Lecture 6: Intro to Entropy Reading: Zumdahl 0., 0., 0.3 Outline: Why enthalpy isn t enough. Statistical interpretation of entropy Boltzmann s Formula Heat engine leads to Entropy as a state function Problems:Z0.-4, Z0.7, Z0.6-7, Z0.9-, Z0.5-6.
Enthalpy and Spontaneous Rxns Early on in the development of thermodynamics, it was believed that if a reaction was exothermic, it was spontaneous. But there are plenty of reactions that go that do not give off heat. Consider the following reaction: H O(s) H O(l) ΔH rxn = +6.0 kj Endothermic..yet spontaneous! Above 0C, but not spontaneous below 0C. Why? Consider the hot packs and cold packs again as well.
Enthalpy and Spontaneous Reactions: Joule s original experiment Consider the following Experiment: Mixing of a gas inside a bulb adiabatically (q = 0). q = 0, The bulbs are insulated w = 0, External Pressure is zero. ΔΤ=0, Experimental Result. ΔE = 0, and ΔH = 0.but it still happens, the gas expands (spontaneously) to fill the two bulbs. Why? 3
Entropy and Statistical Probability The connection between weight (Ω) and entropy (S) is given by Boltzmann s Formula: S = k ln Ω B Boltzmann s constant: = R =.38 0 N 3 For a given state (where E is fixed) Ω is the number of ways of arranging the molecules (or microstates). The difference in entropy between two states then is: k Δ S = kbln Ω kbln Ω = kbln Ω B A Ω J K 4
Molecules Move by Chance Between Bulbs What is the chance that a molecule will be on the left: ½ What is the chance that two molecules will be on the left: ¼ And N molecules on the left: For Large N, very small chance: The molecules move in a statistically independent way. How does the entropy change for the gas if we double the volume? How many ways to put a molecule in either side vs? Twice as many. Ω is the number of ways of arranging a molecule in one bulb, then Ω is the number of ways of arranging it in both bulbs. All particles are statistically independent so get a product of arrangements. Apply Boltzmann s definition of entropy: N N Ω Ω N Bln Bln ( B A) ln ln Ω Ω N A Δ S = k = k = k N = nr 5
Expansion by any amount: For each independent particle Original Weight = Ω Final Weight = Ω In the previous example the statistical weight was directly proportional to volume. Now generalizing to any two different volumes: N Ω f Ω f V f Δ S = kb ln = nrln = nrln I I V Ω Ω I 6
The absolute entropy for a gas in a box Avagadro s principle: mole occupies.4 liters at STP. From this we will compute the absolute entropy for a gas in a box. Imagine the the box of volume V (.4L) is divided up (conceptually anyway) into a number of sub-boxes each of which is capable of holding one molecule at a time. Assume a sub-box is about 0.05 nanometer in length (which is the Bohr radius of an atom). This give us a number of sites (sub-boxes) to put molecules in, one at a time. 3 3 V.4 0 m 9 Λ= = =.8 0 3 V sub ( 50 m) Now count the way to put Avagadro s number of molecules in there: Ω= Λ N A N A Compute the entropy: This is the number of sites each molecule can occupy, and they are all statistically independent. So the total Weight is the product of individual weights. 9 Λ.8 0 3 N A 6.0 0 S = kbln Ω= kbnaln = Rln = R = 05 J This is close to the experimental values for simple gasses (like He or H), and 7 shows that the entropy is proportional to the log of the Volume. K
Example: Crystal of CO One can flip Each arrangement of CO is possible. Consider the depiction of crystalline CO. There are two possible arrangements for each CO molecule. For a mole of CO: Ω = N A Same result as for the Joule Experiment where the volume doubled. S = k ln Ω= k ln = k N ln = Rln = 0.7 R= 5.6 J ( N A ) ( ) ( ) B B B A 8 K
What is Entropy Good For? All other thing being equal, the most probable way to distribute the energy is the most likely way the system will arrange itself. Entropy is a measure of what is probable. The larger the entropy, the more probable the event. An increase in entropy is a more probably event. An increase in volume results in an increase in entropy For a chemical reaction, if entropy increases it ought be more probably (i.e. spontaneous). (See problems -4) In general then a reaction that creates a gas is going to increase its entropy, and that can make a reaction happen. Temperature can help if the entropy (change) is positive If can t get a gas, going to a liquid from solid is next best. In a reaction A+B goes to AB, what is entropy change? 9
Heat Engines (and Entropy) We will study the gas in a box, and determine what the entropy change is from a thermodynamic point of view, and will find that we arrive at the same answer as Boltzmann, and that entropy is a state function and a measurable quantity as well. We have examined isochoric and isobaric changes in a gas in a box. Both of these are done by raising or lowering the temperature (which can be done slowly). Now we consider isothermal processes, many of which cannot be done slowly. We need isothermal, reversible expansion/contraction process because this tells us what entropy is from thermodynamics. 0
Example: Isothermal Expansion Consider a mass connected to a ideal gas contained in a piston. Piston is submerged in a constant T bath such that ΔT = 0 during expansion.
