lecture notes 2014march 13 on thermodynamics a. first …...lecture notes 2014march 13 on...

Dr. W. Pezzaglia Physics 8C, Spring 2014 Las Positas College Lecture #14, 15 & 16 on HEAT 2014Mar18

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Lecture Notes 2014March 13 on Thermodynamics

A. First Law: based upon conservation of energy

1. Work


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(c) More general definition of thermodynamics work, allowing for pressure changing during the process

Work for an Isothermal Process


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2. First Law of Thermodynamics (caution, my sign convention differs from text)

Three simplifications of the first law:

Apply to an ideal gas, isovolumetric process vs. isobaric process


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Derive from above the relationship between constant pressure and constant volume heat capacity of an ideal gas.

So we can write (constant volume) heat capacity as: 1−

=γnR

Cv

Adiabatic Process: Derive constraint equations


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Adiabatic constraint on PV diagram:

Adiabatic constraint in terms of temperature

You could actually derive this more simply by starting with γPV and substituting for pressure using the ideal gas law.


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Enthalpy (1875): In chemistry you used the first law of thermodynamics in terms of a different thermodynamic potential called the “enthalpy”.

Note if you have a process that is isobaric and adiabatic, the enthalpy is conserved!

This is what makes enthalpy useful in chemistry. You are usually having reactions occur in an open beaker, so the atmospheric pressure is constant. If instead you were using the first law in terms of the internal energy: PdVdQdU −= , then you would have to worry about the “work” done by the fluid expanding during the chemical reaction (e.g. as ice froze, it would do work on the environment). For enthalpy, you don’t have to worry about it!

Shortly we will define “S” called the entropy. You can show that while internal energy is a function of just entropy and volume: U(S,V), that enthalpy is function of entropy and pressure H(S,P).

Note, the “trick” of how to change from PdV to VdP is called a Legendre Transformation.


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B. Entropy (2nd Law of Thermodynamics):

1. Definition of Entropy

Without much motivation, we define entropy (for reversible process): T

dQdS = .

This allows us to re-express the first law of thermodynamics in the following form:

Hence it follows, that if we integrate, that if we know the Temperature as a function of entropy (from TS diagram) and Pressure as function of volume (from PV diagram) then the internal energy is a function of entropy and volume: U(S,V).

If we have two systems in contact, such that heat flows from hot system (B) to cold system (A), the total entropy of the system will increase.

Note that entropy is extensive, total entropy change is the sum of entropy changes of the parts.


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For an instant in time, the net change in entropy will be:

Clearly the net entropy change is positive. So one can restate the law that heat flows from high temperature to low (and not the other way) as that the total entropy of the system increases.


Introduced the idea of TS diagrams (as well as PV diagrams). Both are needed to fully describe the first law of thermodynamics.

Clausius Inequality: T

dQdS ≥ . The equality only holds for reversible process. An example of

the inequality would be the free expansion of a gas in a vacuum. No heat is added (dQ=0), but the disorder (entropy) increases.

Also, the work done is: PdVdW ≤ . Again, in the case of a free expansion of a gas in a vacuum, the volume changed but there was no work done. We COULD have utilized the expansion to get some work done, but we didn’t.


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Gibbs Free Energy (1873): Another potential which is used in Chemistry to determine if a reaction will go forward or not. G(T,P)

Again, we used the trick of a Legendre Transformation to go from TdS to SdT. So if you are doing Isobaric chemistry, dG=-SdT.

In brief, in chemistry you determine whether or not a reaction (at constant pressure) will go forward or not.


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2. Entropy of an Ideal Gas

Consider a container of gas which is opened in space and allowed to expand freely. The molecules will still maintain their “average” kinetic energy, so the temperature of the gas is unchanged. The pressure of course will decrease as the gas expands. From first law of thermodynamics we get a relationship between entropy and volume.

Now we do it more general, using an ideal gas, we use the first law of thermodynamics to get an integrable differential for the entropy.

Note: the comment in the lower right means that the resulting entropy does not depend upon in which order we integrate. We can do the temperature integral first and the volume integral second, or in the opposite order.


