# Carnot Cycle | Thermal Physics Lecture Notes

The only fundamental difference between them is the difference in the entropy transfer.  To see this, let dU be the energy change of the system during a reversible process.  Then the heat received by the system can be defined to be

(8.1)

Here the slash on the derivative is to remind us that this is not an exact differential.  From conservation of energy, we have

or

(8.2)

So we see that if ds= 0, then we get pure work.  Similarly, if dU = tds, we get pure heat.  An immediate consequence of this is that cyclic processes cannot produce as much energy as was put into them.  To see this, recall that since work is a form of energy, any form of work can be converted into mechanical work.  Similarly, we can convert work completely into heat since work does not involve entropy.  However, we cannot convert heat completely into work, since heat enters the system with a certain amount of entropy, which the work does not take away.  Thus entropy builds up in the process.  This cannot be allowed to continue indefinitely; rather the entropy must eventually be removed from the system.  The only way to do this is to remove it as heat, but by conservation of energy that requires us to put in an equal amount of extra heat at the start.

## Work Engines

Consider a device which inputs an amount of heat, converts it into work, expels the residual heat and uses some additional work to return to its original state.  This process then repeats again.  Such a cyclic device is called a heat engine.  If the heat engine cycles through as a reversible process, the total energy change is zero.  Now consider the entropy associated with the different parts of the cycle.  When we are putting heat into the system, we are introducing an amount of input entropy equal to .  Similarly, when we extract the heat from the system, we extract an amount of entropy equal to .  Since the process is reversible, we must have that the entropy that enters the system equals the entropy that leaves, and so

(8.3)

Now, recall that the total energy change is zero.  This means that the total work done must equal the difference between the heat inputted and the heat extracted

or, dividing by the heat input,

(8.4)

This is called the Carnot efficiency (or energy conversion efficiency) for a reversible process.  In a real situation, the processes going on in a heat engine are not actually reversible, so the true efficiency is always less than the Carnot efficiency.  Notice that even in the case of a reversible process, we can attain complete conversion of heat into energy only in the case that .

## Refrigerators

If we reverse the process, we can create a refrigerator.  Here we input a certain amount of work in order to extract heat from a lower temperature region and eject it into the higher temperature region.  We still have that the entropy must not change, so that the ratio of the heats is the ratio of the temperature, but now, instead of taking the ratio W/Qh, we are interested in the ratio Ql/W.  This ratio is called the coefficient of refrigerator performance.  Using the fact that the work is still

we get

(8.5)

We can draw a Carnot engine   schematically as

To show that the Carnot engine is an ideal engine, “make” the “device”

where X is our device and C is a Carnot engine.  Assume efficiency hX> hc.  Then

From this, we have that Qh’Qh = Ql’Ql which means that we are transferring heat from the lower to the higher temperature without doing any work, which violates the second law of thermodynamics.

What does the  Carnot cycle look like in terms of a pressure‑volume (pV)

diagram?  Consider an ideal gas.  During the first phase, while heat is transferring energy from the hot reservoir, the temperature of the system remains constant and the process is isothermal.  Similarly, when the system is expelling heat into the cool reservoir, the process is also isothermal.  While work is being extracted and performed, the system does not change entropy, so the change is an isentropic one.  Graphically, this looks like

We can easily see that the work done by the gas during one cycle is the area contained within the enclosed region on the pV diagram.  It should be stressed that it is not meaningful to speak of the heat content or of the work content of a system.  Using the pV diagram, we see that when we go around a closed loop in the pV diagram, a net amount of work is generated by the system, and a net amount of heat is consumed by energy conservation.  But after going around the pV diagram, the system has returned to its original conditions.  This means that there cannot exist two functions Q(s,V) and W(s,V) such that the heat and the work required to take the system from a state (sa,Va) to (sb,Vb) are given by the differences in Q and W.  If such functions existed, the net transfer of heat and work around the loop would have to be zero, and we have already seen that this is not true.  Thus, the transfer of heat and work between to states depends on the path taken in transitioning between the two states.  This path dependence is expressed when we denote heat and work transferred as and .

What are the heat and work transferred for various paths?  Let’s first consider an isothermal process.  In this case we have that the work done is

(8.6)

Thus, for an isothermal process, the Helmholtz function is a more appropriate choice for the energy function than the system energy U.  Also, since F is a minimum in equilibrium, we see that the effect of work performed on a system has the effect of increasing the Helmholtz free energy.

## Gibbs Free Energy

Now consider a process that occurs at a constant pressure. Such a process is called an isobaric process.  In many cases if the pressure remains constant, changes in the system are accompanied by a change in the volume of the system. The work done on the system when it changes it volume by an amount dV is ‑p dV.  If this work is positive, it is provided by the environment and is in a sense “free”.  If the work is negative, the work is delivered to the environment and is not extractable from the system for other purposes.   For this reason it is usually appropriate to subtract ‑d(pV) from the work to get an effective work. Thus

where we have defined a new energy function, the enthalpy, H, as

H = U + pV                                                    (8.7)

The enthalpy plays the same role in isobaric processes as the normal energy U does in processes at constant volume.  There are two types of constant pressure processes which are especially important.  The first one are those in which there is no effective work done.   In these processes, the heat transfer, , is equal to the change in the enthalpy, dH.  An example of this is the evaporation of a liquid from an open vessel.  Thus the heat of vaporization is just the enthalpy difference between the two phases.  The other processes are those at constant pressure and temperature.  In this case, the heat transferred is

and so

where G is another new energy function, called the Gibbs free energy.  In terms of the normal energy, it is defined as

G = U + pV – ts                                                (8.8)

Finally, what happens if we are transferring particles to the system?  Recall

that the energy of the system depends on entropy, volume and number of particles, U = U(s,V,N).  In this case, for a reversible process, the change in the energy is

dU = tds- pdV + mdN                                          (8.9)

where mis defined to be the chemical potential of the system.  We will discuss the chemical potential more later in the semester.  The work done on the system is given

by

We call the term mdN the chemical work.

Share.