How Does Entropy Change with Pressure at Constant Temperature?

[prehisto]

Homework Statement

Hi!
My problem :
There is given substance for which (dV/dT)_p =ap^2,where V-volume,T-temperature,p-pressure,a-const.
How much the entropy S will increase if T=const and pressure changes from p1 to p2?

Homework Equations

The Attempt at a Solution

So
1) Mathematically the question of this excersie could be represented as
dS=Int [(dS/dp)_T dp] from p1 to p2
2) The next step should be to use Gibs potential : dG=Vdp-SdT
From Gibbs potential i should derive something like this (dV/dT)_p = -(dS/dp)_T (*)

But i can go as far as dG=(dG/dp)dp -(dG/dT)dT -> (dG/dp)_T=V ;(dG/dT)_p=-S
I think that the problem is mathematical: i can't figure out how from dG i can get (*)

Could someone can,please help me?

P.S. I'm sorry about the equations, i can't find any option for equation builder.

 

[Chestermiller]

You're very close to having the answer. Here's what you have so far:
$$\left(\frac{\partial G}{\partial P}\right)_T=V$$
$$\left(\frac{\partial G}{\partial T}\right)_P=-S$$
Now, what happens if you take the partial of the first equation with respect to T and the partial of the second equation with respect to P?

Chet

 

[prehisto]

You're very close to having the answer. Here's what you have so far:
$$\left(\frac{\partial G}{\partial P}\right)_T=V$$
$$\left(\frac{\partial G}{\partial T}\right)_P=-S$$
Now, what happens if you take the partial of the first equation with respect to T and the partial of the second equation with respect to P?

Chet

it s dG/dpdT=dV/dP and dG/dTdp=-dS/dp
I assume there is property of dG when dG/dpdT=dG/dTdp and so -dS/dp=dV/dP! Which allows me to substitute dS/dp=-dV/dp in integral and i can now easily integrate.
Thanks for the help!

P.S. Still having problems with formulas ,do you use "latex"code in insert "code" section ? I does not work for me :(

 

[Chestermiller]

it s dG/dpdT=dV/dP and dG/dTdp=-dS/dp
I assume there is property of dG when dG/dpdT=dG/dTdp and so -dS/dp=dV/dP! Which allows me to substitute dS/dp=-dV/dp in integral and i can now easily integrate.

Yes. That's a property of partial differentiation. The second partial with respect to two independent variables is independent of the order of the partial differentiation.

Thanks for the help!

P.S. Still having problems with formulas ,do you use "latex"code in insert "code" section ? I does not work for me :(

I've just been doing the latex in the original message. When you preview the actual message, it comes out looking the way it is supposed to look. Also, it comes out looking right if, after you post the reply, you click on the refresh "circle" next to the back arrow (next to the url).

chet

 

[paranormal]

Hi there,

Great job on trying to approach the problem using the Gibbs potential! Your equation (*), which relates the change in entropy to the change in pressure at constant temperature, is correct. To get there, you can start with the definition of the Gibbs potential:

dG = Vdp - SdT

Since we are considering a constant temperature (dT = 0), we can simplify this to:

dG = Vdp

Now, to get the equation (*) that you mentioned, we need to use the Maxwell relation:

(dV/dT)_p = -(dS/dp)_T

This relates the partial derivatives of volume and entropy with respect to temperature and pressure, respectively. We can rearrange this equation to solve for the change in entropy:

(dS/dp)_T = -(dV/dT)_p

Substituting this into our previous equation for dG, we get:

dG = - (dS/dp)_T dp

This is almost the same as your equation (*), except for a negative sign. This is because the Maxwell relation gives us the negative of the partial derivative of entropy with respect to pressure. So to get the correct equation, we need to add a negative sign, which gives us:

dS = - (dG/dp)_T dp

Now, we can integrate this equation from p1 to p2 to get the change in entropy between these two pressures:

ΔS = - ∫ (dG/dp)_T dp from p1 to p2

Since we know from the problem statement that (dG/dp)_T = V, we can substitute this in and integrate to get:

ΔS = - ∫ V dp from p1 to p2

Finally, using the given relation (dV/dT)_p = ap^2, we can substitute in for V and solve the integral to get the final expression for the change in entropy:

ΔS = - a∫ p^2 dp from p1 to p2

I hope this helps! Let me know if you have any further questions.