Why is pressure independent of the mass of the particles of the gas?

Question

Why is pressure independent of the mass of the particles of the gas? Wouldn't heavier particles hit the wall harder. Pressure is force per area, and force is related to inertia/mass. Doesn't this suggest that pressure is dependent in a certain sense on the mass of the particles?

This is assuming constant temperature, volume and amount of particles. But not the type of particles (i.e. nitrogen, carbon dioxide, etc)

$$ p = mv$$ $$ F = \frac{dp}{dt}$$ $$ P = \frac{F}{A} $$

Answer 1

In a simplified view, the pressure is proportional to the force on the container. The force is proportional to the rate ($1/\Delta t$) of change of momentum $p$.

Note that:

$$ \Delta p \propto mv $$

while

$$ \Delta t \propto \frac 1 v $$

so

$$ f \propto \frac{\Delta p}{\Delta t} = mv^2 $$

and of course,

$$ \frac 1 2 mv^2 = \frac 3 2 kT $$

so pressure depends on temperature and number density.

Answer 2

Heavier particles hit harder with the same velocity, but hit exactly as hard (per unit time) as a lighter particle with the same kinetic energy (the heavier particles have more momentum in each hit but hit less frequently, see JEB’s comment). Temperature and average kinetic energy are the same quantity, this is baked into the ideal gas law $PV=NRT$, $R$ has units of J per K$\cdot$mol.

So, for two different gases that have the same $P$, $V$, and $T$, the gas with smaller molecules will have the molecules moving faster. But both will have the same amount of average kinetic energy.

Answer 3

If we use the formalism of statistical mechanics we can calculate the [canonical] partition function of a gas, as a function of volume $V$ and temperature $T$:

$$Z(T,V) = \frac{1}{h^3}\int_V\int_{\mathbb{R}^3} e^{-\frac{H(\boldsymbol{x},\boldsymbol{p})}{kT}}\text{d}^3\boldsymbol{p}\text{d}^3\boldsymbol{x}$$

where $H(\boldsymbol{x},\boldsymbol{p})$ is the hamiltonian for a single molecule. The Helmholtz free energy is given by:

$$F(T,V) = -kT \ln \frac{Z^N(T,V)}{N!}$$

being $N$ the number of molecules. The pressure can be computed as the derivative:

$$p = -\left( \frac{\partial F}{\partial V}\right)_T$$

For an ideal gas we have:

$$Z = \frac{V(2\pi mkT)^{3/2}}{h^3}$$

and, consequently:

$$F = -kNT \left( \ln V + \frac{3}{2} \ln T + 3\ln \frac{\ln\sqrt{2\pi mk}}{h} \right), \qquad p = \frac{NkT}{V}$$

For a non-ideal gas, the partition function is more complicated but esentially we have:

$$Z = \frac{f(V,T)(2\pi mkT)^{3/2}}{h^3}$$

which lead to:

$$p = \frac{NkT}{V} + \frac{N^2A(T)}{V^2} + \frac{N^3B(T)}{V^3} + \dots $$

The reason: the mass $m$ mass appears factored apart from the volume in the Helmholtz free energ $F$, so the derivative $p = -\partial F/\partial V$ in independent of $m$.

Answer 4

Consider the ideal gas law. It can be expressed as

$$PV=nRT=NkT$$

Where

$n$ = number of moles

$R$ = universal gas constant

$N$ = number of molecules (particles)

$k$= Boltzmann constant = 1.38066 x 10$^{-23}$ J/K = $R/N_A$

$N_A$ = Avogadro's number = 6.0221 x 10$^{23}$ molecules

So we see that, for a given volume and temperature, pressure only depends on the number of gas molecules, not the molecular weight of the individual molecules (not the type of gas molecule).

Since temperature is a measure of the average translational kinetic energy of the gas molecules, in order for the temperature to be the same for an equal number of higher and lower mass molecules, the speeds of the higher mass molecules must be lower than the speeds of the lower mass molecules so that their average kinetic energy is the same.

Hope this helps.

Answer 5

We can use the work-energy theorem over a molecule that bounces at a collision in a given direction (say $x$). Its velocity in the x-direction goes to zero before changes direction. So during that time the work of electromagnetic forces equals the change of kinetic energy:$$F\delta x = \delta (\frac{1}{2}mv^2) = \frac{1}{2}mv_x^2$$ where $F$ is the average force during the collision. As the force is pressure times area:$$P (\delta y \delta z)\delta x = \frac{1}{2}mv_x^2$$The probability of bouncing at each axis is the same, and for each molecule bouncing on $x$ there is a second one bouncing on $y$ and a third one bouncing on $z$. So, for each group of $3$ molecules:$$3P \delta x \delta y \delta z = 3PdV = \frac{1}{2}m(v_x^2 + v_y^2 +v_z^2) = \frac{1}{2}mv^2$$

Where $|v|$ is the average velocity of each molecule. So, for a finite volume $V$, where there are N molecules:

$$PV = N\left(\frac{\frac{1}{2}mv^2}{3}\right)$$

The fraction at the right side is proportional to the temperature, leading to the ideal gas relation: $PV = nRT$

The conclusion is that if the average velocity is the same, bigger masses leads to bigger pressure. But in this case the temperature also is greater. For the same temperature (same kinetic energy), the average velocity is smaller, and the pressure doesn't change.

Answer 6

If you replace a molecule of hydrogen gas with one of uranium, while keeping the kinetic energy the same, then the uranium atom will have a lower velocity.

This lower velocity will cause the uranium atom to hit the container's walls less often than the H2 that it replaced.

The higher mass of the uranium atom means whenever it hits the container wall, it will hit harder than the H2 that it replaced.

These two effects cancel out perfectly, leaving the pressure on the container unchanged.