Ideal Gases

This problem is particularly open ended in that it can be tackled with varying methods. It should be noted that the techniques illustrated below are only a single way of arriving at sensible solutions:

0. To calculate the amount of oxygen in the box, it is necessary to apply the ideal gas equation:

$pV = nRT$
$n = \frac{pV}{RT}$

Useful data to look up is that the standard pressure (according to IUPAC) is 100kPa, and the standard temperature is 273.15K. Additionally, R is the gas constant 8.3145 JK$^{-1}$mol$^{-1}$.

Therefore: $n = \frac{10^5 \times 1}{8.31\times 273} = 44 moles$


1. The number of molecules in the box is given by the product of the number of moles with Avogadro's number:

$N = 44.04 \times 6.022 \times 10^{23} = 2.65 \times 10^{25}$ molecules

Thus, the volume per molecules is: $\frac{1}{N} = 3.77 \times 10^{-26}$ m$^3$.

Assuming an even distribution of molecules, we can model the situation as each molecule occupying a cube of volume.

The length of each side of the cube is $\sqrt[3]{3.77 \times 10^{-26}} =3.35nm$

Thus, if each molecule is at the centre of its cube of volume, then the distance between molecules is the same as the length of the cube.


2. The average speed of molecules in the box can be roughly estimated as the speed of sound $\approx 330 ms^-1$. A more rigorous approach is to use an equation from the kinetic gas theory, which states that:

$\text{Kinetic Energy} = \frac{1}{2}mv^2 = \frac{3}{2}kT$

The mass of a molecule of oxygen can be calculated from knowing that it's molecular mass is 36 g/mol.

$\therefore \sqrt{ \frac{3kT}{m}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times {273}}{\frac{36 \times 10^{-3}}{6.022\times 10^{23}}}} = 435 ms^{-1}$


3.The time taken to cross the box can be calculated by the width of the box divided by the speed of the particle:

$t = \frac{1}{435} = 2.30 ms$


4. This is a particularly difficult question to model accurately, and so we will adopt a particularly easy model. We will assume that all the particles other than the one under investigation are stationary. We have calculated that the distance between particles is 3.35nm, and we can calculate the number of 'particle cubes' that the moving particle must move through when passing from one side of the box to the other:


$\text{Number to pass through} = \frac{1}{3.35 \times 10^{-9}} = 2.99 \times 10^{10}$

Thus, the moving particle must move through ~ 3 x 10$^{10}$ particle cubes when traversing the box. However, when passing through each particle cube, there is only a small probability of actually colliding with the resident particle. Modelling the oxygen molecule as a cube of side 100pm, a simple calculation shows that ~38000 of molecules could fit inside each particle cube. Thus, the probability of this resident particle being in a position where it will collide with the moving particle is somewhat more than $\frac{1}{38000}$.

Since the moving particle must pass through 3 x 10$^{10}$ of these cubes, each with a probability of somewhat more than $\frac{1}{38000} of causing a collision, it can be seen that we clearly do not expect the particle to fully traverse the box without suffering a collision.


5. If the box is oriented such that two of its axes are parallel to the ground, and one is perpendicular, the following will be observed: for particles travelling along the two parallel axes, gravity will have no effect on their time of flight, since horizontal and vertical motion can be considered independent. For particles travelling along the third axis, gravity will have the biggest effect: without gravity, the time taken for them to cross the box is 2.29864ms, and the time taken for them to travel with gravity (after a quick calculation) is 2.29859 ms and against gravity is 2.29870ms. Thus the effect of gravity is small enough that it can be neglected.

However although these effects are small, they are nonetheless present. The presence of gravity would mean that there would be a pressure gradient, as there would be a greater accumulation of particles towards the bottom of the container, as opposed to the top. The volume of the box would be unchanged, since the volume is fixed.

WOULD T CHANGE?

6. What other assumptions....?


7. One would expect the ideal gas equation to break down once the volume became sufficiently small that the particles became close to touching. At this point, the pressure would increase sharply for a given decrease in volume, since the electron densities would be repelling each other.To roughly estimate the volume at which this would be significant, we can calculate the volume of all the molecules in the box.

We modelled each particle previously as a cube of side 100pm. Given that there are $2.65 \times 10^{25}$ molecules in the box, the volume occupied in total is $2.65 \times 10^{-5} m^3 = 26.5cm^3$. Thus, at this volume, the gas will be exceptionally difficult to compress. However, it is more than likely that there will be a deviation from the ideal gas equation at much larger volumes than this.

8. Without a detailed treatment using the Maxwell-Boltzmann velocity distribution, this solution can at best be a broad estimate. We can model the scenario by assuming that all the particles in the box are moving head-on towards one of the six walls of the container. Therefore, when one of the walls is removed, a maximum of 1/6 of the particles could escape. However, because the wall is removed for only a brief time, only particles in a certain volume of the container near to the wall can possibly escape. This volume can be estimated by multiplying the average molecular speed by the time that the wall is removed for and also by the cross-sectional area of the box.

Thus, the number of moles of gas that escape are calculated by the fraction of the volume of the box which gas can escape from, multiplied by 1/6 and also by the total number of moles of gas.

Thus, the number escaping:

$\text{Number escaping} = \frac{n}{6} \times \frac{v_{av} \times time \times area}{volume} = \frac{44.04}{6} \times \frac{435 \times 10^{-9}}{1} = 3.2 \times 10^{-6} moles$

The number of moles of gas in the container has dropped and so the pressure will reduce too. The temperature will remain constant, as no work has been done.