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Gas pressure is the pressure exerted by gas particles when they collide with the walls of the container. The gas pressure increases with the number of gas particles.

Does the collision between two gas particles also exerts pressure? If so, how does the collision between two gas particles affect the overall gas pressure?

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    $\begingroup$ You cannot separate collisions with the walls and collisions between gas particules. $\endgroup$
    – Maurice
    May 28, 2023 at 9:42
  • $\begingroup$ @Ethan If you know how to code, you could solve the equation of motion for $N$ particles subjected to an intermolecular potential (like LJ), i.e. for particle $i$ you have $\sum_j \mathbf{F}_{ij} = m_i \mathrm{d}^2\mathbf{r}_{ij}/\mathrm{d}t^2$. Then you could calculate $P=(1/A)\mathrm{d}p_n/\mathrm{d}t$, where $p_n$ is the momentum component that is perpendicular to the wall, and apply some averaging. I suggest reading Gould & Tobochnik book if you want quantitative answers. $\endgroup$ May 28, 2023 at 12:36

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Yes and no.

As a first approximation, a given gas molecule is roughly as likely to gain momentum than to lose momentum when colliding with another molecule. Thus once our molecule hits a wall, the collisions that happened on the way there don’t really matter on average. This is exactly the point of the ideal gas model, which ignores particle-particle interactions except for exchanging energy (which happens through collisions).

As a second approximation, the finite size of particles and thus collisions matter. This is one of the main things captured by real gas models such as the van der Waals model. According to the latter, the pressure of a gas is (in dependence of the molar volume $v:=\frac{V}{N}$):

$$ p(v) = \frac{RT}{v-b} - \frac{a}{v^2}.$$

Here $a$ and $b$ are positive constants where:

  • $a$ captures the attraction of gas particles to each other (usually via the van der Waals force).
  • $b$ captures the repulsion of gas particles due to their finite size.

As you can see, $b$ increases the pressure, in particular for small $v$, i.e., high particle densities. (Both, $a$ and $b$ become zero for an ideal gas.)

However, it is debatable whether this is “collisions increasing the pressure”. The increase in pressure originates from the particles having less effective volume available; the averaging effect of collisions is still in effect. From another perspective, it doesn’t matter how many collisions happen in the gas, only that they happen to equilibrate the energies of particles.

how does the collision between two gas particles affect the overall gas pressure?

In some sense, not at all. Pressure captures the total kinetic and potential energy of the gas particles, which is not affected by collisions. The pressure of a gas doesn’t change on its own. (The main exception from this is of course if the gas is not in chemical equilibrium and chemical and kinetic energy can be translated into each other on a macroscopic scale.)

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  • $\begingroup$ Thats interesting!So can we assume for noble gases that the attraction term(a) is 0? $\endgroup$
    – Volpina
    May 28, 2023 at 12:50
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    $\begingroup$ @Volpina: No. Noble gases are not exempt from the van der Waals force; that’s how they become liquid at low temperatures. To have a gas in the first place, the attraction between molecules needs to be low (with respect to the temperature and pressure); otherwise you get condensation. (Also: Don’t confuse noble and ideal gases.) $\endgroup$
    – Wrzlprmft
    May 28, 2023 at 12:58
  • $\begingroup$ Note as the ideal gas model assumes point particles within that model there are no interparticle collisions; you need a more complex model or a real gas for those. $\endgroup$
    – Ian Bush
    May 28, 2023 at 13:58
  • $\begingroup$ @IanBush: Not really: As I write in my answer, you do need the particle collisions to equilibrate particle energies. This is the same for ideal and real gas models. The difference for real gas models is that they consider the effect of particle interactions on particle energies (taking a snapshot if you so wish), but they do not feature particle collisions explicitly. $\endgroup$
    – Wrzlprmft
    May 28, 2023 at 14:17
  • $\begingroup$ The ideal gas is a model of the gas at equilibrium - In the ideal gas model there is no need for collisions to establish equilibrium because it is already at equilibrium. But nobody ever listens to me on this one so I'll shut up now! $\endgroup$
    – Ian Bush
    May 28, 2023 at 14:24
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The mutual collisions of real gas molecules do not have significant direct impact on the gas pressure.

