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

Ideal Monatomic Gas Cv = 3/2R Cp = Cv + R = 5/2 R Polyatomic Gas Cv > 3/2R Cp > 5/2 R SUMMARY OF THERMO FORMULAS 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

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

Thermodynamic Pathways: • Example 9.2. Let 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

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

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

Pathway 1 (A to C to B) Step 1 (A to B) Constant pressure expansion (P = 2 atm) from Vi = 10.0 l to Vf = 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 Use ideal gas law PV =nRT to calculate temperature change DT = PDV/nR = 4.05 x 103 J/(2 mol)(8.314 J/mol.K) = 243.6 K

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

Step 2 (C to B) Constant volume cooling until pressure is reduced from 2.00 atm to 1.00 atm: First, calculate DT from ideal gas law: DT = (DP)V/nR (note: P changes, not V) = (-1.00 atm)(30.0 L)/ (2 mol)(.0821 L atm/mol K) = -182.7 K q2 = qv = nCvDT = (2 mol)(3/2R)(-182.7 K) = - 4.6 x 103 J DE2 = nCvDT = -4.6 x 103 J DH2 = nCpDT = -7.6 x 103 J w2 = 0 ( since DV = 0)

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

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

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

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

Pathway 2 (A-->D-->B) Summary Thermodynamic totals for this second pathway are again the sum of values for steps 1 and 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

Pathway 1 q=5.5 x 103 J w=-4.0 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 Comparing Paths 1 and 2 Note: Changes in Energy and Enthalpy are the same for paths 1 and 2, but heat and work are not the same!

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.

Thermodynamic State Functions • Thermodynamic State Functions: Thermodynamic properties which are dependent on the state of the system only, independent of path (Examples: DE and DH) • 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.

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