One-dimensional one-atom gas

1. Feb 18, 2012

johann1301

When we make a gaschamber smaller, the kinetic energy increases <--> temperature increases <--> pressuere increases.

Does this always happen? Isn't it possible to make a gaschamber smaller without applying work to the gas?

Look at this: http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/Piston/jarapplet.html [Broken]

If the pistonhead is not moving during collison with the atom, the velocity of the atom doesnt change.

Is it theoretically possible to make a gaschamber smaller, without any increase in kinetic energy, which means same pressure and temprature, but in a smaller volume?

Last edited by a moderator: May 5, 2017
2. Feb 19, 2012

Rap

Yes, its possible. You have to understand the difference between reversible compression and irreversible compression. The whole kinetic energy increases <--> temperature increases <--> pressure increases idea happens for a reversible process. A reversible process is VERY slow and not specially tailored to the condition of the molecules in the gas. Its called reversible, because if you do a reversible compression, and then a reversible expansion (again, VERY slow), you get back to the original state. If you have an irreversible compression (very fast, or very special, like you mentioned) then you won't get back to the same state with a reversible expansion, and if its an irreversible expansion, who knows where you end up?

3. Feb 19, 2012

Staff: Mentor

I really don't know, but I don't think the statistical nature of thermodynamics is the best option for looking at a one atom. Usually it is used when the number of atoms or molecules is extremely large, as most real life situations are. Could someone clarify?

4. Feb 19, 2012

256bits

By your proposal the velocity of the particle will not change, and thus neither will its kinetic energy. Since temperature of a gas ( in this case one atom ) is related to the kinetic energy of the particles, the temperature will be constant.

On the other hand pressure is related to the number of particles hitting the piston during a certain time period, or rather the time rate of collisions of the particles with the piston.

If you are start with a volume V, then lets say that the atom hits the piston once every time period t, and the atom has moved a distance of ct, where c is its velocity, and x= d, the original distance between the piston and opposite wall, would be ct/2. As you decrease the volume, you are decreasing the distance x, but ct will remain the same.

Say you stop at a position x=d/2 ( half of the original volume ). The number of collision on the piston is now velocity/distance = ct/(d/2) = ct/ ct/2) = 2 times the number at the start.
doubling the collisions means that the pressure has doubled when the volume has halved.

PV = a constant still applies for an ideal gas even if one atom.

5. Feb 19, 2012

256bits

The kinetic theory of gases is formulated from the analysis of atomic collisions with the walls of a container. Derivations involve using F=ma, the mass of the particle(atom, molecule) and its velocity. Using Avogadro's number, one can come up with the basic formula PV=nRT.

Thermodynamics for ideal gases as a whole is based on assuming that all the particles N behave in like fashion, with no intermollecular forces present between particles and that the particles take up no volume themselves in the overall contained volume. When that is not the case, adjustments have to be made to the ideal gas equation.

Statistically thermodynamics follows emperical formulas. But one could also calculate the chances of all the air in your room moving to one corner. Rather a remote chance with the billions and billions of atoms in an air space, but a chance none the less, which is why it is assumed that all particles behave in a like fashion if you can understand what is meant by that. ( ie two containers of a gas undergoing identical thermodynamic cycles will have the the same state before and after and will be indistinguishable from one another )