## PV = nRT

### Pressure, Volume, Temperature, Moles

We know that temperature is proportional to the average kinetic energy of a sample of gas. The proportionality constant is (2/3)R and R is the gas constant with a value of 0.08206 L atm K-1 mol-1 or 8.3145 J K-1 mol-1.

(KE)ave = (2/3)RT

As the temperature increases, the average kinetic energy increases as does the velocity of the gas particles hitting the walls of the container. The force exerted by the particles per unit of area on the container is the pressure, so as the temperature increases the pressure must also increase. Pressure is proportional to temperature, if the number of particles and the volume of the container are constant.

What would happen to the pressure if the number of particles in the container increases and the temperature remains the same? The pressure comes from the collisions of the particles with the container. If the average kinetic energy of the particles (temperature) remains the same, the average force per particle will be the same. With more particles there will be more collisions and so a greater pressure. The number of particles is proportional to pressure, if the volume of the container and the temperature remain constant.

What happens to pressure if the container expands? As long as the temperature is constant, the average force of each particle striking the surface will be the same. Because the area of the container has increased, there will be fewer of these collisions per unit area and the pressure will decrease. Volume is inversely proportional to pressure, if the number of particles and the temperature are constant.

There are two ways for the pressure to remain the same as the volume increases. If the temperature remains constant and so the average force of the particle on the surface, adding additional particles could compensate for the increased container surface area and keep the pressure the same. In other words, if temperature and pressure are constant, the number of particles is proportional to the volume.

Another way to keep the pressure constant as the volume increases is to raise the average force that each particle exerts on the surface. This happens when the temperature is increased. So if the number of particles and the pressure are constant, temperature is proportional to the volume. This is easy to see with a balloon filled with air. A balloon at the Earth's surface has a pressure of 1 atm. Heating the air in the ballon causes it to get bigger while cooling it causes it to get smaller.

### Partial Pressure

According to the ideal gas law, the nature of the gas particles doesn't matter. A gas mixture will have the same total pressure as a pure gas as long as the number of particles is the same in both.

For gas mixtures, we can assign a partial pressure to each component that is its fraction of the total pressure and its fraction of the total number of gas particles. Consider air. About 78% of the gas particles in a sample of dry air are N2 molecules and nearly 21% are O2 molecules. The total pressure at sea level is 1 atm, so the partial pressure of the nitrogen molecules is 0.78 atm and the partial pressure of the oxygen molecules is 0.21 atm. The partial pressures of all of the other gases add up to a little more than 0.01 atm.

Atmospheric pressure decreases with altitude. The partial pressure of N2 in the atmosphere at any point will be 0.78 x total pressure.

### Gas Molar Volume at Sea Level

Using the ideal gas law, we can calculate the volume that is occupied by 1 mole of a pure gas or 1 mole of the mixed gas, air. Rearrange the gas law to solve for volume:

V = nRT/P

The atmospheric pressure is 1.0 atm, n is 1.0 mol, and R is 0.08206 L atm K-1 mol-1. Let's assume that the temperature is 25 deg C or 293.15 K. Substitute these values:

V = (1.0 mol)(0.08206 L atm K-1 mol-1)(298.15 K)/(1.0 atm) = 24.47 L = 24 L (to 2 sig. fig.)

### Gas Velocity and Diffusion Rates

Kinetic molecular theory can derive a quantity related to the average velocity of of a gas molecule in a sample, the root mean square velocity. You can see the derivation in the appendix to Zumdahl's textbook or read about it on an online source. The calculations are beyond the scope of this course.

This velocity quantity is equal to the square root of 3RT/M where M is the mass of the particle.

The relative rate of two gases leaking out of a hole in a container (effusion) as well as the rate of two gases moving from one part of a container to another (diffusion) depends on the ratio of their root mean square velocities.

Can apply this to isotope separation for nuclear reactors? Remember that uranium fuel for commercial reactors must be enriched to 3-5% U-235. Its natural abundance is only about 0.7% with the remainder U-238. The uranium is converted to a volatile form, UF6. Let's calculate the rate at which the lighter 235UF6 would pass through a small hole from one gas centrifuge to the next relative to the heavier gas 238UF6.
mass of 235UF6 = (6)(18.9984 g) + (235.0439 g) = 349.0343 g

mass of 238UF6 = (6)(18.9984 g) + (238.0508) = 352.0412 g

rate of effusion of 235UF6/238UF6 = 352.0412/349.0343 = 1.0086

Now you can see why row-after-row of gas centrifuges are necessary for isotope separation!

Professor Patricia Shapley, University of Illinois, 2011