Relation between heat capacities and van der Waals equation

[patrickmoloney]

Homework Statement

Find the expression for c_p - c_v for a van-der-waals gas, with the equation of state

\Bigg{(}p+\dfrac{a}{V^2}\Bigg{)}(V-b)=RT

Homework Equations

The Attempt at a Solution

Basically I've proved

c_p - c_v = \Bigg{[} p + \Bigg{(}\dfrac{\partial E}{\partial V}\Bigg{)}_T \Bigg{]}\Bigg{(}\dfrac{\partial V}{\partial T}\Bigg{)}_p

(E and V are the energy and volume of one mole).

The question states that

p + \Bigg{(}\dfrac{\partial E}{\partial V}\Bigg{)}_T = T \Bigg{(}\dfrac{\partial p}{\partial T}\Bigg{)}

I don't need to prove this I just need to use it.

So far the only thing that I can come up with is rearranging the van der waals equation in terms of p which gives

p = \dfrac{RT}{(V-b)}-\dfrac{a}{V^2}

Then

\Bigg{(}\dfrac{\partial p}{\partial T}\Bigg{)}_V = \dfrac{R}{(V-b)}

Which can be substituted into the equation c_p - c_v but rearranging the van der waals formula for V is very tedious. Which leads to believe there is a an easier step. It's an exam question so there much be a neat way to solve this. I want to ask here rather than look up someone else's long winded solution. I'm not looking for the solution I genuinely want to be able to tackle this problem or another problem like it should it come up in the exam

 

[Chestermiller]

Why can't you just differentiate the left side of the equation by the product rule?

 

[patrickmoloney]

Why can't you just differentiate the left side of the equation by the product rule?

The van der waals equation?

 

[Chestermiller]

The van der waals equation?

Sure.

 

[patrickmoloney]

Sure.

Differentiating the LHS with respect to T and keeping the volume constant I get

\Bigg{(}\dfrac{\partial p}{\partial T}\Bigg{)}_V= \dfrac{R}{V-b}

Substituting that into my desired equation I get

c_p - c_v = \Bigg{[}T\Bigg{(}\dfrac{R}{V-b}\Bigg{)}\Bigg{]}\Bigg{(}\dfrac{\partial V}{\partial T}\Bigg{)}_p

 

[TSny]

You might be able to avoid solving for V by using the "reciprocal rule" for partial derivatives $$\left( \frac{\partial x}{\partial y} \right)_z = \frac{1}{\left( \frac{\partial y}{\partial x} \right)_z}$$

 

[patrickmoloney]

You might be able to avoid solving for V by using the "reciprocal rule" for partial derivatives $$\left( \frac{\partial x}{\partial y} \right)_z = \frac{1}{\left( \frac{\partial y}{\partial x} \right)_z}$$

Would that mean I could do \Bigg{(}\dfrac{\partial T}{\partial V}\Bigg{)}_p to solve the VDW equation?

 

[TSny]

Would that mean I could do \Bigg{(}\dfrac{\partial T}{\partial V}\Bigg{)}_p to solve the VDW equation?

Yes, but I'm not sure what you mean when you say "solve the VDW equation".

 

[patrickmoloney]

Yes, but I'm not sure what you mean when you say "solve the VDW equation".

Van der waals equation of state. Like I did with \Bigg{(}\dfrac{\partial p}{\partial T}\Bigg{)}_V

 

[TSny]

Van der waals equation of state. Like I did with \Bigg{(}\dfrac{\partial p}{\partial T}\Bigg{)}_V

Yes, you use the VdW equation of state to find an expression for the partial derivative.

 

[patrickmoloney]

I got \Bigg{(}\dfrac{\partial T}{\partial V}\Bigg{)}_p = -\Bigg{(}\dfrac{2ab}{V^3} +\dfrac{a}{V^2}- P\Bigg{)} \cdot \dfrac{1}{R}

Then inverting this equation yields

\Bigg{(}\dfrac{\partial V}{\partial T}\Bigg{)}_p = -\Bigg{(}\dfrac{V^3}{2ab} +\dfrac{V^2}{a}-\dfrac{1}{P}\Bigg{)} \cdot R

 

[TSny]

I got \Bigg{(}\dfrac{\partial T}{\partial V}\Bigg{)}_p = -\Bigg{(}\dfrac{2ab}{V^3} +\dfrac{a}{V^2}- P\Bigg{)} \cdot \dfrac{1}{R}

Then inverting this equation yields

\Bigg{(}\dfrac{\partial V}{\partial T}\Bigg{)}_p = -\Bigg{(}\dfrac{V^3}{2ab} +\dfrac{V^2}{a}-\dfrac{1}{P}\Bigg{)} \cdot R

The first equation looks correct. It involves the three variables P, V, and T. You could use the VdW equation of state to express one of these variables in terms of the other two so that you reduce the number of variables in the expression. For example, you can express P in terms of V and T if you want the final answer to be in terms of V and T.

[EDIT: Sorry. Your first equation does not involve T. But if you use the first equation as you wrote it, your final answer for the difference in specific heats will involve the three variables P, V, and T. You can use the VdW equation to eliminate one of these variables.]

You did not invert the equation correctly. The inverse of the expression $\frac{2}{3} + \frac{4}{5}$ is not $\frac{3}{2} + \frac{5}{4}$. Rather, it would be $\frac{1}{\frac{2}{3} + \frac{4}{5}}$.

 

[patrickmoloney]

The first equation looks correct. It involves the three variables P, V, and T. You could use the VdW equation of state to express one of these variables in terms of the other two so that you reduce the number of variables in the expression. For example, you can express P in terms of V and T if you want the final answer to be in terms of V and T.

