Determing equation of state from thermodynamic coefficients

In summary, the isobaric expansion coefficient and isothermal compressibility have been experimentally determined to be given by the equations: $$\alpha_p = \frac{1}{T} + \frac{3a}{VT^3} \quad \kappa_T = \frac{1}{p}\left(1+\frac{a}{VT^2}\right)$$ However, when trying to determine the equation of state using these equations, there is a discrepancy in the second mixed partial derivatives. This suggests that there may be inconsistency in the experimental data, unless a value of the constant a can be found to ensure the equality of the two mixed derivatives.
  • #1

Homework Statement


The isobaric expansion coefficient and the isothermal compressibility are given by:
$$\alpha_p = (1/V)(\partial V/\partial T)_p \quad \kappa_T = -(1/V)(\partial V / \partial p)_T$$
Suppose they have experimentally been determined to be: $$ \alpha_p = \frac{1}{T} + \frac{3a}{VT^3} \quad \kappa_T = \frac{1}{p}\left(1+\frac{a}{VT^2}\right)$$ with some constant a. Try to determine the equation of state.

Homework Equations


$$dV(T,p) = \frac{\partial V}{\partial T}dT + \frac{\partial V}{\partial p}dp$$
$$\frac{\partial^2 F}{\partial V^2} = \frac{1}{\kappa_T V}$$ (Not sure if the last one is useful)

The Attempt at a Solution


I calculated the free energy by plugging in ##\kappa_T## and then solving the equation for ##F## giving: ##F(V,p,T) = pV ln(a+T^2V)+\left(\frac{ap}{T^2}\right)ln(a+T^2V)+c_2V+c_1##
Using the first two equations for ##\alpha_p## and ##\kappa_T## I got $$V(T,p)=\frac{-a~ln~p}{(1+ln~p)T^2}+c,\quad V(T)=\frac{-3a}{2T^2(1-ln~T)}+c$$ by integrating over ##T## for ##\alpha_p## and over ##T## for ##\kappa_T##.

But this approach doesn't seem to make much sense since the two equations for V differ.
 
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  • #2
I'm a bit curious about the source of this problem. Could you share the source, please?
 
  • #3
Useful nucleus said:
I'm a bit curious about the source of this problem. Could you share the source, please?

I don't have the source but I attached a screenshot of the problem sheet as it was handed out.

unknown.png
 

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  • #4
Thank you. I'm a bit confused about how your calculation of F leads to the expressions you wrote for V. But the first relevant equation you wrote is very useful because you can replace ∂V/∂P by -VκT and replace ∂V/∂T by Vα. Now the question in the problem sheet asks about something first before trying to evaluate V(T,P) which is whether these experimentally determined expressions make sense. One needs to recall a condition on the second derivatives of V(T,P) (or any thermodynamic state function)

The second part of the question, evaluating V(T,P) should straightforward integration of the partial differential expression you wrote:

AwesomeTrains said:
$$dV(T,p) = \frac{\partial V}{\partial T}dT + \frac{\partial V}{\partial p}dp$$
 
  • #5
Useful nucleus said:
Thank you. I'm a bit confused about how your calculation of F leads to the expressions you wrote for V. But the first relevant equation you wrote is very useful because you can replace ∂V/∂P by -VκT and replace ∂V/∂T by Vα. Now the question in the problem sheet asks about something first before trying to evaluate V(T,P) which is whether these experimentally determined expressions make sense. One needs to recall a condition on the second derivatives of V(T,P) (or any thermodynamic state function)

The second part of the question, evaluating V(T,P) should straightforward integration of the partial differential expression you wrote:

My calculation for ##F## didn't really seem to go anywhere, added it to the question just in case it could have been useful.

##dV(T,p) = \frac{\partial V}{\partial T}dT + \frac{\partial V}{\partial p}dp \implies \int \frac{dv}{v} = -\int \kappa_T dp + \int \alpha_p dT \implies## $$ln~V = -\left(1+\frac{a}{VT^2}\right)ln~p + ln~T-\frac{3a}{2VT^2}+c$$ but I can't seem to solve this for V, does this mean that they really don't make sense as you said?
 
  • #6
You cannot integrate this way, in the presence of V(T,P) in the right hand side. You need to have T,P only in the right hand side.

To check the validity of the expressions you should invoke the symmetry of the mixed second derivatives. Again when you calculate the second derivative make sure that you treat V as a function of T,P.
 
  • #7
Useful nucleus said:
You cannot integrate this way, in the presence of V(T,P) in the right hand side. You need to have T,P only in the right hand side.

To check the validity of the expressions you should invoke the symmetry of the mixed second derivatives. Again when you calculate the second derivative make sure that you treat V as a function of T,P.

I cannot seem to get the differential fiddled into parts like: ##f(V)dV = g(T)dT + h(p)dp## the best I could do is something like: $$dV+V\left(\frac{dp}{p}-\frac{dT}{T}\right) = \frac{-adp}{T^2}+\frac{3adT}{T^3}$$ or $$\frac{1}{V}\left(dV+\frac{adp}{T^2}-\frac{3adT}{T^3}\right)= \frac{-dp}{p}+\frac{dT}{T}$$

For the second mixed partial derivatives I got: $$\frac{\partial^2 V}{\partial p \partial T} = \frac{\partial \alpha_p V}{\partial p} = 0 + \alpha_p\frac{\partial V}{\partial p} = -\alpha_p V\kappa_T$$ and $$ \frac{\partial^2 V}{\partial T \partial p} = -\frac{\partial V \kappa_T}{\partial T} = -V\alpha_p\kappa_T + \frac{2a}{pT^3} $$

which means they differ by ##\frac{2a}{pT^3}##.
In the lecture we just always assume they are equal, e.g the order doesn't matter. But does this mean then that there is no equation of state?
 
  • #8
If there is no value for the constant a that ensures the equality of the two mixed derivative, then it looks to me that there is inconsistency in the experimental data. Note that a = 0, leads to the equation of state of the ideal gas as you can check by integration of dV(T,P).
 
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1. How do you determine the equation of state from thermodynamic coefficients?

To determine the equation of state from thermodynamic coefficients, you must use the fundamental thermodynamic relation which relates the equation of state to the thermodynamic coefficients. This relation can be obtained by manipulating the first and second laws of thermodynamics.

2. What are the thermodynamic coefficients?

Thermodynamic coefficients are quantities that describe the behavior of a thermodynamic system. These coefficients include the heat capacity, compressibility, thermal expansion coefficient, and others.

3. Why is it important to determine the equation of state from thermodynamic coefficients?

The equation of state is a fundamental relationship that describes the behavior of a thermodynamic system. By determining the equation of state from thermodynamic coefficients, we can understand the properties and behavior of a system, which is crucial for many applications in science and engineering.

4. What experimental methods can be used to determine thermodynamic coefficients?

There are several experimental methods that can be used to determine thermodynamic coefficients, such as calorimetry, dilatometry, and pressure-volume-temperature measurements. These methods involve measuring the response of a system to changes in temperature, pressure, or volume, and using this data to calculate the coefficients.

5. Can the equation of state be determined for all thermodynamic systems?

The equation of state can be determined for most thermodynamic systems, but there are some exceptions. For example, in systems with phase transitions or chemical reactions, the equation of state may not be well-defined or may change depending on the conditions. In these cases, other approaches must be used to describe the behavior of the system.

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