# Solutions 1

- Page ID
- 47368

### Q1.1

Describe what happens to the average kinetic energy of ideal gas molecules when the conditions are changed as follows:

* \[E_{rms} = \frac{3VP}{2N} (1)\] *

- The pressure of the gas is increased by reducing the volume at constant temperature.
- The pressure of the gas is increased by increasing the temperature at constant volume.
- The average velocity of the molecules is increased by a factor of 2.

*a) If the pressure is increased with the volume being decreased in equal measure, from equation 1 we can see the average kinetic energy remains unchanged. *

*b) Given the ideal gas law PV = nRT if T goes up and V is constant then the pressure P must increase. If P increases then by equation 1 the average kinetic increases. *

*c) the kinetic energy is related to the velocity by this equation. \[E_{rms} = \frac{1}{2}mv^2_{rms}\] so if the velocity doubles the kinetic energy increases by four. *

### Q1.2

What assumptions are made when applying the kinetic molecular theory to gases? Are all of these assumptions necessary? Why or why not?

The assumptions are:

*1. The gas consists of objects with a defined mass and zero volume.*

*2. Gas particles move in straight lines unless acted upon by collision with another particles or container walls. *

*3. All collisions between gas particles and each other and the container walls are elastic*

*4. Gas particles do not interact aside from colliding with each other. *

*5. The average kinetic energy of the gas is proportional to the temperature. *

*These assumptions allow us to approximately calculate the macroscopic properties of a gas without dealing with a lot of complications. *

### Q1.3

The root-mean-square speed of 6 particles is 2.47 ms^{-1}. The speed of 5 of the particles are 1.0, 2, 1.5, 3.0 and 2.5. Calculate the unknown speed of the 6^{th} particle. Find the average speed of the 6 particles.

\[V^2_{rms} = \frac{V^2_{1} + V^2_{2} + V^2_{3} + ... + V^2_{N}}{N}\]

\[6.1009 = \frac{1 + 4 + 2.25 + 9 + 6.25 + V^2_{6}}{6}\]

\[V^2_{6} = 36.6 - 22.5 = 14.1054\]

\[V_{6} = 3.76\]

The average speed of the particles is just equal to all the speeds added together and divided by six.

\[V_{avg} = \frac{1 + 2 + 1.5 + 3 + 2.5 + 3.76}{6} = 2.29\]

### Q1.4

Consider the distribution of molecular velocities in a sample of helium. If the sample is cooled, will the distribution of velocities look more like that of H_{2} or of H_{2}O? Explain your answer.

\[V_{rms} = \sqrt{\frac{3RT}{M}}\]

*It should look more like water because as far as velocity goes decreasing the temperature could be modelled equally as well by increasing the mass of the gas particles. *

### Q1.5

Find the \(v_{rms}\) of N_{2(g)} at 25ºC. What temperature must \(Cl_{2(g)}\) be to have the same \(v_{rms}\)?

\[v^2_{rms} = \frac{3RT}{M}\]

For nitrogen with at 298 K with a molar mass of .028 kg/mol

\[v^2_{rms} = 265454 \frac{m^2}{s^2}\]

\[v_{rms} = 515.22 \frac{m}{s}\]

For chlorine with a molar mass of .0709 kg/mol the temperature at which its root-mean-square velocity is equal to nitrogen's at 298 K is

\[T = \frac{v^2_{rms}*T}{3R}\]

\[T = \frac{265454 * .0709}{24.942} = 754.5 K \]

### Q1.6

Use the Maxwell formula

\[f(v)=4\pi v^2 \left (\dfrac{m}{2\pi k_b T} \right)^{3/2}exp\left [\frac{-mv^2}{2k_bT}\right]\]

to derive an expression for \(c_{mp}\) of a gas.

\(c_{mp}\) is determined by finding the value of v where the derivative of f(v) with respect to v is zero.

\[\frac{df(v)}{dv} = 0 = 8\pi v \left (\dfrac{m}{2\pi k_b T} \right)^{3/2}exp\left [\frac{-mv^2}{2k_bT}\right] - (\frac{mv}{k_bT})4\pi v^2 \left (\dfrac{m}{2\pi k_b T} \right)^{3/2}exp\left [\frac{-mv^2}{2k_bT}\right]\]

Then rearranging the terms we have:

\[8\pi v \left (\dfrac{m}{2\pi k_b T} \right)^{3/2}exp\left [\frac{-mv^2}{2k_bT}\right] = (\frac{mv}{k_bT})4\pi v^2 \left (\dfrac{m}{2\pi k_b T} \right)^{3/2}exp\left [\frac{-mv^2}{2k_bT}\right]\]

Cancelling out terms on both sides

\[2 = v^2 * (\frac{m}{k_bT})\]

\[v_{mp} = \sqrt{\frac{2k_bT}{m}} = \sqrt{\frac{2RT}{M}} = c_{mp}\]