# Calculating Thermal Conductivity & Mean Free Path of Argon

• mitch_1211
In summary: According to this information, the density of solid Argon is 1.6e3 kg/m^3 and 0.74 of the volume is filled with atoms. Using the density formula, we can calculate the effective atomic radius:\rho = \frac{m}{V}=\frac{N_{A}M}{V}Where:\rho = densitym = massV = volumeN_A = Avogadro's numberM = molar massSubstituting the given values, we get:1.6e3 = \frac{(6.022e23)(40)}{V}(0.74)Solving for V, we get:V=2.48e-
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

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

## 1. What is thermal conductivity?

Thermal conductivity is a measure of a material's ability to conduct heat. It is the rate at which heat energy is transferred through a material per unit of distance and temperature difference.

## 2. How is thermal conductivity of a gas like argon calculated?

The thermal conductivity of a gas like argon can be calculated using the kinetic theory of gases, which takes into account the mass, velocity, and collision frequency of the gas particles.

## 3. What is the mean free path of a gas?

The mean free path of a gas is the average distance that a gas particle can travel between collisions with other particles. This is an important factor in determining the thermal conductivity of a gas.

## 4. How is mean free path related to thermal conductivity?

The mean free path is inversely proportional to the thermal conductivity of a gas. This means that as the mean free path increases, the thermal conductivity decreases, and vice versa.

## 5. What is the SI unit for thermal conductivity?

The SI unit for thermal conductivity is watts per meter per Kelvin (W/mK). This means that it is measured in the amount of heat energy (in watts) that can be transferred through a material with a thickness of 1 meter and a temperature difference of 1 Kelvin.

Replies
10
Views
1K
Replies
8
Views
1K
Replies
1
Views
2K
Replies
9
Views
2K
Replies
5
Views
1K
Replies
1
Views
2K
Replies
2
Views
1K
Replies
6
Views
2K
Replies
3
Views
1K
Replies
3
Views
897