Derive the energy equation for a van der Waal gas.

In summary, the conversation revolved around deriving the energy equation for a van der Waal gas with T and v as the independent variables. The person started with a random approach but then used the canonical partition sum and the Helmholtz free energy to derive the internal energy. They also discussed using Maxwell relations to obtain ∂S/∂V at constant T, which can be used to determine the second term in the energy equation, which is not zero for a real gas.
  • #1
McAfee
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1

Homework Statement



Derive the energy equation for a van der Waal gas when T and v are the independent variables.


Homework Equations



I'm required to start from: [itex]du=\frac{\partial u}{\partial T} dT +\frac{\partial u}{\partial v} dv[/itex]

The Attempt at a Solution



We'll when I first started this problem. I just started doing random derivatives that made sense to me but after looking it over I had no clue what I was doing.

Does anyone know what the answer I should be getting after I fully derive it?
 
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  • #2
I don't know, what's given. So I can't help to find an answer to your question. One way to derive the caloric equation of state, i.e., the internal energy [itex]U[/itex] is to use the canonical partition sum
[tex]Z=\frac{z^N}{N!}[/tex]
with the single-particle partition function
[tex]z=\frac{V^*}{\Lambda^3} \exp \left (-\frac{\phi}{2 T} \right),[/tex]
where [itex]V^*=V-2 \pi N d^3/3[/itex] is the available volume (geometrical volume minus excluded volume due to the hard-sphere model for the molecules), and [itex]\phi[/itex] is the interaction energy of the particles
[tex]\phi=\frac{N}{V} \int_d^\infty \mathrm{d} r \; U_{\text{rel}}(r) 4 \pi r^2,[/tex]
with the effective two-body potential
[tex]U_{\text{rel}}(r)=\begin{cases}
\infty & \text{for} \quad r \leq d ,\\
-\alpha (d/r)^6 & \text{for} \quad r>d.
\end{cases}
[/tex]
The Helmholtz free energy is given by
[tex]A(T,V,N)=-T \ln Z,[/tex]
and from this you can derive all thermodynamical quantities from the usual thermodynamic relations like
[tex]p=-\left (\frac{\partial A}{\partial V} \right)_{T,N}[/tex]
etc. The internal energy is given by the usual Legendre transformation
[tex]U=A+T S.[/tex]
 
  • #3
[tex]dU=TdS-PdV=T\left(\frac{∂S}{∂T}\right)dT+T\left(\frac{∂S}{∂V}\right)dV-PdV=
C_VdT+(T\left(\frac{∂S}{∂V}\right)-P)dV[/tex]
The next step is to determine ∂S/∂V at constant T. This can be obtained from a Maxwell relation, starting from the equation dA=-SdT-PdV.

S=-∂A/∂T
P=-∂A/∂V

So [tex]\frac{∂S}{∂V}=\frac{∂P}{∂T}[/tex]

Therefore,
[tex]dU=C_VdT+(T\left(\frac{∂P}{∂T}\right)-P)dV[/tex]

The second term in this equation is zero for an ideal gas, but not for a real gas. Just substitute the van der Waals equation into the second term of this equation.

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FAQ: Derive the energy equation for a van der Waal gas.

What is the van der Waals equation?

The van der Waals equation is a thermodynamic equation that describes the behavior of real gases, taking into account the intermolecular forces and the volume occupied by the gas particles. It is an improvement upon the ideal gas law, which does not account for these factors.

What is the derivation of the van der Waals equation?

The van der Waals equation can be derived from the ideal gas law by introducing two correction terms for intermolecular forces (a) and volume (b). These terms account for the attractive forces between gas particles and the finite size of the gas particles, respectively.

What is the energy equation for a van der Waals gas?

The energy equation for a van der Waals gas is given by U = Uideal + Ucorrection, where Uideal is the internal energy calculated from the ideal gas law and Ucorrection is the correction term for intermolecular forces.

How does the van der Waals equation account for real gas behavior?

The van der Waals equation accounts for real gas behavior by considering the attractive forces between gas particles and the finite size of the gas particles. These factors cause the gas to deviate from ideal gas behavior at high pressures and low temperatures.

What are the limitations of the van der Waals equation?

The van der Waals equation is not accurate for all gases, as it assumes a spherical shape for gas particles and does not account for other factors such as dipole-dipole interactions. It also becomes less accurate at high temperatures and pressures. However, it is a useful approximation for many real gases.

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