Lecture 2: Heat Capacities/State functions

1. Lecture 2: Heat Capacities/State functions • Reading: Zumdahl 9.3 • Outline • Definition of Heat Capacity (Cv and Cp) • Calculating DE and DH using Cv and Cp • Example of Thermodynamic Pathways • State Functions

2. Heat Capacity at Constant V • Recall from Chapter 5 (section 5.6): (KE)ave = 3/2RT (ideal monatomic gas) • Temperature is a measure of molecular speed • In thermodynamic terms, an increase in system temperature corresponds to an increase in system kinetic energy ( i.e., q is proportional to <KE>)

3. Heat Capacity at Constant V • (KE)ave = 3/2RT (ideal monatomic gas) • How much energy in the form of heat is required to change the gas temperature by an amount DT? Heat required = 3/2R DT = 3/2R (for DT = 1K) • Therefore, Cv = 3/2 R is the heat required to raise one mole of an ideal gas by 1K. Cv is called constant volume molar heat capacity.

4. Heat Capacity at Constant P • What about at constant pressure? In this case, PV type work can also occur: PDV = nRDT = RDT (for 1 mole) = R (for DT = 1 K) • Cp = “heat into translation” + “work to expand the gas” = Cv + R = 5/2R (for ideal monatomic gas)

5. Cv for Monatomic Gases • What are the energetic degrees of freedom for a monatomic gas? • Just translations, which contribute 3/2R to Cv.

6. Cv for Polyatomics • What are the energetic degrees of freedom for a polyatomic gas? Translations, rotations, and vibrations. All of which may contribute to Cv (depends on T).

7. Cv for Polyatomics • When heat is provided , molecules absorb energy and the translational kinetic energy increases • In polyatomic gases, rotational and vibrational kinetic energies increase as well (depending on T).

8. Cv for Polyatomics • T measures the average translational kinetic energy • Increases in rotational and vibrational kinetic energies do not increase T directly • It takes more heat to increase T by the same amount (Cv/Cp larger)

9. Variation in Cp and Cv • Monatomics: • Cv = 3/2 R • Cp = 5/2 R • Polyatomics: • Cv > 3/2 R • Cp > 5/2 R • But….Cp = Cv + R

10. Energy and Cv • Recall from Chapter 5: Eave = 3/2 nRT (average translational energy) DE = 3/2 nR DT DE = n CvDT (since 3/2 R = Cv) • Why is Cv=DE/DT When heating our system at constant volume, all heat goes towards increasing E (no work).

11. Enthalpy and Cp • What if we heated our gas at constant pressure? Then, we have a volume change such that work occurs: q p = n CpDT = n (Cv + R) DT = DE + nRDT = DE + PDV = DH or DH = nCpDT

12. Ideal Monatomic Gas Cv = 3/2R Cp = Cv + R = 5/2 R Polyatomic Gas Cv > 3/2R Cp > 5/2 R All Ideal Gases • DE = nCvDT • DH = nCpDT Keeping Track

13. Example • What is q, w, DE and DH for a process in which one mole of an ideal monatomic gas with an initial volume of 5 l and pressure of 2.0 atm is heated until a volume of 10 l is reached with pressure unchanged? Pinit = 2 atm Pfinal = 2 atm Vinit = 5 l Vfinal = 10 l Tinit = ? K Tfinal = ? K

14. Example (cont.) • Since PDV = nRDT (ideal gas law) we can determine DT • Then DV = (10 l - 5 l) = 5 l • And:

15. Example (cont.) • Given this:

16. Ideal Monatomic Gas Cv = 3/2R Cp = Cv + R = 5/2 R Polyatomic Gas Cv > 3/2R Cp > 5/2 R To Date…. All Ideal Gases • DE = q + w • w = -PextDV (for now) • DE = nCvDT = qV • DH = nCpDT = qP • If DT = 0, then DE = 0 and q = -w

17. State Functions • If we start in Seattle and end in Chicago, but you take different paths to get from one place to the other .. • Will the energy/enthalpy, heat/work we spend be the same along both paths?

