Find thermal capacity of a Van der Waals gas

[MatinSAR]

Homework Statement

Find thermal capacity of vander waals gas.

Relevant Equations

Tds equations.

Can anyone guide me how can I find $C_P$ and $C_V$?
There are two equations in my book which I think can help but I'm not sure if I can use them to find $C_P$ and $C_V$ of a vander waals gas.

Spoiler: Equation 1

Spoiler: Equation 2

I know equation of state of this gas.

 

[MatinSAR]

From first equation I find out that : $$C_P-C_V = \dfrac {R}{1 - \dfrac {2a(V-b)^2}{RTV^3}}$$
But I am not sure if I can use 2nd equation to find another relation between these to ...

I've checked some books but I didn't find anything about heat capacities of Van der Waals gas. Does anyone knows a book or a source that discuss this thing ?

 

[pines-demon]

You cannot calculate
$$\beta=\left(\frac{\partial P}{\partial V}\right)_S$$
because you do not have enough information about $S$, the equation of state does not suffice here.

Imagine that you have $PV=nRT$ instead, how do you calculate $\beta$? Well, you can't.

 

[MatinSAR]

You cannot calculate
$$\beta=\left(\frac{\partial P}{\partial V}\right)_S$$
because you do not have enough information about $S$, the equation of state does not suffice here.

Imagine that you have $PV=nRT$ instead, how do you calculate $\beta$? Well, you can't.

You are right. My professor told us that this is a hard question ... So I guess I cannot solve it using provided equations in my book. Thank you.

 

[MatinSAR]

You cannot calculate
$$\beta=\left(\frac{\partial P}{\partial V}\right)_S$$
because you do not have enough information about $S$, the equation of state does not suffice here.

Imagine that you have $PV=nRT$ instead, how do you calculate $\beta$? Well, you can't.

As you mentioned the professor told that I cannot use those equations to find thermal capacities ...I've checked many books but I did not find anything about thermal capacity of a Van der Waals gas. Does anyone know an article or source related to this topic? Thanks.

 

[MatinSAR]

My current idea:
Find internal energy of Van der Waals gas:
$$dU= \frac {\partial U} {\partial T}dT + \frac {\partial U} {\partial V}dV$$$$U(T,V)=\int C_V dT-\dfrac {n^2a}{V}$$
Then finding heat capacities using: $$C_V= \frac {\partial U} {\partial T}$$$$C_P= C_V + ([\frac {\partial U} {\partial V}]_T +P)[\frac {\partial V} {\partial T}]_P$$

The problem us that it gives $C_V=C_V$ and $C_P=C_V + sth$ and it doesn't help.

 

[pines-demon]

My current idea:
Find internal energy of Van der Waals gas:
$$dU= \frac {\partial U} {\partial T}dT + \frac {\partial U} {\partial V}dV$$$$U(T,V)=\int C_V dT-\dfrac {n^2a}{V}$$
Then finding heat capacities using: $$C_V= \frac {\partial U} {\partial T}$$$$C_P= C_V + ([\frac {\partial U} {\partial V}]_T +P)[\frac {\partial V} {\partial T}]_P$$

Again if you do not define $U$ then you cannot calculate $C_V$ and $C_p$ independently. Please try first with an ideal gas with $PV=nRT$, you can't. Unless you postulate that $U=\frac32 nRT$, but you will get a different answer if $U=\frac52 nRT$. The same goes for the van der Waals gas, if you do not specify the $U$ or $S$ you cannot proceed with the calculation. The equation of state is not enough.

 

[MatinSAR]

Again if you do not define $U$ then you cannot calculate $C_V$ and $C_p$ independently. Please try first with an ideal gas with $PV=nRT$, you can't. Unless you postulate that $U=3/2 nRT$, but you will get a different answer if $U=5/2 nRT$. The same goes for the van der Waals gas, if you do not specify the $U$ or $S$ you cannot proceed with the calculation. The equation of state is not enough.

So at first I need to find something like $U=3/2nRT$ for Van der Valse gas. Yes? And find heat capacities related to this $U$ ...

 

[MatinSAR]

if you do not specify the U or S you cannot proceed with the calculation. The equation of state is not enough.

Now, I understand that why equation of state is not enough. Thanks ...

 

[pines-demon]

So at first I need to find something like $U=3/2nRT$ for Van der Valse gas. Yes?

Exactly. Either define $U$ or an equation that relies $U$ or $S$ to the other variables.

 

[Chestermiller]

Actually, for a van Der Waals gas, $C_v$ is independent of volume and depends only on T. This means that it is the same as in the ideal gas limit: $$C_v=C_v^{IG}$$If does not depend on volume.

