# Adiabatic Processes in the Carnot Cycle

1. Apr 18, 2004

### Moose352

I'm not sure I understand why the Carnot cycle involves adiabatic processes at all. I can't seem to find a reason. Also on a related note, how exactly does a Carnot heat pump (refrigerator type) work? Is the sequence the same as a Carnot heat engine?

2. Apr 19, 2004

### kuengb

The Carnot cycle has to be all reversible (by definition). And basically the only way to "switch between temperature levels" reversibly is the adiabatic process. On the other hand, the only way to exchange heat with the environment in a reversible way is the isotherm expansion/compression which also appears in the Carnot process.

I think it's all the same like the heat engine but the other way around. So the heat pump receives an amount of heat at the lower temperature level and throws out (more) heat at the upper level.

Bruno

3. Apr 19, 2004

### Kalimaa23

Aha, I might be able to help you here, I have a thermodynamics exam coming up...

The reason why you have adiabatic processes is that the pressure will drop more rapidly with expanding volume in an adiabatic proces then in a isothermic proces.

Isothermically you have P.V = constant
Adiabatically you have P.V^(gamma) = constant. Never mind where this comes from, but just now that gamma > 1

Without adiabatics, the cycle would basically do nothing

1) Expand the gas isothermically (through contact with a heat reservoir), this gives you an about of work -W = Q1
2) Compress the gas. This requires an amount of work of at least W. So you have done nothing.

Now see what happens if we do it correctly :

1) Expand the gas isothermically (through contact with a heat reservoir), this gives you an about of work -W = Q1
2) Expand the gas adiabatically. This will decrease the pressure more rapidly
3) Compress the gas isothermically. This requires an about of work W = Q2, the generated heat Q2 is drained away in a cold reservoir (so the gas stays at the same temperature)
4) Compress the gas adiabatically the bring it back in its original state.

The gas is back in it's original state, so the internal energy U does not change.
By the 1st law : dU = dQ + dW, so -W = Q

In steps 2 and 4 no heat is exchanged, so -W = Q1 - Q2, and since you expanded it adiabatically Q2 < Q1 and the system has generated work!