• nobosity
In summary: So in this equation, the argument of the logarithm should be 1/T- To, which would be the same as saying x = a*log(1/T-To). But in the equation that you were given, it is x = a*log(1+\beta*T-To).
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.

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.

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

TSny said:
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.

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

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.

Chestermiller said:
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...

nobosity said:
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
Chestermiller said:
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!

Chestermiller said:
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!

nobosity said:
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- To. The argument of a logarithm is generally dimensionless. Also, the equation for S is undefined at the reference state where T = To. It looks to me that the argument of the logarithm should maybe be T/To. The the equation for S will yield S = So at the reference state.

## 1. What is the definition of a gradient in thermodynamics?

A gradient in thermodynamics refers to the rate of change of a certain property, such as temperature or pressure, in a given system. It is represented by a slope on a graph or diagram, and can be used to determine the direction and magnitude of changes in the system.

## 2. How is the gradient of an adiabat represented on a PV diagram?

The gradient of an adiabat is represented by a curve on a PV diagram. The curve shows the relationship between pressure and volume at a constant rate of change of temperature, indicating the direction and magnitude of change in the system.

## 3. What factors affect the gradient of an adiabat in a PV diagram?

The gradient of an adiabat is affected by the specific heat capacity, the amount of gas present, and the work done on or by the system. These factors can change the temperature, pressure, and volume of the gas, thus affecting the slope of the adiabat curve.

## 4. How does the gradient of an adiabat differ from that of an isotherm in a PV diagram?

The gradient of an adiabat is steeper than that of an isotherm, indicating a higher rate of change in temperature for the same change in pressure. This is because an adiabatic process does not allow for the transfer of heat, while an isothermal process keeps the temperature constant through heat exchange.

## 5. Can the gradient of an adiabat ever be negative in a PV diagram?

No, the gradient of an adiabat can never be negative. This is because an adiabatic process follows the first law of thermodynamics, which states that energy cannot be created or destroyed. Therefore, the temperature and pressure of the system will always increase or decrease in the same direction, resulting in a positive gradient.

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