Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 1 Lecture Notes 2014March 13 on Thermodynamics A. First Law: based upon conservation of energy 1. Work 1
Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 2 (c) More general definition of thermodynamics work, allowing for pressure changing during the process Work for an Isothermal Process 2
Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 3 Lecture Notes 2014March 18 on Thermodynamics 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 3
Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 4 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: nr C v = γ 1 Adiabatic Process: Derive constraint equations 4
Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 5 Adiabatic constraint on PV diagram: Adiabatic constraint in terms of temperature You could actually derive this more simply by starting with using the ideal gas law. γ PV and substituting for pressure 5
Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 6 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: du = dq PdV, 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. 6
Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 7 Lecture Notes 2014March 20 on Thermodynamics B. Entropy (2nd Law of Thermodynamics): 1. Definition of Entropy Without much motivation, we define entropy (for reversible process): dq ds =. T 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. 7
Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 8 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. Lecture Notes 2014March 25 on Thermodynamics Introduced the idea of TS diagrams (as well as PV diagrams). Both are needed to fully describe the first law of thermodynamics. dq Clausius Inequality: ds. The equality only holds for reversible process. An example of T 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: dw PdV. 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. 8
Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 9 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. 9
Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 10 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. 10
Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 11 Integrate to get entropy as a function of volume and temperature. The above derivation was for a monoatomic gas. Below we generalize it. 11
Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 12 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. 12
Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 13 Isobaric Process of Gas on PV and TS diagram 3. Reversibility and 2 nd 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] 13
Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 14 Lecture Notes 2014March 27 on Thermodynamics 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 T h and the cool part at T c. The system approaches an equilibrium temperature of T e. We calculate the total change in entropy of the system: 2 e Note that since T > TcT h, 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. 14
Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 15 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 2 nd 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 dq dq ds = (rather than ds > ). Note however that this efficiency is never 100% unless the T T cold reservoir is at absolute zero. 15
Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 16 Carnot Engine Cycle: is made of isothermal and isentropic (adiabatic) processes. By the first law of thermodynamics the net heat into the system Q=(Q h -Q c ) 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. 16
Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 17 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. 17
Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 18 Refrigerators and the 3 rd 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. 18
Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 19 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 12 th decimal place, which is pretty negligible. 2. Boltzmann s formula for entropy: S = nk ln(ω). 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? 19
Dr. W. Pezzaglia Physics 8C, Spring 2014 Page 20 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: S = nk ln(v ) -end of thermodynamics- 20