Relation between heat capacities and van der Waals equation

In summary: I think I'm getting somewhere.In summary, we can find an expression for c_p - c_v for a van-der-waals gas by using the van der waals equation of state and the reciprocal rule for partial derivatives. This involves the variables P, V, T, a, and b.
  • #1
patrickmoloney
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Homework Statement


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

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

Homework Equations

The Attempt at a Solution


Basically I've proved

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

([itex]E[/itex] and [itex]V[/itex] are the energy and volume of one mole).

The question states that

[tex]p + \Bigg{(}\dfrac{\partial E}{\partial V}\Bigg{)}_T = T \Bigg{(}\dfrac{\partial p}{\partial T}\Bigg{)}[/tex]

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 [itex]p[/itex] which gives

[tex]p = \dfrac{RT}{(V-b)}-\dfrac{a}{V^2}[/tex]

Then

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

Which can be substituted into the equation [itex]c_p - c_v[/itex] but rearranging the van der waals formula for [itex]V[/itex] 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
 
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  • #2
Why can't you just differentiate the left side of the equation by the product rule?
 
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  • #3
Chestermiller said:
Why can't you just differentiate the left side of the equation by the product rule?
The van der waals equation?
 
  • #4
patrickmoloney said:
The van der waals equation?
Sure.
 
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  • #5
Chestermiller said:
Sure.
Differentiating the LHS with respect to [itex]T[/itex] and keeping the volume constant I get

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

Substituting that into my desired equation I get

[tex]c_p - c_v = \Bigg{[}T\Bigg{(}\dfrac{R}{V-b}\Bigg{)}\Bigg{]}\Bigg{(}\dfrac{\partial V}{\partial T}\Bigg{)}_p[/tex]
 
  • #6
You might be able to avoid solving for V by using the "reciprocal rule" for partial derivatives $$\left( \frac{\partial x}{\partial y} \right)_z = \frac{1}{\left( \frac{\partial y}{\partial x} \right)_z}$$
 
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  • #7
TSny said:
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 [tex]\Bigg{(}\dfrac{\partial T}{\partial V}\Bigg{)}_p[/tex] to solve the VDW equation?
 
  • #8
patrickmoloney said:
Would that mean I could do [tex]\Bigg{(}\dfrac{\partial T}{\partial V}\Bigg{)}_p[/tex] to solve the VDW equation?
Yes, but I'm not sure what you mean when you say "solve the VDW equation".
 
  • #9
TSny said:
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 [tex]\Bigg{(}\dfrac{\partial p}{\partial T}\Bigg{)}_V[/tex]
 
  • #10
patrickmoloney said:
Van der waals equation of state. Like I did with [tex]\Bigg{(}\dfrac{\partial p}{\partial T}\Bigg{)}_V[/tex]
Yes, you use the VdW equation of state to find an expression for the partial derivative.
 
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  • #11
I got [tex]\Bigg{(}\dfrac{\partial T}{\partial V}\Bigg{)}_p = -\Bigg{(}\dfrac{2ab}{V^3} +\dfrac{a}{V^2}- P\Bigg{)} \cdot \dfrac{1}{R}[/tex]

Then inverting this equation yields

[tex]\Bigg{(}\dfrac{\partial V}{\partial T}\Bigg{)}_p = -\Bigg{(}\dfrac{V^3}{2ab} +\dfrac{V^2}{a}-\dfrac{1}{P}\Bigg{)} \cdot R[/tex]
 
  • #12
patrickmoloney said:
I got [tex]\Bigg{(}\dfrac{\partial T}{\partial V}\Bigg{)}_p = -\Bigg{(}\dfrac{2ab}{V^3} +\dfrac{a}{V^2}- P\Bigg{)} \cdot \dfrac{1}{R}[/tex]

Then inverting this equation yields

[tex]\Bigg{(}\dfrac{\partial V}{\partial T}\Bigg{)}_p = -\Bigg{(}\dfrac{V^3}{2ab} +\dfrac{V^2}{a}-\dfrac{1}{P}\Bigg{)} \cdot R[/tex]
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}}##.
 
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  • #13
TSny said:
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}}##.
[tex]\Bigg{(}\dfrac{\partial T}{\partial V}\Bigg{)}_p = -\Bigg{(}\dfrac{2ab}{V^3} +\dfrac{a}{V^2}- P\Bigg{)} \cdot \dfrac{1}{R}[/tex]

Letting

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

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

[tex]\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}[/tex]

But this still looks hideous even though I have everything in terms of [itex]V, a, b, R[/itex] and [itex]T[/itex]
 
  • #14
patrickmoloney said:
[tex]\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}[/tex]

But this still looks hideous even though I have everything in terms of [itex]V, a, b, R[/itex] and [itex]T[/itex]
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.
 
