Taking the derivative of the van der waals equation

In summary, the student is trying to find the derivative of temperature with respect to molar volume for the van der Waals equation. They are confused about how the result was reached and mention using the product rule for the first term. They are also unsure about where the term [R/(Vm-b)](dT/dVm) came from.
  • #1
Razael
32
0

Homework Statement



I'm trying to find dT/dVm for the van der waals equation. I was looking at http://courses.washington.edu/bhrchem/c456/vdw_jtc.pdf" [Broken] page, but I'm confused about how they reached their result:

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The Attempt at a Solution



I honestly have no idea what they did. dT/dVm of RT(Vm-b) is obviously -RT/(Vm-b)^2, and dT/dVm of -a/Vm^2 is 2a/Vm^3, and dT/dVm of P is 0, but I have no idea where [R/(Vm-b)](dT/dVm) came from. Am I missing something obvious here? It's been a while since calc.
 
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  • #2
The first term is the product of RT and 1/(V_m-b), so you need to use the product rule: (fg)'(x)=f'(x)g(x)+f(x)g'(x).
 

What is the van der Waals equation?

The van der Waals equation is a thermodynamic equation used to describe the behavior of real gases. It takes into account the intermolecular forces between gas molecules, which are typically ignored in the ideal gas law.

Why do we need to take the derivative of the van der Waals equation?

Taking the derivative of the van der Waals equation allows us to determine the critical point of a gas, which is the temperature and pressure at which it undergoes a phase transition from gas to liquid. It also allows us to calculate other important properties such as the compressibility factor and the heat capacity of a gas.

What is the process for taking the derivative of the van der Waals equation?

The process involves using the chain rule and product rule to differentiate the equation with respect to the variables of interest (usually temperature and pressure). The resulting derivative equation will then be solved for the critical point, or other desired properties.

Can the van der Waals equation be used for all gases?

No, the van der Waals equation is only applicable to real gases, which have non-negligible intermolecular forces. Ideal gases, which have negligible intermolecular forces, can be described using the simpler ideal gas law.

What are the limitations of the van der Waals equation?

The van der Waals equation is not accurate for gases at high pressures and low temperatures, as it does not account for the volume occupied by the gas molecules themselves. It also does not take into account other factors such as molecular shape and the effects of temperature on intermolecular forces.

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