwell relations be applied to all thermodynamic systems?

Yes, Maxwell relations can be applied to all thermodynamic systems as long as the four fundamental thermodynamic potentials are defined for that system. However, the specific form of the relations may vary depending on the system and the properties being considered.

## 5. What are some practical applications of Maxwell relations?

Some practical applications of Maxwell relations include calculating the heat capacity of a substance, determining the effect of temperature on a system's properties, and predicting phase transitions. They are also useful in chemical thermodynamics for analyzing and predicting the behavior of chemical reactions.

Replies
34
Views
2K
Replies
4
Views
1K
Replies
19
Views
1K
Replies
4
Views
881
Replies
1
Views
1K
Replies
4
Views
3K
Replies
1
Views
1K
Replies
3
Views
1K
Replies
1
Views
1K
Replies
3
Views
1K