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Integrate to get entropy as a function of volume and temperature.

The above derivation was for a monoatomic gas. Below we generalize it.


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Entropy Change for Adiabatic (Isentropic) process.

Note, an isentropic process must be adiabatic. However, a process can be adiabatic but NOT isentropic because of the inequality in the Clausius definition of entropy. The adiabatic process is only also isentropic if it is reversible.

Note, the reversible adiabatic expansion of a gas would look the following on PV and TS graphs.

Isothermal expansion process of a gas on PV and TS diagram

Note, this was repeated on board #13.


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Isobaric Process of Gas on PV and TS diagram

3. Reversibility and 2nd law of thermodynamics

First we show (again) that heat flowing from a hot system to a cooler one increases the total entropy of the system.

[Board 13 deleted, it was the same as board #10]


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Now consider that the temperatures of the components of the system change as heat is moved from the hot to cold components. Assume a “heat capacity” for each component. To simplify, let us assume both components are identical with same heat capacity.

Initially the warm part is at temperature Th and the cool part at Tc. The system approaches an equilibrium temperature of Te. We calculate the total change in entropy of the system:

Note that since hce TTT >2 , the total entropy change is positive. So entropy increasing is

equivalent to saying that heat flows from the hot to the cold until equilibrium is reached.


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4. Efficiency

Carnot came up with a famous “Carnot diagram” that using the 1s law of thermodynamics can illustrate the efficiency of an “engine”.

Recall the 2nd law of thermodynamics that the total entropy change must be zero or increase,

Hence the “most efficient” possible engine is the “Carnot Cycle”, which is the case in which

T

dQdS = (rather than

T

dQdS > ). Note however that this efficiency is never 100% unless the

cold reservoir is at absolute zero.


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Carnot Engine Cycle: is made of isothermal and isentropic (adiabatic) processes.

By the first law of thermodynamics the net heat into the system ∆Q=(Qh-Qc) will equal the work done. In passing we note that if the processes are not reversible then the inequalities will result in the system’s efficiency being less than the ideal Carnot cycle.

THIS LAST SLIDE HAS TOO MANY ERRORS.

Misnomer, I don’t think this is the OTTO cycle. I think it’s a “Stirling Cycle” (1816).

I don’t quite believe the efficiency calculation at the bottom. Its from some very old notes I have that I haven’t had time to check. Wikipedia says that the efficiency should be close to the Carnot cycle.


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Lecture Notes 2014April01 on Thermodynamics

5. Refrigerators

Note I seem to use a different symbol for efficiency and coefficient of performance than the text.

The biggest coefficient of performance is for a Carnot engine run backwards.


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Refrigerators and the 3rd law of thermodynamics.

C. Statistical Mechanics

1. Review the Binomial Coefficient and Pascal’s Triangle. The numbers represent the statistics of flipping a group of “n” coins, the possibilities of having “i” coins heads up and (n-i) heads down.

Here we calculate the probability.


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Plot of probability: As “N” increases, the chances of deviation from the average (N/2) gets smaller and smaller.

So, if we interpret the statistics as the probability of a gas with “N” molecules having “i” on the left and (n-i) on the right, then the most probable situation is that half will be on each side. Recall that pressure is proportional to density of molecules. With a mole of gas the deviations in pressure would be seen in the 12th decimal place, which is pretty negligible.

2. Boltzmann’s formula for entropy: )ln(Ω= nkS . The entropy for the completely ordered state of all heads would be S=nk Ln(1) = 0. What is the entropy for the most probable situation of the average?


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3. Stirling’s Formula: How to calculate logs of really really big numbers.

We then have the calculation of the entropy of the most probable situation (which is also the macrostate of maximum entropy).

So, if we reinterpret that initially we had all the molecules on the left side (S=0) and then a short time later equilibrium has been reached, then the “volume” of the gas has increased a factor of 2. The entropy has increased proportional to Ln(2). This is consistent with our Clausius definition of entropy of a gas going like: )ln(VnkS =

-end of thermodynamics-

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