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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. To Date…. All Ideal Gases • D E = q + w • w = -P ext D V (for now) • D E = nC v D T = q V • D H = nC p D T = q P • If D T = 0, then D E = 0 and q = -w.

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  1. 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

  2. Lecture 3: State Functions • Reading: Zumdahl 9.3, 9.4 • Outline • Example of Thermo. Pathways • State Functions • (9.4…for laboratory)

  3. Thermodynamic Pathways: an Example • Example 9.2. We take 2.00 mol of an ideal monatomic gas undergo the followng: • 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

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

  6. 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 • And DT = PDV/nR = 4.05 x 103 J/(2 mol)(8.314 J/mol.K) = 243.6 K

  7. 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 )

  8. 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

  9. 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)

  10. 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

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

  12. 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

  13. 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)

  14. 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

  15. 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

  16. 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

  17. 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!

  18. 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. • In this example, start in Seattle, end in Chicago, but you take different paths to get from one place to the other.

  19. 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.

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