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  • #15
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.
 
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  • #16
TSny said:
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 ]
TSny said:
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

[tex]\dfrac{T^2(2a-1)}{V^3(V-b)}[/tex] 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.
 
  • #17
patrickmoloney said:
in the end I got

[tex]\dfrac{T^2(2a-1)}{V^3(V-b)}[/tex] which looks respectable.
I don't think this is correct.

You have
[tex]c_p - c_v = \Bigg{[}T\Bigg{(}\dfrac{R}{V-b}\Bigg{)}\Bigg{]}\Bigg{(}\dfrac{\partial V}{\partial T}\Bigg{)}_p[/tex]
This can be written
[tex]c_p - c_v = \frac{T\Bigg{(}\dfrac{R}{V-b}\Bigg{)}}{\Bigg{(}\dfrac{\partial T}{\partial V}\Bigg{)}_p}[/tex]
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.
 
  • #18
TSny said:
I don't think this is correct.

You have
[tex]c_p - c_v = \Bigg{[}T\Bigg{(}\dfrac{R}{V-b}\Bigg{)}\Bigg{]}\Bigg{(}\dfrac{\partial V}{\partial T}\Bigg{)}_p[/tex]
This can be written
[tex]c_p - c_v = \frac{T\Bigg{(}\dfrac{R}{V-b}\Bigg{)}}{\Bigg{(}\dfrac{\partial T}{\partial V}\Bigg{)}_p}[/tex]
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

[tex]\dfrac{RV(V-b)}{4a(\frac{b}{V}-1)}[/tex]
I've just realized how bad I am at algebra.
 
  • #19
I got $$\frac{R}{1-\frac{2a}{V^3}\frac{(V-b)^2}{RT}}$$
 
  • #20
Chestermiller said:
I got $$\frac{R}{1-\frac{2a}{V^3}\frac{(V-b)^2}{RT}}$$
I get the same result.
 
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  • #21
I got [tex]\dfrac{RT}{1-\frac{2a}{V^3}\frac{(V-b)^2}{RT}}[/tex]. But I'm happy enough that I know how to solve the problem should it come up. Thanks guys.
 
  • #22
patrickmoloney said:
I got [tex]\dfrac{RT}{1-\frac{2a}{V^3}\frac{(V-b)^2}{RT}}[/tex]. 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.
 
  • #23
Chestermiller said:
There should not be a T in the numerator.
I got it from [tex]T \Bigg{(}\dfrac{\partial p}{\partial T}\Bigg{)}_V[/tex]

I couldn't get an [itex]R[/itex] on its own
 
  • #24
patrickmoloney said:
I got it from [tex]T \Bigg{(}\dfrac{\partial p}{\partial T}\Bigg{)}_V[/tex]

I couldn't get an [itex]R[/itex] 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.
 

1. What is the relationship between heat capacities and the van der Waals equation?

The van der Waals equation is a mathematical representation of the behavior of real gases, taking into account intermolecular forces and molecular volume. It describes how the pressure, volume, and temperature of a gas are related. The heat capacities of a gas, on the other hand, refer to the amount of heat energy needed to raise the temperature of the gas. The van der Waals equation includes terms for the heat capacities of gases, allowing for a more accurate prediction of their behavior.

2. How does the van der Waals equation account for intermolecular forces?

The van der Waals equation includes a term for the attractive forces between molecules, known as the van der Waals force. This force is responsible for the deviation of real gases from ideal gas behavior, and the equation takes it into account by subtracting it from the pressure term.

3. Can the van der Waals equation accurately predict the behavior of real gases?

While the van der Waals equation is more accurate than the ideal gas law, it is still an approximation and cannot account for all factors that affect gas behavior. It does not take into account the size and shape of molecules, as well as the variation of intermolecular forces with temperature and pressure. However, it is a useful tool for predicting the behavior of real gases under certain conditions.

4. What is the significance of the heat capacity terms in the van der Waals equation?

The heat capacity terms in the van der Waals equation allow for a more accurate prediction of gas behavior, particularly at high pressures and low temperatures where intermolecular forces play a significant role. These terms take into account the energy needed to overcome intermolecular forces and the effect of molecular volume on the behavior of real gases.

5. How is the van der Waals equation related to other equations of state?

The van der Waals equation is one of many equations of state that describe the behavior of gases. It is an improvement over the ideal gas law, which does not take into account intermolecular forces. Other equations of state include the Peng-Robinson equation and the Redlich-Kwong equation, which also incorporate terms for intermolecular forces and molecular volume. Each equation has its own strengths and limitations, and the choice of which one to use depends on the specific conditions and properties of the gas being studied.

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