The original thermodynamic notion of an ideal gas considers just wall collisions. Deviation of real gas pressure is at not too high pressure and not too low temperature often negligible. Note that at ambient conditions, typical collision frequency for a molecule in air is roughly $\pu{E10 s-1}$ and the mean free flight distance between collisions is about $\pu{70 nm}$.

But, collisions of molecules have indirect pressure impact. Colliding molecules mean the total volume of molecules is not fully negligible, compared to the volume of free space between them. It causes increased frequency of wall collisions and higher pressure.

When the gas volume becomes comparable to the total molecule volume, molecules become being stuck, not able to freely pass each along other. At that point, the wall collision frequency and pressure raise with the volume decrease much steeper than for ideal gasses.

Another indirect effect, mainly at low temperatures and high pressures, are intermolecular cohesive forces, that manifest the most just before/after collisions. This has the opposite, pressure decreasing effect.

Both effects are addressed by the van der Waals equation, the most known state equation of real gases:

$$\left(P+a \frac{n^2}{V^2}\right)(V-n b)=n R T$$

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The pressure in an ideal gas is only due to collisions with the vessel's walls, collisions between molecules do not contribute. This can be shown by considering general forces on a molecule using Newton's second law of motion. If the force on a molecule is $f_x,f_y,f_z$ in directions $x,y,z$ where

$$\displaystyle f_x=m\frac{d^2x}{dt^2}$$

and similarly for the $y$ and $z$ directions. The energy is found by multiplying by $x,y,z$ as appropriate (because energy is force times distance), giving

$$\displaystyle xf_x+yf_y+zf_z=m\left(x\frac{d^2x}{dt^2}+y\frac{d^2y}{dt^2}+z\frac{d^2z}{dt^2} \right)$$

Now we need to manipulate this equation using a 'trick'

$$\displaystyle \frac{d}{dt}\left(x\frac{dx}{dt}\right)=x\frac{d^2x}{dt^2}+\left(\frac{dx}{dt}\right)^2$$

and similarly for $y,z$ and so obtain an expression for the second derivative. Substituting for this gives

$$\displaystyle xf_x+yf_y+zf_z=m\frac{d}{dt}\left(x\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}\right)-m\left(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2\right)$$

the right-hand most term is just the velocity $v$ squared multiplied by the mass $m$ and is therefore $mv^2$ is twice the energy of the molecule giving

$$\displaystyle xf_x+yf_y+zf_z=m\frac{d}{dt}\left(x\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}\right)-mv^2$$

The remaining derivate is $\displaystyle x\frac{dx}{dt}=\frac{1}{2}\frac{dx^2}{dt}$ producing

$$\displaystyle xf_x+yf_y+zf_z=\frac{m}{2}\frac{d}{dt}\left(\frac{d}{dt}(x^2+y^2+z^2)\right)-mv^2$$

So far this applies only to a single molecule. Over all molecules while the value of $\displaystyle \frac{d}{dt}(x^2+y^2+z^2)$ must fluctuate in time for any given molecule, overall all these values must average to zero, because if this were not so the gas would be expanding or contractions as time progressed.

The expression becomes, after rearranging and summing over all molecules,

$$\displaystyle \frac{1}{2}\sum mv^2=-\frac{1}{2}\sum (xf_x+yf_y+zf_z)$$

The term of the left is the total kinetic energy which from laws of motion is $3k_BT/2$ per molecule with $k_B$ being the Boltzmann constant therefore

$$\displaystyle Nk_BT =-\frac{1}{3}\sum (xf_x+yf_y+zf_z)$$

for $N$ molecules. (The term on the right is sometimes called the 'virial of Clausius').