[EDIT: Sorry. Your first equation does not involve T. But if you use the first equation as you wrote it, your final answer for the difference in specific heats will involve the three variables P, V, and T. You can use the VdW equation to eliminate one of these variables.]

You did not invert the equation correctly. The inverse of the expression $\frac{2}{3} + \frac{4}{5}$ is not $\frac{3}{2} + \frac{5}{4}$. Rather, it would be $\frac{1}{\frac{2}{3} + \frac{4}{5}}$.

\Bigg{(}\dfrac{\partial T}{\partial V}\Bigg{)}_p = -\Bigg{(}\dfrac{2ab}{V^3} +\dfrac{a}{V^2}- P\Bigg{)} \cdot \dfrac{1}{R}

Letting

p = \dfrac{RT}{(v-b)}-\dfrac{a}{V^2}

and substituting into my equation for \Bigg{(}\dfrac{\partial T}{\partial V}\Bigg{)}_p gives

\Bigg{(}\dfrac{\partial T}{\partial V}\Bigg{)}_p= -\Bigg{(}\dfrac{2ab}{V^3} +\dfrac{2a}{V^2}- \dfrac{RT}{(V-b)}\Bigg{)} \cdot \dfrac{1}{R}

But this still looks hideous even though I have everything in terms of V, a, b, R and T

 

[TSny]

\Bigg{(}\dfrac{\partial T}{\partial V}\Bigg{)}_p= -\Bigg{(}\dfrac{2ab}{V^3} +\dfrac{2a}{V^2}- \dfrac{RT}{(V-b)}\Bigg{)} \cdot \dfrac{1}{R}

But this still looks hideous even though I have everything in terms of V, a, b, R and T

It looks like maybe the sign of the first term is wrong. (I should have caught that earlier.)

It won't be that bad when you put it all together.

 

[TSny]

Somehow I overlooked @Chestermiller 's post #2. His suggestion is a nice way to get $\left(\frac{\partial V}{\partial T} \right)_p$ without needing to use the reciprocal rule. It's a method that is well worth keeping in mind since it can be used in many different circumstances.

At this point, you are almost done with the problem. After finishing it off, if you would like some guidance in this alternate method, let us know.

 

[patrickmoloney]

It looks like maybe the sign of the first term is wrong. (I should have caught that earlier.)

It won't be that bad when you put it all together.

I was thinking cause ]

Somehow I overlooked @Chestermiller 's post #2. His suggestion is a nice way to get $\left(\frac{\partial V}{\partial T} \right)_p$ without needing to use the reciprocal rule. It's a method that is well worth keeping in mind since it can be used in many different circumstances.

At this point, you are almost done with the problem. After finishing it off, if you would like some guidance in this alternate method, let us know.

Yeah I would love that in the end I got

\dfrac{T^2(2a-1)}{V^3(V-b)} which looks respectable. It's not a homework problem anyway but it was from an exam paper which means it could be something that could come up in the exam. Would you be able to teach the method chester was talking about?

Thanks very much for all the help. You've been more than helpful.

 

[TSny]

in the end I got

\dfrac{T^2(2a-1)}{V^3(V-b)} which looks respectable.

I don't think this is correct.

You have
c_p - c_v = \Bigg{[}T\Bigg{(}\dfrac{R}{V-b}\Bigg{)}\Bigg{]}\Bigg{(}\dfrac{\partial V}{\partial T}\Bigg{)}_p
This can be written
c_p - c_v = \frac{T\Bigg{(}\dfrac{R}{V-b}\Bigg{)}}{\Bigg{(}\dfrac{\partial T}{\partial V}\Bigg{)}_p}
After substituting for the denominator, I don't see how you would get your result.

Would you be able to teach the method chester was talking about?

Yes, certainly.

 

[patrickmoloney]

I don't think this is correct.

You have
c_p - c_v = \Bigg{[}T\Bigg{(}\dfrac{R}{V-b}\Bigg{)}\Bigg{]}\Bigg{(}\dfrac{\partial V}{\partial T}\Bigg{)}_p
This can be written
c_p - c_v = \frac{T\Bigg{(}\dfrac{R}{V-b}\Bigg{)}}{\Bigg{(}\dfrac{\partial T}{\partial V}\Bigg{)}_p}
After substituting for the denominator, I don't see how you would get your result.

Yes, certainly.

Ok I've tried it using your fraction and got

\dfrac{RV(V-b)}{4a(\frac{b}{V}-1)}
I've just realized how bad I am at algebra.

 

[Chestermiller]

I got $$\frac{R}{1-\frac{2a}{V^3}\frac{(V-b)^2}{RT}}$$

 

[TSny]

I got $$\frac{R}{1-\frac{2a}{V^3}\frac{(V-b)^2}{RT}}$$

I get the same result.

 

[patrickmoloney]

I got \dfrac{RT}{1-\frac{2a}{V^3}\frac{(V-b)^2}{RT}}. But I'm happy enough that I know how to solve the problem should it come up. Thanks guys.

 

[Chestermiller]

I got \dfrac{RT}{1-\frac{2a}{V^3}\frac{(V-b)^2}{RT}}. But I'm happy enough that I know how to solve the problem should it come up. Thanks guys.

There should not be a T in the numerator.

 

[patrickmoloney]

There should not be a T in the numerator.

I got it from T \Bigg{(}\dfrac{\partial p}{\partial T}\Bigg{)}_V

I couldn't get an R on its own

 

[Chestermiller]

I got it from T \Bigg{(}\dfrac{\partial p}{\partial T}\Bigg{)}_V

I couldn't get an R on its own

Too bad. Check the units on your expression for the difference between the heat capacities. It doesn't have the correct units.