Maxwell relations with heat capacity.

[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.

 

[anathema]

First, let's start with the Euler chain relation:

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

 

[Moetasim]

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