 

[pines-demon]

Actually, for a van Der Waals gas, $C_v$ is independent of volume and depends only on T. This means that it is the same as in the ideal gas limit: $$C_v=C_v^{IG}$$If does not depend on volume.

Monoatomic gas? Diatomic gas? Still some additional detail about the gas has to be provided. The three equations provided are not enough.

 

[MatinSAR]

Actually, for a van Der Waals gas, $C_v$ is independent of volume and depends only on T. This means that it is the same as in the ideal gas limit: $$C_v=C_v^{IG}$$If does not depend on volume.

Thanks. When $C_V$ doesn't depend on volume, Can I use following equation to find $C_P$?

Because $C_V=C^{IG}_{V}$ is knwon ...

Monoatomic gas? Diatomic gas? Still some additional detail about the gas has to be provided. The three equations provided are not enough.

This details should be provided by the professor(problem setter) ? Or should be found by me?

 

[pines-demon]

Thanks. When $C_V$ doesn't depend on volume, Can I use following equation to find $C_P$?
View attachment 339584

Because $C_V=C^{IG}_{V}$ is knwon ...

This details should be provided by the professor(problem setter) ? Or should be found by me?

If you have $C_V$ you can get $C_p$, and viceversa, from Mayer's relation (your eq1). But you need the internal energy $U$ to have some details about your gas.

Yes, somebody has to give you that. Maybe in your book there is a specific "internal energy of the van der Waals gas" somewhere?

 

[MatinSAR]

Yes, somebody has to give you that. Maybe in your book there is a specific "internal energy of the van der Waals gas" somewhere?

The professor just told us find heat capacities if you can ... He did not provide any other information ;)
I remember that he said it is difficult. What's your opinion? Isn't there any other source except my book to look for internal energy of this gas?

 

[Chestermiller]

Monoatomic gas? Diatomic gas? Still some additional detail about the gas has to be provided. The three equations provided are not enough.

The ideal gas behavior is available in the literature, regardless of how many atoms the molecule has.

 

[MatinSAR]

@Chestermiller @pines-demon
I've found a picture of my professor note's in my phone. He said that in constant temperature $C_V$ does not depend on $V$ for Van der Waals gas. So I want to use this assumption to solve this example.

This means that it is the same as in the ideal gas limit:

Can't I prove this somehow? I cannot think of any proof.

 

[Chestermiller]

Thanks. When $C_V$ doesn't depend on volume, Can I use following equation to find $C_P$?
View attachment 339584
Yes.

Because $C_V=C^{IG}_{V}$ is knwon ...

This details should be provided by the professor(problem setter) ? Or should be found by me?

You have $$du=C_vdT+\frac{a}{v^2}dv$$So, $$\frac{\partial^2 u}{\partial T\partial v}=\frac{\partial C_v}{\partial v}=\frac{\partial (a/v^2)}{\partial T}=0$$

 

[MatinSAR]

You have $$du=C_vdT+\frac{a}{v^2}dv$$So, $$\frac{\partial^2 u}{\partial T\partial v}=\frac{\partial C_v}{\partial v}=\frac{\partial (a/v^2)}{\partial T}=0$$

A clever idea to prove!
Thanks for your help @Chestermiller
But I cannot understand sth else:

This means that it is the same as in the ideal gas limit:

Can't I prove it somehow?

 

[pines-demon]

The ideal gas behavior is available in the literature, regardless of how many atoms the molecule has.

But $C_V$ does depend on the number of atoms per molecule, that's why you cannot get a specific value for $C_V$ of an ideal gas just from the equation of state you need some information about $U$

 

[Chestermiller]

But $C_V$ does depend on the number of atoms per molecule, that's why you cannot get a specific value for $C_V$ of an ideal gas just from the equation of state you need some information about $U$

Who cares if it can look it up in a handbook.

 

[pines-demon]

Who cares if it can look it up in a handbook.

The question is clear: from the three equations above (two relations between $C_V$ and $C_P$) and the equation of state, determine $C_V$ and $C_P$. The answer to that is no, it is not possible. The problem is incomplete.

You may propose that $C_V$ is the same as that of an ideal gas (but then you only need two of the equations above). You may also propose that it is specifically a diatomic gas. But it could be something different. Both of these additional formulas are just that, suggestions. The real answer would depend on the specifics of the problem, there is no unique answer.

 

[Chestermiller]

The question is clear: from the three equations above (two relations between $C_V$ and $C_P$) and the equation of state, determine $C_V$ and $C_P$. The answer to that is no, it is not possible. The problem is incomplete.

