Thermodynamics: Gradient of Adiabat in PV diagram

[nobosity]

Homework Statement

The Equation of State and the expression for the entropy for a sample of salt water is given by:

V = V_{0}(1 + \beta(T - T_{0}) - \gamma(P - P_{0})) S = S_{0} + C_{v}ln(T - T_{0}) + \frac{\beta}{\gamma}(V - V_{0})

where the subscript 0 denotes a reference state, the coefficients \beta and \gamma are constants and C_{v} is the heat capacity of the salt water at constant volume.

Derive an expression for the gradient of an adiabat in a PV diagram.

Homework Equations

Listed above.

The Attempt at a Solution

I struggle to write the attempts I've made trying to answer this question. I understand that in this case we have V(T,P) and S(T,V), the gradient will be (dP/dV) and using the fact that an adiabat occurs when there is no change in heat energy. Also aware of the fact that Cv can be written as a differential in terms of (dU/dT), which is possibly relevant.

The real issue is I have no idea of the best way to put all of this information together and find a logical pathway to answer. Do I want to get to: dP = \frac{\partial{P}}{\partial{T}}dT + \frac{\partial{P}}{\partial{V}}dV and substitute a concoction of the above information to get to the gradient?

Been banging my head against this problem for a couple of weeks, and would be very grateful for someone to point me in the right direction!

EDIT:
Probably should have included that in my explanation. I get that for an adiabatic change, dS is 0 assuming reversibility. Similarly, I got to the point where I have:

dV = \frac{\partial{V}}{\partial{T}}dT + \frac{\partial{V}}{\partial{P}}dP

and

dS = \frac{\partial{S}}{\partial{T}}dT + \frac{\partial{S}}{\partial{V}}dV = 0

However, none of these approaches seems to lead to a place where I can rearrange to get to dP/dV. Thats the part I'm a bit stuck on.

 

[TSny]

Construct dV from the first given equation and dS from the second equation. "Adiabatic" tells you something about dS.

 

[nobosity]

Construct dV from the first given equation and dS from the second equation. "Adiabatic" tells you something about dS.

Probably should have included that in my explanation. I get that for an adiabatic change, dS is 0 assuming reversibility. Similarly, I got to the point where I have:

dV = \frac{\partial{V}}{\partial{T}}dT + \frac{\partial{V}}{\partial{P}}dP

and

dS = \frac{\partial{S}}{\partial{T}}dT + \frac{\partial{S}}{\partial{V}}dV = 0

However, none of these approaches seems to lead to a place where I can rearrange to get to dP/dV. Thats the part I'm a bit stuck on.Thanks for the reply though.

 

[TSny]

Can you solve your dS equation for dT and sub into the first equation?

 

[Chestermiller]

Solve the S equation for $T-T_0$ and substitute into the V equation. Then take the derivative of P with respect to V at constant S.

 

[nobosity]

Solve the S equation for $T-T_0$ and substitute into the V equation. Then take the derivative of P with respect to V at constant S.

So solving dS gives me:

dS = \frac{dT}{T-T_{0}} + \frac{\beta}{\gamma} = 0

and subbing into dV gives:

-\frac{\gamma}{\beta}\frac{dT}{T-T_0} = \frac{\partial{V}}{\partial{T}}dT + \frac{\partial{V}}{\partial{P}}dP

But from here, and I'm probably being thick, I find myself with a lot of differentials and not a lot of directions. Calculating the partials gives:

-\frac{\gamma}{\beta}\frac{dT}{T-T_0} =\beta{V_{0}}dT - \gamma{P_{0}}dP

but then there's no Vs anywhere in sight...

 

[Chestermiller]

So solving dS gives me:

dS = \frac{dT}{T-T_{0}} + \frac{\beta}{\gamma} = 0

and subbing into dV gives:

-\frac{\gamma}{\beta}\frac{dT}{T-T_0} = \frac{\partial{V}}{\partial{T}}dT + \frac{\partial{V}}{\partial{P}}dP

But from here, and I'm probably being thick, I find myself with a lot of differentials and not a lot of directions. Calculating the partials gives:

-\frac{\gamma}{\beta}\frac{dT}{T-T_0} =\beta{V_{0}}dT - \gamma{P_{0}}dP

but then there's no Vs anywhere in sight...

Who said anything about differentiating the S equation? Not me.

 

[nobosity]

Who said anything about differentiating the S equation? Not me.

It all dawned on me at once, I was leading myself down the garden path. Thanks!

 

[nobosity]

Who said anything about differentiating the S equation? Not me.

And just for future reference for anyone who stumbles across a similar problem, I ended up with:

P = P_{0} + \frac{1}{\gamma{V}} + \frac{\beta}{\gamma{V}}e^{-C_{v}}e^{-\frac{\beta}{\gamma}(V-V_{0})}

dP/dV of that should give the slope of the adiabat!

 

[TSny]

It all dawned on me at once, I was leading myself down the garden path.

You have been given two different approaches. My approach is to use the equation for S and the fact that dS = 0 to get an expression for dT in terms of dV. Then when you take the differential of the V equation, you can express it in terms of just the differentials dV and dP. From that you can get an expression for dP/dV in terms of V, P, and T. But you can eliminate T by using the V equation.

The equation that you were given for S doesn't look correct to me. You have a logarithm of the dimensional quantity T- T_o. The argument of a logarithm is generally dimensionless. Also, the equation for S is undefined at the reference state where T = T_o. It looks to me that the argument of the logarithm should maybe be T/T_o. The the equation for S will yield S = S_o at the reference state.