Isothermal Expansion: Start Δh Δ V = AΔh Δh PV = PV = PV = nrt Initially, V = V Finally V = V P = P P = P Pressure of gas is equal to that created by mass: P = force/area = M g/a where A = piston area g = gravitational acceleration (9.8 m/s )
Isothermal Expansion: One Step (Not Reversible; it is Irreversible) One-Step Expansion. We change the weight to M =M /4, then P ext = P =(M /4)g/A = P /4 Internal pressure is larger than the external pressure so the gas will expand. The work done is defined by the external pressure. The mass will be lifted until the internal pressure equals the external pressure. In this case V = 4V Because nrt = PV = PV = PV q=-w = P ext ΔV= P /4 (4V -V ) = 3/4 P V 3
PV Diagram: One Step vs Reversible Work at Constant Temperature Expansion (green line) raises mass M = M in one step. 4 Compression (red line) lowers mass M in a single step and returns the system to state. w= 3PV = 3nRT 3 3 w= PV = nrt 4 4 4
One Step Compression State of the system after the Step Expansion: Vacuum 4 M V = 4 V P = P initial 4 initial 3 4 M What do we have to do to get the system back to its original state before -step expansion? [Pick up Weight] System did work: Raised M/4; to restore it we have to raise 3M/4 the same distance. We must do more work to restore the system than we got from it in the first place. 5
Reversible Isothermal Expansion: (Infinite number of steps; can go back and forth) Add up areas (integrate) from V to V V w total = PdV = nrt V dv V V V V = nrt(ln(v )) V = nrt ln V V rev V = rev = ln V q w nrt 6
A Thermodynamic Engine: Defining Entropy Imagine doing all steps reversibly, then compression is simpler to understand. The system variables (P,T) are always only incrementally different from the external values. Use two different heat reservoirs at two different temperatures. 4 Let s consider the four-step cycle illustrated: : Isothermal expansion : Isochoric cooling 3: Isothermal compression 3 4: Isochoric heating V V Volume 7
First Step of Four Cycle 4 Step : Isothermal Expansion at T = T high from V to V Now ΔT = 0; therefore, ΔE = 0 and q = -w 3 V V Volume Do expansion reversibly. Then: q = w = nrt high ln V V 8
4 Step 3 V V Volume Step : Isochoric Cooling to T = T low. Now ΔV = 0; therefore, w = 0 q = q V =ΔE = nc v ΔT = nc v (T low -T high ) 9
Step 3 4 Step 3: Isothermal Compression at T = T low from V to V. Now ΔT = 0; therefore, ΔE = 0 and q = -w 3 V V Volume Do compression reversibly, then q 3 = w 3 = nrt low ln V V 0
4 Step 4 3 V V Volume Step 4: Isochoric Heating to T = T high. Now ΔV = 0; therefore, w = 0 q 4 =q V = ΔE = nc v ΔT= nc v (T high -T low ) = -q
4 Add together: All 4 steps of the engine Δ E =Δ H = 0 Total Total (State Function) q total = q + q + q 3 + q 4 3 V V Volume q total = q + q 3 V V qtotal = nrthigh ln + nrtlow ln V V V V q = nrt ln nrt ln V qtotal = nrδ Tln = w 0 V total high low V V total
Joule s Experiment (Z0.6) One mole of and ideal gas at liter, and 5 Atm. expands isothermally into an evaluated bulb to reach a total volume of liters. Calculate q, w and q rev for this change of state. How do we approach this? Write down first law(s) and the formula for work. Δ E = nc Δ T = q + w but Δ T = 0 V w= PextΔ V and Pext = 0 Now that q=w=0, how do we determine the reversible heat? It is still isothermal, and we must find a reversible path. Δ E = nc Δ T = 0 = q + w q = w V rev rev rev rev V wrev = nrtln = ( PV ) ln = 5 0 ln = 500 0.7J = 350J V qrev qrev =+ 350J Δ S = = nrln = 8.3 0.7 = 5 J T K 3