You may propose that $C_V$ is the same as that of an ideal gas (but then you only need two of the equations above). You may also propose that it is specifically a diatomic gas. But it could be something different. Both of these additional formulas are just that, suggestions. The real answer would depend on the specifics of the problem, there is no unique answer.

I have no idea what you are talking about. Have you ever solved a real thermo problem.

 

[MatinSAR]

The question is clear: from the three equations above (two relations between $C_V$ and $C_P$) and the equation of state, determine $C_V$ and $C_P$. The answer to that is no, it is not possible. The problem is incomplete.

You may propose that $C_V$ is the same as that of an ideal gas (but then you only need two of the equations above). You may also propose that it is specifically a diatomic gas. But it could be something different. Both of these additional formulas are just that, suggestions. The real answer would depend on the specifics of the problem, there is no unique answer.

I think I can assume these because the professor didn't gave us any details ...

 

[pines-demon]

I think I can assume these because the professor didn't gave us any details ...

I think that if you want to proceed you can do one the following:

• Derive everything in terms of $\beta$ and leave it at that.
• Copy the internal energy (or entropy) of a van der Waals gas from another book (there are several but you can choose the first you find)
• Choose to postulate $C_V$ equal to that of the ideal gas and just leave a constant for all the different cases (monoatomic, diatomic, and so on)

My problem with the last one is that the objective is to calculate $C_V$, but you would be already using it as assumption and you won't be needing all the equations that you suggested at first.

 

[MatinSAR]

I think that if you want to proceed you can do one the following:

• Derive everything in terms of $\beta$ and leave it at that.
• Copy the internal energy (or entropy) of a van der Waals gas from another book (there are several but you can choose the first you find)
• Choose to postulate $C_V$ equal to that of the ideal gas and just leave a constant for all the different cases (monoatomic, diatomic, and so on)

My problem with the last one is that the objective is to calculate $C_V$, but you would be already using it as assumption and you won't be needing all the equations that you suggested at first.

What if I prove that $C_V$ does not depend on $V$ using post #18 then I claim that because of this $C_V=C_{V}^{IG}$. Then I assume that molecules have 3 degrees of freedom so for Ideal gas we have $C_{V}^{IG}=\dfrac 3 2 R$ . Now I know what is $C_V$ so I can find $C_P$ using Mayer's relation.

• Choose to postulate $C_V$ equal to that of the ideal gas and just leave a constant for all the different cases (monoatomic, diatomic, and so on)

My problem with the last one is that the objective is to calculate $C_V$, but you would be already using it as assumption and you won't be needing all the equations that you suggested at first.

I want to use this assumption because the professor told us that $C_V$ doesn't depend on $V$ here ...

 

[pines-demon]

What if I prove that $C_V$ does not depend on $V$ using post #18 then I claim that because of this $C_V=C_{V}^{IG}$. Then I assume that molecules have 3 degrees of freedom so for Ideal gas we have $C_{V}^{IG}=\dfrac 3 2 R$ . Now I know what is $C_V$ so I can find $C_P$ using Mayer's relation.

I want to use this assumption because the professor told us that $C_V$ doesn't depend on $V$ here ...

If that is what the professor told you then go for it. If you want to assume monoatomic gas go for it.

However how is post#18 helpful again? How does that prove that $C_V$ does not depend on $V$? You are assuming that, the argument is circular.

Once you have assumed that $C_V=\frac32 nR$ then you use Mayer's relation and that's it.

Edit: Actually from the equation of state you can prove that $C_V$ does not depend on $V$. See https://scienceworld.wolfram.com/physics/vanderWaalsGas.html but as you see you get to equation of post#18 and you still have no indication of the full form of $C_V$

 

[MatinSAR]

If that is what the professor told you then go for it. If you want to assume monoatomic gas go for it.

Professor's note:

However how is post#18 helpful again? Also shouldn't P be more complicated?

Using that post I can prove: $$[\frac {\partial C_V} {\partial V}]_T=0$$

Once you have assumed that $C_V=\frac32 nR$ then you use Mayer's relation and that's it.

So I'm going to assume $C_V=\frac 3 2 nR$ then using Mayer's relation I can find $C_P$:
$$C_P=\frac 3 2 nR + \dfrac {R}{1 - \dfrac {2a(V-b)^2}{RTV^3}}$$

Thanks for your help and time @pines-demon and @Chestermiller.

 

[pines-demon]

Using that post I can prove: $$[\frac {\partial C_V} {\partial V}]_T=0$$

See my edit of post#27

 

[MatinSAR]

See my edit of post#27

So for showing that $[\frac {\partial C_V} {\partial V}]_T=0$ I can use the link you've sahred ...
Do you see any other problem in my final answer?