2. Lecture 2: Heat Capacities/State functions Reading: Zumdahl 9.3 Outline –Definition of Heat Capacity (C v and C p ) –Calculating  E and  H using C v and C p –Example of Thermodynamic Pathways –State Functions

3. 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 )

4. 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  T? Heat required = 3/2R  T = 3/2R (for  T = 1K) Therefore, C v = 3/2 R is the heat required to raise one mole of an ideal gas by 1K. C v is called constant volume molar heat capacity.

5. Heat Capacity at Constant P What about at constant pressure? In this case, PV type work can also occur: P  V = nR  T = R  T (for 1 mole) = R (for  T = 1 K) C p = “heat into translation” + “work to expand the gas” = C v + R = 5/2R (for ideal monatomic gas)

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

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

8. C v 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).

9. C v 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 (C v /C p larger)

10. Variation in C p and C v Monatomics: –C v = 3/2 R –C p = 5/2 R Polyatomics: –C v > 3/2 R –C p > 5/2 R –But….C p = C v + R

11. Energy and C v Recall from Chapter 5: E ave = 3/2 nRT (average translational energy)  E = 3/2 nR  T  E = n C v  T (since 3/2 R = C v ) Why is C v =  E/  T When heating our system at constant volume, all heat goes towards increasing E (no work).

12. Enthalpy and C p What if we heated our gas at constant pressure? Then, we have a volume change such that work occurs: q p = n C p  T = n (C v + R)  T =  E + nR  T =  E + P  V =  H or  H = nC p  T

13. Keeping Track Ideal Monatomic Gas C v = 3/2R C p = C v + R = 5/2 R Polyatomic Gas C v > 3/2R C p > 5/2 R All Ideal Gases  E = nC v  T  H = nC p  T

14. Example What is q, w,  E and  H 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? P init = 2 atm V init = 5 l T init = ? K P final = 2 atm V final = 10 l T final = ? K

15. Example (cont.) Since P  V = nR  T (ideal gas law) we can determine  T Then  V = (10 l - 5 l) = 5 l And:

16. Example (cont.) Given this:

17. To Date…. Ideal Monatomic Gas C v = 3/2R C p = C v + R = 5/2 R Polyatomic Gas C v > 3/2R C p > 5/2 R All Ideal Gases  E = q + w w = -P ext  V (for now)  E = nC v  T = q V  H = nC p  T = q P If  T = 0, then  E = 0 and q = -w

18. 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?

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

20. 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

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

22. Pathway 1 Step 1. Constant pressure expansion (P = 2 atm) from 10.0 l to 30.0 l. –P  V = (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 10 3 J = -w (the system does work) –And  T = P  V/nR = 4.05 x 10 3 J/(2 mol)(8.314 J/mol.K) = 243.6 K (from the ideal gas law)

23. Pathway 1 (cont.) Step 1 is isobaric (constant P); therefore, –q 1 = q P = nC p  T = (2mol)(5/2R)(243.6 K) = 1.0 x 10 4 J =  H 1 –And  E 1 = nC v  T = (2mol)(3/2R)(243.6 K) = 6.0 x 10 3 J (check:  E 1 = q 1 + w 1 = (1.0 x 10 4 J) -(4.0 x 10 3 J) = 6.0 x 10 3 J )

24. Pathway 1 (cont.) Step 2: Isochoric (const. V) cooling until pressure is reduced from 2.00 atm to 1.00 atm. First, calculate  T: –Now  T =  PV/nR (note: P changes, not V) = (-1.00 atm)(30.0 l)/ (2 mol)(.0821 l.atm/mol K) = -182.7 K

25. Pathway 1 (cont.) q 2 = q v = nC v  T = (2 mol)(3/2R)(-182.7 K) = - 4.6 x 10 3 J and  E 2 = nC v  T = -4.6 x 10 3 J and  H 2 = nC p  T = -7.6 x 10 3 J Finally w 2 = 0 (isochoric…no V change, no PV-type work)

26. Pathway 1 (end) Thermodynamic totals for this pathway are the sum of values for step 1 and step 2 q = q 1 + q 2 = 5.5 x 10 3 J w = w 1 + w 2 = -4.0 x 10 3 J  E =  E 1 +  E 2 = 1.5 x 10 3 J  H =  H 1 +  H 2 = 2.5 x 10 3 J

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

28. Pathway 2 Step 1: Isochoric cooling from P = 2.00 atm to P = 1.00 atm. First, calculate  T:  T =  PV/nR = (-1.00 atm)(10.0 l)/ (2 mol)R  = -60.9 K

29. Pathway 2 (cont.) Then, calculate the rest for Step 1: q 1 = q v = nC v  T = (2 mol)(3/2 R)(-60.9 K) = -1.5 x 10 3 J =  E 1  H 1 = nC P  T = (2 mol)(5/2 R)(-60.9 K) = -2.5 x 10 3 J w 1 = 0 (constant volume)

30. Pathway 2 (cont.) Step 2: Isobaric (constant P) expansion at 1.0 atm from 10.0 l to 30.0 l.  T = P  V/nR = (1 atm)(20.0 l)/(2 mol)R = 121.8 K

31. Pathway 2 (cont.) Then, calculate the rest: q 2 = q p = nC P  T = (2 mol)(5/2 R)(121.8 K) = 5.1 x 10 3 J =  H 2  E 2 = nC v  T = (2 mol)(3/2 R)(121.8 K) = 3.1 x 10 3 J w 1 = -P  V = -20 l.atm = -2.0 x 10 3 J

32. Thermodynamic totals for this pathway are again the sum of values for step 1 and step 2: q = q 1 + q 2 = 3.6 x 10 3 J w = w 1 + w 2 = -2.0 x 10 3 J  E =  E 1 +  E 2 = 1.5 x 10 3 J  H =  H 1 +  H 2 = 2.5 x 10 3 J Pathway 2 (end)

33. Comparison of Path 1 and 2 Pathway 1 q = 5.5 x 10 3 J w = -4.1 x 10 3 J  E = 1.5 x 10 3 J  H = 2.5 x 10 3 J Pathway 2 q = 3.6 x 10 3 J w = -2.0 x 10 3 J  E = 1.5 x 10 3 J  H = 2.5 x 10 3 J Note: Energy and Enthalpy are the same, but heat and work are not the same!

34. 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

35. Thermodynamic State Functions Thermodynamic State Functions: Thermodynamic properties that are dependent on the state of the system only. (Example:  E and  H) 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.