# Total pressure in a gaseous system

The kinetic theory of gases assumes that all collisions between gas molecules are completely elastic. So kinetic energy is conserved in collisions between molecules. Thus the average value of velocity remains constant for the gas. Pressure is caused by the change in momentum associated with collisions of gas molecules against the walls of the container. So as the average value of velocity remains constant it is safe to ignore the effects of the collisions between the molecules themselves, when calculating the pressure of the system. Is this reasoning correct?

If it's correct, PV=nRT successfully calculates the true pressure of the gas. Consider my thought experiment. Two ideals gases are sealed in a container. There would be some temperature and total pressure associated with the system. Now, if we can successfully ignore the effects of the collisions between the gas molecules themselves, then this system is equivalent to having the two gases separate, in similar containers ( they just add up, that's all) If so, The individual pressures of the gases are going to be equal to their partial pressure, which is a measure of how much a given gas contributes to the total pressure.

But the gases are in equilibrium are they not? So their pressures must equal the total pressure. Where does the additional pressure come from? Is there is something wrong in the above reasoning? Can somebody point out where?

Thanks for any help offered

## 2 Answers

"So as the average value of velocity remains constant it is safe to ignore the effects of the collisions between the molecules themselves, when calculating the pressure of the system. Is this reasoning correct?"

Whether it is safe to ignore collisions between the molecules themselves has nothing to do with constancy of average velocity of the molecules. If the gas is in thermodynamic equilibrium, mean velocity of molecules is constant (no further conditions are needed). Collisions between molecules, however, may matter or may not - it depends on how the mean free path of molecules compares to size of the molecules.

"But the gases are in equilibrium are they not? So their pressures must equal the total pressure. Where does the additional pressure come from?"

I do not know what you mean here. Are you asking if each gas individually has the same pressure the original system of mixed gases had? Why would that be so? If the gases got separated while maintaining the temperature constant, the pressure of any of the two gases will be lower than the original pressure. In case the gases interact only negligibly, the sum of the pressure after separation is equal to the original pressure of the mix.

• What is the actual pressure that a gas experiences? Is it it's partial pressure? Then why is partial pressure said to be hypothetical? See en.m.wikipedia.org/wiki/Partial_pressure – SNB Sep 1 '17 at 9:07
• Additional question,In every source I found the pressure of a gas is defined to be the pressure that the gas exerts on the walls of it's container. But a gas need not necessarily be in a container. In such occasions what is the pressure of a gas? – SNB Sep 1 '17 at 9:10
• > "the pressure of a gas is defined to be the pressure that the gas exerts on the walls of it's container. But a gas need not necessarily be in a container. In such occasions what is the pressure of a gas? " Total pressure of a gas is the force with which the gas acts on any flat face of a test body exposed to the gas, divided by the face area. If not such body is exposed to the gas, no actual forces are acting, but based on experience the pressure does not depend on the quality of the test body, so we can define pressure as force per unit area that would be present if the body was there. – Ján Lalinský Sep 1 '17 at 10:29
• This total, measurable pressure is not always equal to sum of partial pressures, because partial pressures are quantities determined in a different physical system - one where only one gas is present (in the same volume and with the same temperature as the mixed system had), so there are no interactions between different kinds of gas. – Ján Lalinský Sep 1 '17 at 10:49

(1) There is no single kinetic theory. There is the simplest, and there are increasingly sophisticated versions.

(2) In practice, with all molecules except monatomic ones, some translational KE is usually transferred to rotational or vice versa during collisions. But on average, the translational KE does stay the same, so you're right that "it is safe to ignore the effects of the collisions between the molecules themselves, when calculating the pressure of the system".

(3) "If it's correct, PV=nRT successfully calculates the true pressure of the gas." No, there are other things, apart from whether or not the average velocity of the molecules is affected by collisions, that must apply in order for $pV=nRT$ to hold. In particular, (a) the molecules themselves must occupy a negligible fraction of the container volume and (b) forces between molecules must be negligible except during collisions. In fact, (a) and (b) apply quite well in a gas at low density. But for real gases at moderate pressures they don't apply. More sophisticated versions of kinetic theory can make allowances for their not applying.

(4) For gases at low densities, because (a) and (b) apply quite well, Dalton's Law of partial pressures also applies. There will then be no "additional pressure"; the pressure you get is the sum of the pressures that each gas would exert if it were the sole occupant of the container.

• Yes, but what is the pressure that the gas itself has in the mixture? Is it still the partial pressure? Then why not just call it the pressure of the gas then? – SNB Sep 1 '17 at 9:13
• The partial pressure, $p_A$, of gas A is the pressure that it would exert in the container at a given temperature, if it were the sole occupant of the container. $p_B$ is similarly defined. Dalton's law of partial pressures (which works well at lowish densities) states that the total pressure when A and B are both put in the same container is simply $p_A+p_B$. So, if you want to, you can say that the pressure that A exerts in the mixture is its partial pressure. That's certainly the way we regard it in kinetic theory. – Philip Wood Sep 1 '17 at 13:39