# Heat Capacity of a Van Der Waals' gas in an Open system

• Astrocyte
In summary, I'm so embarrassed when I meet a problem that look like easy but actually hard. But this problem also.
Astrocyte
Homework Statement
Find Cp-Cv for Van Der Waals' Gas
Relevant Equations
Grand Partition Function, Grand potential.
In our class, we're using Wassermann's Thermal physics as textbook.
I always try to solve all question which included in Text book.
But sometime when I meet a problem that look like easy but actually hard, I'm so embarrassed.
This problem do also.

First, in the textbook grand potential for van der Waals' gas is

Next, I calculated routinely, but it's so complex.
The internal energy come from previous problem. I guess it's right.

From definition of each Heat capacities. And, Entropy from Grand potential Omega.
I used previous calculation on average particle number <N>.
<N>=exp( \beta (\mu + <N>a/V))*n_Q*V
Also, n_Q is quantum concentration, which come from Ideal gas partition function, and define (m/(2*\pi*\hbar^2*\beta)^3/2
So, n_Q also depends on \beta or Temperature T.

Entropy is so complex.

And I given up, because it's so complex for calculating by hand.
Where am I wrong?

Well, one thing you're doing wrong is assuming N is a function of T.

Astrocyte
To find where i wrong, I tried to solve isobaric heat capacity in ideal gas.
So, I got entropy. it's the Sackur-Tetrode equation.
$$S=\braket{N} k[\frac{5}{2}-\beta\mu], when \;\mu \; is\; chemical\; potential.$$
it's same with
$$S=\braket{N}k[5/2-\ln{\frac{ \braket{N}}{(n_Q*V)} } ].$$
The definition of isobaric heat capacity is
$$C_{p}=T(\frac{\partial {S}}{\partial {T}})_{p},$$.
As we know, isobaric heat capacity in ideal gas is
$$C_{P}=\frac{5}{2} Nk.$$
But, I cannot prove it also...
I tried
$$C_{p}=T(\frac{\partial {S}}{\partial {T}})_{p,N}$$
It didn't work...

Are you familiar with the following equation for an arbitrary equation of state:

$$C_p-C_v=T\left(\frac{\partial V}{\partial T}\right)_P\left(\frac{\partial P}{\partial T}\right)_V$$

Astrocyte
Thanks!

I have checked all of my equation... And, I found several mistakes.

By your feedback, I tried it without derivative of N respective of T.

And I got a right Cp value. 5/2 Nk.

And also I will try the equation you gave me.

Really Really Thank you. :D

Astrocyte said:
Thanks!

I have checked all of my equation... And, I found several mistakes.

By your feedback, I tried it without derivative of N respective of T.

And I got a right Cp value. 5/2 Nk.

And also I will try the equation you gave me.

Really Really Thank you. :D
That Cp value is only for an ideal gas, right?

Astrocyte
Chestermiller said:
That Cp value is only for an ideal gas, right?

Yes. I only tried it about Cp value, because the van Der Waals gas's one is so complicated.
In addition, that Cp value come from
$$\ln{\frac{⁡⟨N⟩}{nQV}} = \ln{\frac{⁡⟨N⟩βP}{nQ⟨N⟩}} = ln⁡{\frac{βP}{n_Q}}$$

in Entropy term.
And fortunately, In my midterm, my professor didn't ask about heat capacity, and Van der Waals' gas.

## 1. What is the definition of heat capacity in a Van Der Waals' gas?

The heat capacity of a Van Der Waals' gas is a measure of the amount of heat energy required to raise the temperature of a given amount of gas by one degree.

## 2. How does the heat capacity of a Van Der Waals' gas differ from an ideal gas?

The heat capacity of a Van Der Waals' gas is typically higher than that of an ideal gas due to the attractive forces between particles in the gas, which require additional energy to overcome.

## 3. How does the heat capacity of a Van Der Waals' gas change with temperature?

The heat capacity of a Van Der Waals' gas decreases with increasing temperature, as the attractive forces between particles become weaker and require less energy to overcome.

## 4. What is the relationship between heat capacity and pressure in a Van Der Waals' gas?

In a Van Der Waals' gas, the heat capacity at constant pressure is typically higher than at constant volume, as the gas can expand and do work against the external pressure, requiring more energy.

## 5. How does the heat capacity of a Van Der Waals' gas change with the addition of intermolecular forces?

The heat capacity of a Van Der Waals' gas increases with the addition of intermolecular forces, as these forces require additional energy to overcome and therefore contribute to the overall heat capacity of the gas.

Replies
6
Views
2K
Replies
23
Views
9K
• Thermodynamics
Replies
4
Views
338
Replies
7
Views
2K
Replies
6
Views
3K
Replies
8
Views
2K
Replies
1
Views
2K
Replies
8
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
839