1. 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 SUMMARY OF THERMO FORMULAS

2. Lecture 3: State Functions Reading: Zumdahl 9.3, 9.4 Outline: Example of Thermodynamic Pathways State Functions

3. Thermodynamic Pathways: Example 9.2. Let 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

4. 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

5. Thermodynamic Path: The series of steps a system takes in going from an initial state to a final state const v isobaric

6. Pathway 1 (A to C to B) Step 1 (A to B) Constant pressure expansion (P = 2 atm) from V i = 10.0 l to V f = 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 Use ideal gas law PV =nRT to calculate temperature change  T = P  V/nR = 4.05 x 10 3 J/(2 mol)(8.314 J/mol.K) = 243.6 K

7. 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

8. Step 2 (C to B) Constant volume cooling until pressure is reduced from 2.00 atm to 1.00 atm: First, calculate  T from ideal gas law:  T = (  P)V/nR (note: P changes, not V) = (-1.00 atm)(30.0 L)/ (2 mol)(.0821 L atm/mol K) = -182.7 K q 2 = q v = nC v  T = (2 mol)(3/2R)(-182.7 K) = - 4.6 x 10 3 J  E 2 = nC v  T = -4.6 x 10 3 J  H 2 = nC p  T = -7.6 x 10 3 J w 2 = 0 ( since  V = 0)

9. Pathway 1(A-->C-->B) Summary 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

10. The other Pathway (A-->D-->B)

11. Step 1(A-->D): Constant volume cooling from P = 2.00 atm to P = 1.00 atm. First, calculate  T from ideal gas law:  T =  PV/nR = (-1.00 atm)(10.0 L)/ (2 mol)R  = -60.9 K 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 ( since constant volume)

12. Step 2(D-->B): Isobaric (constant P) expansion at 1.0 atm from V= 10.0 l to 30.0 l.  T = P  V/nR = (1 atm)(20.0 L)/(2 mol)R = 121.8 K 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

13. Thermodynamic totals for this second pathway are again the sum of values for steps 1 and 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 (A-->D-->B) Summary

14. Comparing Paths 1 and 2 Pathway 1 q = 5.5 x 10 3 J w = -4.0 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: Changes in Energy and Enthalpy are the same for paths 1 and 2, but heat and work are not the same!

15. 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.

16. Thermodynamic State Functions Thermodynamic State Functions: Thermodynamic properties which are dependent on the state of the system only, independent of path (Examples:  E and  H) Other variables will be dependent on pathway (Examples: q and w). These are NOT state functions. The pathway from one state to the other must be defined.