3Rd Set of Notes

3rd Set of Notes

Chapter 4

The First Law tells us if a process if possible based on

Does not tell us whether the process is spontaneous or not.

Spontaneous?

Second Law will answer these questions.

Several Different Ways to State the Second Law:

Historically defined based on observations that:

No Engine can be constructed that can convert heat ______

into Work.

Where was this observed?

Power Plants

Automobiles

Some heat must be ______.

Talked about this previously, said that heat was a ______form of energy transfer than work.

Second way of stating this is:

There is a thermodynamic state property called the entropy, and:

“THE ENTROPY OF UNIVERSE MUST ______or will stay the

______for the case of a reversible process.

Entropy is associated with the ______.

Note that this applies to BOTH the system and surroundings and therefore may somewhat difficult to apply.

One place where it can be applied directly is ______.

Why?

Follow the Historic route.

Consider a heat engine, a device to take heat and covert it into usable work.

You are familiar with some. Consider a power plant:

Look at the system as just the water. Follow the processes being carried out on the system.

Start off with liquid water in the boiler:

1. Turn the water into steam (expand the system by putting in heat.)

2. Allow the steam to do work by hitting the turbines. (work due to expansion)

3. The steam is recondensed. (heat is released, and compression to liquid occurs)

4. The system is pumped back into the boiler. (further compression, no heat transfer)

Note that this is a cycle, that is, the system is changed but returns back to its original state, and then is changed again and returns back and so on and so on.

This represents the real world case:

The Rankine Cycle most closely resembles the process by of steam-operated power plants or heat engines. That is a power generating plant that converts heat to work using water as the working fluid. The two most common processes used in these power plants are the combustion of fossil fuels such as natural gas and coal, and nuclear fission.

There is another cycle that provides the maximum theoretical efficiency that can be obtained when converting heat to work (efficiency = Wnet/Q). It is called the Carnot cycle. The main difference between the Carnot Cycle and the Rankine Cycle is that the heat addition (in the boiler) and rejection (in the condenser) are isobaric in the Rankine Cycle and isothermal in the Carnot Cycle.

IS THIS the Rankine or Carnot Cycle?

Also, the ideal case, the most efficient, the one that would provide maximal net work out from the heat in would be the case where each step is ______.

These four steps, this ______, carried out in an ideal manner is called the ______, or the ______Engine.

Define the CARNOT Cycle. Assume the working fluid is an ideal gas. Steps:

1. Isothermal Reversible Expansion

P1V1 -> P2V2 Thot

Q1 is W1 is

2. Adiabatic Reversible Expansion

P2V -> P3V3 Thot -> Tcold

Q2 is W2 is

3. Isothermal Reversible Compression

P3V3 -> P2V2 Tcold

Q3 is W3 is

4. Adiabatic Reversible Compression

P4V4 -> P ? V? T?

Q4 is W4 is

What do we know about DU for the four steps inclusive?

Also the total work is W1 + W2 + W3 + W4 = Wtot

And the total heat is Q1 + Q2 + Q3 + Q4 = Qtot

Qtot = - Wtot

Calculate Q and W for each step

First, we need to derive an additional relation for an adiabatic reversible process.

The process can be described by a PV diagram. Isotherms on a PV diagram we have seen before, adiabatic reversible steps are steeper than isotherms.

Now calculate each

Q1 and W1:

Q2 and W2:

Q3 and W3:

Q4 and W4

Thus the total work is:

And the Total Heat is:

Carnot found that although the sum of all of the Qs was not zero, that

the sum of ______was equal to zero. (remember this is the reversible case).

For the isothermal steps we find:

And for the other two adiabatic steps, of course:

Setting the -W/T equal to Q/T

But

So

Q1 / Thot + Q2 / Tcold =

Since this is equal to zero, surmise that Qrev / T is a state variable.

Is this only true for the case of using an ideal gas as the working fluid.

Rationalized by considering the efficiency of the heat engine.

Efficiency = - Wout / Qin

Can rearrange this to solve for efficiency in terms of the temperatures of the reservoirs.

From experience we find that heat only flows from a hotter reservoir to a colder reservoir.

If any two heat engines operating in reversible cycles between the hot and the cold reservoirs have different efficiencies, then we could operate the less efficient heat engine in reverse as a heat pump and use the more efficient one as a heat engine, the work output of which is used to drive the heat pump. The combination of the two engines would contradict the fact that heat only flows in only one direction.

Thus all reversible heat engines have the same efficiency, so this is independent of using an ideal gas and thus Qrev / T is truly a new state function, the entropy.

Example:

Also consider this eqn. as defining the absolute thermodynamic temperature scale, the same as the one in PV = nRT, since the eqn. works in the case that we use a ideal gas as well.

SECOND LAW STATEMENT Based on the Entropy

As said before: statements of the second law.

1. Heat transfer occurs from a hotter body to a colder body.

2. Cannot convert heat completely into work. That is the efficiency of a heat engine cannot be 100 %.

3. The entropy of the universe must be greater than or equal to zero for any process. The equal sign is the case for the reversible system. The disorder of the universe must increase.

DS(sys) + DS(surr) ≥ 0

Thus application of this equation involves both the system and the surroundings.

Remember that DS = Qrev / T (only for the reversible case)

Examples:

How can one measure the entropy?

Relies on the idea that DS = ∫ dQrev / T or DS > ∫ dQirrev / T

Application to Chemical Rxns.