18. Thermodynamic Pathways: an Example • Example 9.2. We take 2.00 mol of an ideal monatomic gas undergo the following: • Initial (State A): PA = 2.00 atm, VA = 10.0 L • Final (State B): PB = 1.00 atm, VB = 30.0 L • We’ll do this two ways: Path 1: Expansion then Cooling Path 2: Cooling then Expansion

19. Thermodynamic Jargon • When doing pathways, we usually keep one variable constant. The language used to indicate what is held constant is: • Isobaric: Constant Pressure • Isothermal: Constant Temperature • Isochoric: Constant Volume

20. Thermodynamic Path: A series of manipulations of a system that takes the system from an initial state to a final state isochoric isobaric

21. Pathway 1 • Step 1. Constant pressure expansion (P = 2 atm) from 10.0 l to 30.0 l. • PDV = (2.00 atm)(30.0 l - 10.0 l) = 40.0 l.atm = (40.0 l.atm)(101.3 J/l.atm) = 4.0 x 103 J = -w (the system does work) • And DT = PDV/nR = 4.05 x 103 J/(2 mol)(8.314 J/mol.K) = 243.6 K (from the ideal gas law)

22. Pathway 1 (cont.) • Step 1 is isobaric (constant P); therefore, • q1 = qP = nCpDT = (2mol)(5/2R)(243.6 K) = 1.0 x 104 J = DH1 • And DE1 = nCvDT = (2mol)(3/2R)(243.6 K) = 6.0 x 103 J (check: DE1 = q1 + w1 = (1.0 x 104 J) -(4.0 x 103 J) = 6.0 x 103 J )

23. Pathway 1 (cont.) • Step 2: Isochoric (const. V) cooling until pressure is reduced from 2.00 atm to 1.00 atm. • First, calculate DT: • Now, DT = DPV/nR (note: P changes, not V) = (-1.00 atm)(30.0 l)/ (2 mol)(.0821 l.atm/mol K) = -182.7 K

24. Pathway 1 (cont.) • q2 = qv = nCvDT = (2 mol)(3/2R)(-182.7 K) = - 4.6 x 103 J • and DE2 = nCvDT = -4.6 x 103 J • and DH2 = nCpDT = -7.6 x 103 J • Finally w2 = 0 (isochoric…no V change, no PV-type work)

25. Pathway 1 (end) • Thermodynamic totals for this pathway are the sum of values for step 1 and step 2 q = q1 + q2 = 5.5 x 103 J w = w1 + w2 = -4.0 x 103 J DE = DE1 + DE2 = 1.5 x 103 J DH = DH1 + DH2 = 2.5 x 103 J

26. Next Pathway • Now we will do the same calculations for the green path.

27. Pathway 2 • Step 1: Isochoric cooling from P = 2.00 atm to P = 1.00 atm. • First, calculate DT: • DT = DPV/nR = (-1.00 atm)(10.0 l)/ (2 mol)R • = -60.9 K

28. Pathway 2 (cont.) • Then, calculate the rest for Step 1: q1 = qv = nCvDT = (2 mol)(3/2 R)(-60.9 K) = -1.5 x 103 J = DE1 DH1 = nCPDT = (2 mol)(5/2 R)(-60.9 K) = -2.5 x 103 J w1 = 0 (constant volume)

29. Pathway 2 (cont.) • Step 2: Isobaric (constant P) expansion at 1.0 atm from 10.0 l to 30.0 l. DT = PDV/nR = (1 atm)(20.0 l)/(2 mol)R = 121.8 K

30. Pathway 2 (cont.) • Then, calculate the rest: q2 = qp = nCPDT = (2 mol)(5/2 R)(121.8 K) = 5.1 x 103 J = DH2 DE2 = nCvDT = (2 mol)(3/2 R)(121.8 K) = 3.1 x 103 J w1 = -PDV = -20 l.atm = -2.0 x 103 J

31. Pathway 2 (end) • Thermodynamic totals for this pathway are again the sum of values for step 1 and step 2: • q = q1 + q2 = 3.6 x 103 J • w = w1 + w2 = -2.0 x 103 J • DE = DE1 + DE2= 1.5 x 103 J • DH = DH1 + DH2 = 2.5 x 103 J

32. Pathway 1 q=5.5 x 103 J w=-4.1 x 103 J DE= 1.5 x 103 J DH=2.5 x 103 J Pathway 2 q = 3.6 x 103 J w = -2.0 x 103 J DE = 1.5 x 103 J DH = 2.5 x 103 J Comparison of Path 1 and 2 Note: Energy and Enthalpy are the same, but heat and work are not the same!

33. State Functions • A State Function is a function in which the value only depends on the initial and final state….NOT on the pathway taken

34. Thermodynamic State Functions • Thermodynamic State Functions: Thermodynamic properties that are dependent on the state of the system only. (Example: DE and DH) • Other variables will be dependent on pathway (Example: q and w). These are NOT state functions. The pathway from one state to the other must be defined.