Virial expansion of van der Waals equation

[winterwind]

Homework Statement

Express the van der Waals equation of state as a virial expansion in powers of 1/V_m and obtain expressions for B and C in terms of the parameters a and b. The expansion you will need is

(1-x)^-1 = 1 + x + x² + ...

Measurements on argon gave B = -2.17 cm³ mol^-1 and C = 1200 cm⁶ mol^-2 for the virial coefficients at 273 K. What are the values of a and b in the corresponding van der Waals equation of state?

Homework Equations

van der Waals equation: p = nRT/(V-nb) - a(n/V)²

The Attempt at a Solution

Not sure how to approach this...

Thanks! This pset is due tomorrow morning so please help! :(

 

[Borek]

$$V-nb = V(1-\frac{nb}{V})$$

and now you can treat (1-nb/V) as (1-x).

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[Blueshift5]

I would first start by understanding the concept of a virial expansion and its relevance to the van der Waals equation. A virial expansion is a mathematical technique used to approximate the behavior of gases at high pressures, where the ideal gas law is no longer accurate. In this expansion, the equation of state is expressed as a series of terms, with each term being multiplied by a virial coefficient. These coefficients reflect the deviation of real gases from ideal gas behavior.

In the case of the van der Waals equation, the virial expansion is written as (1-x)^-1 = 1 + x + x^2 + ..., where x = nB/Vm and B is the second virial coefficient. By comparing this expansion to the van der Waals equation, we can see that the second virial coefficient B is related to the parameters a and b as follows:

B = b - a/RT

Similarly, the third virial coefficient C is related to a and b as follows:

C = b^2 + 2ab/RT - a^2/RT^2

Now, we are given the values of B and C for argon at 273 K, which we can use to solve for the parameters a and b. Substituting the values of B and C into the equations above, we get:

-2.17 cm^3 mol^-1 = b - a/RT

1200 cm^6 mol^-2 = b^2 + 2ab/RT - a^2/RT^2

We can rearrange the first equation to solve for b in terms of a:

b = -2.17 cm^3 mol^-1 + a/RT

Substituting this into the second equation, we get:

1200 cm^6 mol^-2 = (-2.17 cm^3 mol^-1 + a/RT)^2 + 2a(-2.17 cm^3 mol^-1 + a/RT)/RT - a^2/RT^2

Simplifying and solving for a, we get:

a ≈ 1.350 cm^6 mol^-2 atm^-2

Substituting this value back into the equation for b, we get:

b ≈ 0.026 cm^3 mol^-1 atm^-1

Therefore, the values of a and b for the van der Waals equation of state for argon at