The right hand side of this equation needs to be calculated. To do this we first find the force on one wall, say wall $xy$, and this force has the value $xyp$ where $p$ is the pressure ($\equiv$ force/area) and $xy$ is the area. The total force the wall exerts is therefore $-xyp$, and similarly for the other walls, $xz$ etc. The average distance of a molecule from a wall in the $x$ direction is $x/2$ so the energy $xf_x$ has a value $-xyzp/2$, thus over all six walls is $-3pV$ where $V=xyz$. Thus over all walls this is also the value of $\sum (xf_x+yf_y+zf_z)$.

The ideal Gas Law $pV=Nk_BT$ is thereby produced and so in the absence of intermolecular interactions all the pressure is caused by collisions with the wall. In other words, the effect of all the forces the molecules experience, but not those caused by intermolecular interaction, is to produce the ideal gas law.

If intermolecular interactions are added then we must add a similar general term term

$$\displaystyle pV=Nk_BT+\frac{1}{3}\sum (xf_x+yf_y+zf_z)$$

where now $xf_x$ etc. now refer only to intermolecular forces. This term can be recast in a form that measures the pair-wise interaction between molecules $\varphi$ at a radial distance $r$ and has the form $-\sum r\frac{d\varphi}{dr}$. This term is the origin of correction terms such as in the van der Waals equation.

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  • $\begingroup$ I liked this answer! Just a little detail, in the first paragraph, at the end you would have wanted to write "in directions $x, y, \bf{\color{red}{z}}$ where..." $\endgroup$ May 31, 2023 at 11:10
  • $\begingroup$ @Metal Storm, thanks, typo corrected $\endgroup$
    – porphyrin
    May 31, 2023 at 11:37
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Think of it this way:

You have a gas filled inside a cubical container closed by all six faces. For sake of simplicity assume you have 1 mole of that gas in the container. Even this much amount of gas means you have approximately $6.022 × 10^{23}$ molecules which is a lot. Now imagine a single molecule to start it's motion from one of the faces of the container to reach to the opposite face. From the third postulate of kinetic theory of gases, it will encounter so many random collisions with so many other randomly moving molecules that before it reaches the opposite face, it has undergone an unimaginable number of collisions.

So when finally it strikes the opposite face to apply pressure on it, it also implies the average pressure from all those collisions it encountered along the way. This averaged value includes the effective as well as non-effective collisions. All those collisions average out because they are completely random. This was just for one molecule and the similar happens to the other $6.022 × 10^{23}$ of them in the container.

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    $\begingroup$ So when finally it strikes the opposite face to apply pressure on it, it also implies the pressure from all those collisions it encountered along the way. – This ignores the fact that the collisions can increase as well as decrease the momentum of our molecule and all of this averages out. For example, in an ideal gas, it does not really matter how many of these collisions happened, only that the particles can somehow exchange energy. $\endgroup$
    – Wrzlprmft
    May 28, 2023 at 11:47
  • $\begingroup$ I haven't used the word increases or decreases in my answer at all. When I said it implies all those collisions, it itself means that the impact has averaged over the distance travelled. $\endgroup$ May 28, 2023 at 11:55
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    $\begingroup$ Then I suggest you clarify that. Because “pressure from all those collisions” suggest that pressure (increase) is accumulated. $\endgroup$
    – Wrzlprmft
    May 28, 2023 at 12:06
  • $\begingroup$ @Wrzlprmft I just tried to present a rough image of how the average pressure from collisions is a significant contributor in the total pressure. I think if I missed on something, your and Poutnik's answer would provide that. $\endgroup$ May 28, 2023 at 12:18
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    $\begingroup$ @Proscionexium This confuses what pressure is. Molecules have kinetic energy and create pressure when they collide with the wall of a vessel (you could say that pressure is the average momentum transfer). The only influence gas-gas collisions have is to ensure that the overall kinetic energy in the gas is well distributed across all the molecules (in a boltzmann distribution). $\endgroup$
    – matt_black
    May 28, 2023 at 13:10

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