Developing a model for the behavior of gases

In this section we will develop the ideal gas law from first principles, that is, by considering the behavior of individual particles in an ensemble of gas particles. "Particles" will be our generic term for atoms and/or molecules.

Before we begin, we have to realize that this is a pretty complicated task, so we have to make some approximations. That's the way we often have to move forward in science: We make approximations to come up with a simple model, then ask, if we remove the constraints of our approximations (like adding friction in projectile motion), how does our model change, and can we then add that change to it?

Our assumptions are pretty simple →

1. Gases consist of spherical particles so small the distance between gas particles (atoms or molecules) is always much smaller than the distance between them.

Force on container walls – pressure

Consider some kind of container holding gas "particles." We'll call them particles, though they could be atoms, like He or Ar, or molecules, like CO₂ or O₂. The particles are in constant motion in three dimensions. They have a range of velocities, but there are enough of them that an average velocity is a meaningful number. Here we're looking at the container from one side.

If we look at just one gas particle with average velocity $v$, we can decompose that velocity vector into velocity $v_x$, toward the wall, and $v_y$, parallel to it. For the purpose of calculating the force on the wall, we only care about the component of $v$ perpendicular to the wall, $v_x$

If we use that velocity to calculate the momentum of the particle in the x-direction, $p_x = mv_x$, we have this simplified picture.

Now we're interested in how much force each particle exerts on the container wall as it hits it, so we'll have to let it bounce off the wall. Remember that one of our approximations is that this will be a perfectly-elastic collision — no attractive forces between particles and walls. In the picture below, the inward and outward momentum vectors are angled just a bit so they won't overlap, but it's easy to see that the change in momentum is the forward momentum minus the backward momentum (a negative number).

So the momentum change perpendicular to the wall upon bouncing off of it is $\Delta p_x = 2 p_x$.

Time between collisions with the walls

In order to calculate the total force that any particle exerts on the container walls, it's necessary to know how often it hits a wall. More hits in a given time means more force over that time which means more pressure (remember that pressure is force divided by area).

The particle will hit any one side wall, say the left one in the figure above, after traveling a distance of $2L$, where $L$ is the distance between walls. Using the velocity definition $v$ = distance/time, we can calculate $\Delta t$.

$$\Delta t = \frac{2L}{v_x}$$

Force on the walls

Now from our studies of momentum and impulse ($\Delta p$), we know that

$$F = \frac{\Delta p}{\Delta t}$$

where $\Delta p$ is the impulse and $\Delta t$ is the time between collisions. If $p_x = mv_x$ is the momentum in the $x$-direction we get the force of a particle on the wall as a function of its velocity, mass and the length of the container:

The ideal gas law & average KE

In the early days of exploring the properties of gases, Boyle, Charles, Gay-Lussac and Clapeyron developed a few important empirical laws of gas behavior, relating pressure, temperature, volume and number of particles. The most important relationship, due to Clapeyron, showed that the product of pressure and volume $(PV)$ is proportional to the number of particles and the temperature:

$$PV = NkT$$

where $k$ is a proportionality constant known as the Boltzmann constant, $k = 1.381 \times 10^{-23} \; J/K$

The constant can be viewed merely as being there to get the units right. The product $kT$ has units of energy (Joules, J)

Connection between the microscopic and macroscopic worlds

The empirically-derived ideal gas law, $PV = NkT$ and the kinetic molecular theory of an ideal gas give us a connection between the measurable properties of gases and the average kinetic energy of the gas particles.

If we let $R$ be the Boltzmann constant multiplied by Avogadro's number,

$$ \begin{align} \left(1.38065 \times 10^{-23} \, \frac{J}{K}\right)&(6.02215 \times 10^23/mol) \\[5pt] &= 8.314 \frac{J}{mol\cdot K} \end{align}$$

Practice problems

1. Calculate the average velocity of an H₂ molecule at 0˚C if the average velocity of an O₂ molecule at that temperature is 490 m/s.

   Solution

   The kinetic energy of a molecule of gas is $KE = \frac{1}{2} mv^2$. The average KE of H₂ and O₂ molecules is the same at a given temperature, so we have

   $$ \require{cancel} \begin{align} \cancel{\frac{1}{2}} m_{H_2} v_{H_2}^2 &= \cancel{\frac{1}{2}} m_{O_2} v_{O_2}^2 \\[5pt] m_{H_2} v_{H_2}^2 &= m_{O_2} v_{O_2}^2 \\[5pt] v_{H_2}^2 &= \frac{m_{O_2}}{m_{H_2}} v_{O_2}^2 \\[5pt] v_{H_2} &= \sqrt{\frac{m_{O_2}}{m_{H_2}}} v_{O_2} \\[5pt] &= \sqrt{\frac{32}{2}} \cdot 490 \; \frac{m}{s} \tag{*} \\[5pt] &= 1960 \; \frac{m}{s} \end{align}$$

   Equation (*) contains the ratio of the masses of the molecules in grams per mole, but the mass unit doesn't matter because they cancel in this ratio anyway.




