# The ideal gas law

1. Feb 13, 2007

### pixel01

Hi every body,

We are taught in school that the gas law : pv=nRT can not be applied at high pressure, say a hundred atm. Now I just think of the fact that can we use that equation at very low pressure? Several mTorr for example.

Thanks

2. Feb 13, 2007

### dmoravec

Yes, the ideal gas law works at low pressures (as long as the temperatures aren't really low)

3. Feb 13, 2007

### rdx

why not?

Why does the "law" fail at high pressures? I think it's because the molecules have volume which is negligible at normal or low pressures but comes into play when the molecules are crowded. If there is another reason then the logic might be different and low pressures might also be affected.

4. Feb 13, 2007

### Cyrus

To answer this question, we have quantitatively ask what is a 'high' and 'low' temperature and pressure?

The ideal gas law, Pv=RT, will deviate when the temperature or pressure of a substance is above or below its critical temperature or pressure.

1. At very low pressures $$(P_r <<1)$$ gases behave as an ideal gas regardless of temperature.
2. At high temperatures $$(T_r >2)$$ ideal-gas behavior can be assumed with good accuracy regardless of pressure (except when $$P_r <<1$$ )
3. The deviation of a gas from ideal-gas behavior is greatest in the vicinity of the critical point.

Where $$T_r = \frac{T}{T_{critical}}$$ and $$P_r = \frac{P}{P_{critical}}$$.

As the substance approaches these points, you have to make a correction using a compressiblity factor, Z, defined as: Pv=ZRT from charts.

rdx, is correct but only half way there. The Van der Waals equation of state takes into account two factors. The one mentioned by rdx, the volume, and in addition the molecular attraction forces.

This corrected equation is given by:

$$(P +\frac{a}{\nu^2})(\nu-b )=RT$$

If you want to read this you can open up Thermodynamics Cegnel, Boles -p. 137 & p.144

Last edited: Feb 13, 2007
5. Feb 13, 2007

### FredGarvin

It definitely has something to do with that. It also has to do with how molecules attract and repel as their relative distances change. At large distances, the molecules really don't see each other. At moderate distances, they tend to attract. At close packing they repel. If you look at the van Der Waals equation of state, there are two constants that are provided as a "tweak" to the ideal gas law. There are also factors added in for molecules that are not perfectly spherical, i.e. water (has a healthy dipole moment). If you look in some thermo tables you will see a correction known as the "acentric" factor.

Last edited: Feb 14, 2007
6. Feb 14, 2007

### pixel01

Thank you all for replying. But now I begin to confuse other thing: In fact what is the pressure of gas on the wall. Is this the hit of individual gas molecules when they move, so why pressure does not depend on molecule weight?. Naturally, the heavier the molecule, the harder it hit the wall. Can anypone explain this?

7. Feb 14, 2007

### Gokul43201

Staff Emeritus
The heavier the molecules, the slower they move (at any given temperature), and the slower they hit the walls.

8. Feb 14, 2007

### pixel01

You are right all right. But I am still confused.
At the same temperature, the molecule's weight and its velocity is quadratically propotional : mv^2=3kT while the molecule's momentum is just linearly: p=mv. So when the weight increases, the momentum also increases because velocity reduces more slowly. If we consider the pressure is caused by the momentum of the moving gas molecules transfered, this is not true!

9. Feb 14, 2007

### Staff: Mentor

Yes, the momentum of the individual molecules will increase. But the pressure also depends on the rate at which the molecules impact the walls. The rate at which molecules impact the walls depends on their speed, which has decreased. It balances out.

10. Feb 15, 2007

### pixel01

Well, you say "the momentum of the individual molecules will increase. But the pressure also depends on the rate at which the molecules impact the walls". My understanding is that the pressure depends on the molecule's momentum which consists of mass and velocity already.

Let's take an example :

At 77oF (25oC)--> T=298.15K, a hydrogen molecule (m=2) has velocity of 1928m/s (RMS), so its momentum equals to 1928*2 = 3856.
In case of nitrogen (m=28) an average molecule's velocity will be 515m/s ==> the momentum = 515*28= 14420, very much higher than that of hydrogen.
And according to the gas law, at the same T, the same V and the same number of the gas mole, the pressure will be the same regardless of what kind of gas it is.
So the pressure of gas on a cylinder's wall may not only because of the momentum (velocity) ?

Last edited: Feb 15, 2007
11. Feb 15, 2007

### Staff: Mentor

Pressure is a result of gazillions of individual collisions per second:
(1) The faster the molecules, the harder they hit per collision due to their greater momentum;
(2) The faster the molecules, the more collisions per second, since they hit the walls more often.​
These factors combine to make pressure proportional to momentum*velocity, or mass*velocity squared. Thus for the same temperature, different molecules produce the same pressure.

12. Feb 15, 2007

### Cyrus

Is gazillions in SI or US customary units? I forget

bajillion < gazillions < kajillion

1.21 Jiggawatts! :surprised Great scott!

Last edited: Feb 15, 2007
13. Feb 15, 2007

### pmb_phy

I go by a very vauge approximation - If the temper and temperature are near STP then the ideal gas law works well.

The reason it doesn't work for high pressure gasses is that the intermolecular forces become effective and the results start to skew at these pressures when temperature etc. is measured.

Pete

14. Feb 15, 2007

### Gokul43201

Staff Emeritus
Pressure is the mean force exerted upon a wall of unit area. For the sake of simplicity, assume all molecules have the same speed, v (the RMS speed for the given gas, at the given temperature). Also assume that the molecules happen to move only along the 6 orthogonal directions (normal to the walls of a cuboidal container).

Over some small time dt, only those molecules within a distance vdt from a wall of the container (and that also happen to be traveling towards the wall) will experience a collision with the wall. If the wall has area A, the number of these molecules is simply one-sixth of the number density, n, times the volume within which they are found at the beginning of that interval = Avdt. If the mass of each molecule is m, the total mass contained in this volume of interest is (1/6)nmAvdt. All of these molecules will elastically bounce off the wall and experience a change in velocity given by 2v. Therefore, the total momentum change during the interval dt, is simply (1/3)nmAv2dt. The pressure, or force (momentum change per unit time) per unit area, is then nothing but (1/3)nmv2.

Then using the equipartition theorem, (1/2)mv2 = (3/2)kT, and writing n=N/V, you end up with the correct form of the ideal gas equation, P = (N/V)kT ... and all reference to the molecule's mass disappears.

Last edited: Feb 15, 2007
15. Feb 16, 2007

### pixel01

Thank you all for responding the quest. Thanks Al Doc andh Gokul43201 for good explanation about the pressure and RMS speed of gas molecules.
Now I ve got a smaller question: does a liquid exert any pressure on the wall the same way (except from its own gravity)