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

AI Thread Summary
The discussion revolves around the challenges of calculating heat capacity for a Van der Waals' gas using Wassermann's Thermal Physics textbook. The complexity arises from deriving entropy and internal energy, leading to confusion about the dependence of particle number on temperature. After several attempts and corrections, the correct expression for isobaric heat capacity, Cp, was found to be 5/2 Nk, applicable only to ideal gases. The participant acknowledges the difficulties with Van der Waals' gas calculations and expresses relief that the midterm did not include this topic. The conversation highlights the intricate nature of thermodynamic calculations in open systems.
Astrocyte
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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
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Next, I calculated routinely, but it's so complex.
The internal energy come from previous problem. I guess it's right.
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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.
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Entropy is so complex.
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And I given up, because it's so complex for calculating by hand.
Where am I wrong?
 
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Well, one thing you're doing wrong is assuming N is a function of T.
 
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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:

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