# Work done by a gas is PV

1. Apr 26, 2013

### mrcotton

1. The problem statement, all variables and given/known data

2. Relevant equations

PV=nRT
the average KE of the whole gas = 3/2 nRT

3. The attempt at a solution

why can I not say that the change in temperature of 147K gives a change in average kinetic energy.

so if the energy changes by 3/2nR (147) this change in energy is equal to the work?

I know that the real answer is PV= work
(force/area) volume = force times distance
So what is wrong with my logic about equating the change in average kinetic energy of the whole gas with the work done by the gas on the piston?

2. Apr 26, 2013

### Simon Bridge

Consider - in an isothermal process, the temperature does not change.
By your argument there is no work done. How do you explain, then, the work in an isothermal process?

But why is the work not the change in average kinetic energy?

If I have two equal masses m, one slow one fast. The slow one has speed v and the other speed 2v.
What is the average KE of this system. Call it E1.

Now I do some work on the system so that the slow mass speeds up. Now both masses have the same speed - 2v.
What is the average kinetic energy of the system? Call it E2
What is the change in the average kinetic energy? ΔE=E2-E1

But surely the actual work done is the change in the kinetic energy of the slow mass?
What is this? Compare with the result for ΔE.

3. Apr 26, 2013

Imagine that the piston was glued in place so that the gas could not expand. In this case no external work would be done but the temperature of the gas would still increase. All of the electrical energy input would result in an increase of the internal energy. With the moving piston some of the energy input is converted to internal energy and the rest to external work.

4. Apr 28, 2013

### mrcotton

Energy book keeping

So my understanding so far is that we are changing from state A with P(A), V(A), T(A) to state B with P(B), V(B) T(V) by putting energy into the system ΔE. We are keeping P constant. This make the ΔW calculation easier as (F/A)*V is for the work only has a change of Δs.

The energy ΔE that is added to the system can increase the average kinetic energy of the gas and some of this energy will then be used to move the piston through a distance Δs, which is the work done by the gas on the piston ΔW. So the answer to my query is that they are not equatable.

If the force F on the piston is due to the rate of change of momentum as the gas molecules collide with the piston, why then does the pressure not increase as the internal energy rises?

Is it photons leaving the wire that increase the KE of the gas molecules, and if so is the energy transfer by collision or absorption (This is a minor point really I am just interested).

Thanks for any help

5. Apr 28, 2013