2. Calculate the rms speed of nitrogen molecules (N₂) at 30˚C.

   Solution

   The root-mean-squared speed of a molecule of mass $m$ at temperature $T$ is $v_{rms} = \sqrt{\frac{3RT}{m}}$, where $R$ is the molar gas constant, $R = 8.314 \; J/mol \cdot K$.

   $$ \begin{align} v_{rms} &= \sqrt{\frac{3RT}{m}} \\[5pt] &= \sqrt{\frac{3 (8.314 \; \frac{J}{\cancel{mol}\cdot \cancel{K}})(303 \cancel{K})}{28 \; \frac{\cancel{g}}{\cancel{mol}} \left( \frac{1 \; Kg}{1000 \cancel{g}} \right)}} \\[5pt] &= 519 \; \frac{m}{s} \end{align}$$

   In the last step the units are $\frac{J}{Kg}$, which is $\frac{1}{\cancel{Kg}} \frac{\cancel{Kg} \cdot m^2}{s^2}$. The square root gives us the desired speed units.




3. Calculate the rms speed of CO₂ molecules at 25˚C.

   Solution

   The root-mean-squared speed of a molecule of mass $m$ at temperature $T$ is $v_{rms} = \sqrt{\frac{3RT}{m}}$, where $R$ is the molar gas constant, $R = 8.314 \; J/mol \cdot K$.

   $$ \begin{align} v_{rms} &= \sqrt{\frac{3RT}{m}} \\[5pt] &= \sqrt{\frac{3 (8.314 \; \frac{J}{\cancel{mol}\cdot \cancel{K}})(298 \cancel{K})}{44 \; \frac{\cancel{g}}{\cancel{mol}} \left( \frac{1 \; Kg}{1000 \cancel{g}} \right)}} \\[5pt] &= 411 \; \frac{m}{s} \end{align}$$

   In the last step the units are $\frac{J}{Kg}$, which is $\frac{1}{\cancel{Kg}} \frac{\cancel{Kg} \cdot m^2}{s^2}$. The square root gives us the desired speed units.




4. How does an increase in the number of moles of gas at constant volume and temperature affect the pressure?

   Solution

   The ideal gas law is $PV=nRT$, which we can rearrange to show that the pressure is

   $$P = n\frac{RT}{V}$$

   Now it's clear that the pressure (volume and temperature being held constant) is proportional to the number of molecules, thus scales linearly with it.




5. In a mixture of sulfur dioxide (SO₂) and oxygen (O₂) gases, what is the ratio of the average kinetic energy of SO₂ to O₂? What is the ratio of the rms speeds of the two gases?

   Solution

   The average KE of molecules of any kind at a give temperature is the same, so we have

   $$ \begin{align} \cancel{\frac{1}{2}} m_{SO_2} v_{SO_2}^2 &= \cancel{\frac{1}{2}} m_{O_2} v_{O_2}^2 \\[5pt] m_{SO_2} v_{SO_2}^2 &= m_{O_2} v_{O_2}^2 \\[5pt] \frac{v_{SO_2}^2}{v_{O_2}^2} &= \frac{m_{O_2}}{m_{SO_2}} \\[5pt] \frac{v_{SO_2}}{v_{O_2}} &= \sqrt{\frac{m_{O_2}}{m_{SO_2}}} \end{align}$$

   Now the ratio of the rms velocities is

   $$ \begin{align} \frac{v_{rms\; SO_2}}{v_{rms\; O_2}} &= \frac{\sqrt{\frac{\cancel{3RT}}{m_{SO_2}}}}{\sqrt{\frac{\cancel{3RT}}{m_{O_2}}}} \\[5pt] &= \sqrt{\frac{m_{O_2}}{m_{SO_2}}} \end{align}$$

   The ratios are the same.




6. In 2015, a Stanford U. physics research group set a new record for low-temperature of a material. Using laser cooling, they got 100,000 rubidium (Rb) atoms to a temperature of 50 pK. Calculate the rms speed of the atoms and the total internal energy of the ensemble.

   Solution

   At 50 picoKelvins, the average KE of the atoms is

   $$\begin{align} \bar{KE} &= \frac{3}{2}kT \\[5pt] &= \frac{3}{2} (1.381 \times 10^{-23}\frac{J}{\cancel{K}}) 50 \times 10^{-9} \, \cancel{K} \\[5pt] &= 2.072 \times 10^{-30} \, J \end{align}$$

   Now the average velocity is

   $$v = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{2(2.072 \times 10^{-30} \, J)}{85.5 \, \cancel{g} \left( \frac{1 \, Kg}{1000 \cancel{g}} \right)}} = 6.96 \times 10^{-15} \, \frac{m}{s}$$

   The total KE for the ensemble of 100,000 atoms is just the average KE multiplied by 100,000, or $2.072 \times 10^{-24} \, J$