Partial pressure - gases


We were able to build a good understanding of the ideal gas law and the mechanism responsible for its relations by using a model of a gas where the molecules were far apart and interact rarely. The pressure the gas exerts on the wall of its container is interpreted as arising from the molecules bouncing off the wall. Since the molecules reverse their momentum (bounce back) upon hitting the wall, they must feel a force from the wall (using Newton's 2nd law or the Impulse-Momentum Theorem) and therefore, they must exert a force on the wall (by Newton's 3rd law). Averaging over many molecules, the result of many random hits is an (approximately) constant, smooth emergent phenomenon - pressure. 

Be careful! We are going to be talking about both pressure and momentum and in the past we have used the letter "$p$" for both of these. They are NOT the same thing. Momentum is a vector and pressure is a scalar; and they have different units (momentum looks like force x time, pressure like force/area). Usually you can tell from context which is meant. On this page we will ONLY use $p$ to stand for PRESSURE. Even though we are talking and thinking about momentum, we won't use it in any equations. 

Relating pressure and density

In physics, we often focus on forces and so the implication of pressure is taken as leading to effects associated with force (such as buoyancy). In biology and chemistry, the focus is often on how much of a gas you have rather than the force it exerts. The air is everywhere, exerting pressure both inside and outside and organisms evolve to accommodate that experience — and basically take it for granted. But if there is not enough oxygen in the air — or too much carbon dioxide — an organism can be at risk. Since we found from our model of gas as individual particles moving freely (in Kinetic Theory: The ideal gas law), that the pressure is proportional to the number density, we can use it as a stand it for density. Let's see how this works by looking at the equations.

Our ideal gas law for a single gas is

$$pV = Nk_B T$$

 where $N$ is the total number of molecules in the volume $V$. We have chosen the physicist's form (expressed in terms of the number of molecules) instead of the chemist's form (expressed in terms of the number of moles) since we are going to be reasoning in terms of what is happening with individual molecules.

If we divide both sides of the equation by the volume, we get

$$p= \frac{N}{V}k_B T$$

We note that each molecule in the gas contributes the same amount to the pressure: $k_BT/V$. We can conveniently express the result as

$$p = nk_BT$$


$$n = \frac{N}{V}$$

is the number density — number of molecules per unit volume — otherwise known as the concentration

A mixture of gases: Partial pressure

Since each molecule in a volume contributes the same amount to the pressure, if we have a mixture of different kinds of gases in a given volume, we can just add the contributions of each kind of molecule. This makes sense because to get the pressure we are just adding up the forces that all of the molecules exert on a unit area of wall. So if we have multiple types of molecules mixed together, say 1, 2, 3,..., they each contribute a pressure

$$p_1 = n_1k_BT \quad \quad p_2 = n_2k_BT \quad \quad p_3 = n_3k_BT ...$$

and the total pressure the wall will feel is the sum of these:

$$p = p_1 + p_2 + p_3 + ...$$

(The wall doesn't really matter here. The pressure is present everywhere throughout the gas - but we need a wall in order to measure the pressure.) For this reason, the contribution $nk_BT$ of each kind of gas is called that gas's partial pressure. Since at a fixed temperature, the partial pressure is just proportional to the concentration of that gas, we can use a gas's partial pressure as a stand-in for its concentration. Since most chemistry and biochemistry is done at a fixed total pressure ~ atmospheric pressure this works well.

Why does pressure go with number density (concentration) rather than mass density?

If we think about this result a bit it seems a little strange. Since pressure is a result of momentum change why does it only go with concentration (molecules/cm3) and not with mass density (g/cm3)? Don't heavier particles have more momentum and therefore exert more force on the wall? Sure they do -- if they had the same velocity as the lighter particles! 

But we have a constraint that we have to take into account: all the different gases have the same temperature. This means that the average KE of a molecule of each of the gases are the same. Since the average $½mv^2$ of the different gases are the same, the heavier molecules are moving slower than the light ones. The equation shows us how this works. At fixed $T$only the concentration matters. And the derivation of the equation shows us how different mass molecules contribute equally to the pressure. The $mv^2$ comes about from the momentum the molecule has when it hits the wall (proportional to $mv$) times how often it hits the wall (proportional to $v$). So the heavier ones carry more momentum than the light ones, but hit the wall less often. Since $T$ is the same it has to balance out. 

How can we get away with this?
Don't the molecules affect each other?

We're treating the molecules as if they are all independent. They certainly have to collide and exchange energy, otherwise they wouldn't all come to the same temperature. To get a sense of whether this is sensible in air, let's do a quick estimate of how far apart air molecules are, on average.

We know from chemistry that 1 mole of molecules at STP occupies a volume of 22.4 liters. That can tell us how far apart they are on average. Since we know how big an atom is (the diameter of one atom is on the order of 0.1 nm) we can figure out how far apart they are.

Let's do our estimation in equations first. Let the volume occupied by 1 mole of molecules be $V_0$ and the number in one mole be $N_A$ (Avogadro's number). Let's estimate by putting all our molecules on a cubic lattice so it's easy to calculate. Then we'll let them move away at random -- but that shouldn't change the average spacing. 

Each molecule will then occupy a fraction of the total volume equal to $V_0/N_A$. A cube with this volume has the length of its side, L, as being the cube root of this volume. From the figure at the right we can easily see that this distance, $L$, is also the spacing between the molecules on the grid. So this let's us calculate the average spacing of the molecules as 

$$L = \bigg( \frac{V_0}{N_A}\bigg)^{?} = \bigg(\frac{22.4 \times 10^3 \mathrm{cm}^3}{6 \times 10^{23}}\bigg)^{?} $$

$$L \approx (4 \times 10^{-20} \mathrm{cm}^3)^{?} \approx (40 \times 10^{-21} \mathrm{cm}^3)^{?}$$

$$L \approx 3.4 x 10^{-7} \mathrm{cm} \times \frac{1 \mathrm{m}}{100 \mathrm{cm}} = 3.4 x 10^{-9} \mathrm{m}$$

or about 3.4 nm.

If the diameter of an atom is about 0.1 nm, then we might take the diameter of a two-atom molecule like N2 or O2 (which make up most of the earth's atmosphere) to be 0.2 nm. This means the spacing is about 17 times the size of the molecule. This looks something like this (on the grid on the left and randomized on the right).



We see that there is plenty of room between the molecules to sneak in a bit of another gas without their getting in each others' way — and that they will get close enough to interact (when the molecules almost touch) only very rarely. So our model description makes good sense.

Joe Redish 10/26/14


Article 473
Last Modified: May 24, 2019