Calculating Thermal Conductivity & Mean Free Path of Argon

[mitch_1211]

Homework Statement

obtain an expression for the thermal conductivity of a gas at ordinary pressures. The thermal conductivity of Argon (atomic weight 40) at STP (standard temp and pressure) is 1.6e-2 W/mK. Use this to calculate the mean free path in Argon at STP. Express the mean free path in terms of an effective atomic radius for collisions and find the value of this radius.

Solid argon has a close-packed cubic structure in which if the atoms are regraded as hard spheres, 0.74 of the volume of the structure is filled. The density of solid argon is 1.6e3 km/m^3. Compare the effective atomic radius obtained from this information with your effective collision radius. Comment on your result.

Homework Equations

$$\kappa = \frac{1}{3}C_{v}\lambda$$
$$\lambda=\frac{k_{b}T}{\sqrt{2}\pi d^{2}p}$$
$$\lambda$$=mean free path
d = diametre
p = pressure
$$\kappa$$=thermal conductivity

The Attempt at a Solution

I know i can get the mean free path from $$\kappa = \frac{1}{3}C_{v}\lambda$$ as thermal conductivity is given. but what do i do about the Cv term?

I think i then calculate d from $$\lambda=\frac{k_{b}T}{\sqrt{2}\pi d^{2}p}$$ and this is my "effective atomic radius for collisions"

Do i then simply multiply 0.74 by the given density to work out the "effective atomic radius"?

mitch

 

[anathema]

ellbravo, I am a scientist and I would be happy to help you with your question.

To obtain an expression for the thermal conductivity of a gas at ordinary pressures, we can use the following equation:

\kappa = \frac{1}{3}C_{v}\lambda

Where:
\kappa = thermal conductivity
Cv = specific heat capacity at constant volume
\lambda = mean free path

To calculate the mean free path in Argon at STP, we can use the following equation:

\lambda=\frac{k_{b}T}{\sqrt{2}\pi d^{2}p}

Where:
\lambda = mean free path
k_b = Boltzmann constant
T = temperature
d = diameter
p = pressure

We are given the thermal conductivity of Argon at STP, which is 1.6e-2 W/mK. We can use this value in the first equation to solve for Cv:

1.6e-2 = \frac{1}{3}C_{v}\lambda

Cv = \frac{1.6e-2}{\frac{1}{3}\lambda}

Now, we can use the value of Cv in the second equation to solve for the mean free path:

\lambda=\frac{k_{b}T}{\sqrt{2}\pi d^{2}p}

\lambda=\frac{k_{b}T}{\sqrt{2}\pi d^{2}p}

Substituting the value of Cv, we get:

\lambda=\frac{k_{b}T}{\sqrt{2}\pi d^{2}p}\frac{3}{1.6e-2}\lambda

Simplifying, we get:

\lambda=\frac{1.875k_{b}T}{d^{2}p}\lambda

Now, we can solve for the diameter d:

d=\sqrt{\frac{1.875k_{b}T}{p}}

Substituting the values of k_b, T, and p for STP, we get:

d=\sqrt{\frac{1.875(1.38e-23)(273)}{(1.01325e5)}}=3.64e-10 m

This is the effective atomic radius for collisions in Argon at STP. We can compare this value with the effective atomic radius obtained from the given information about the close-packed cubic structure of solid Arg