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Chem 300 - Ch 20/#1 Today’s To Do List

Chem 300 - Ch 20/#1 Today’s To Do List. Start Chapter 20: Spontaneous Processes Entropy. Spontaneous Processes. Mechanical Processes Falling Ball - Decrease in Potential Energy. Spontaneous Processes. Spontaneous reactions: 2H 2 + O 2 --> 2H 2 O  H < 0

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Chem 300 - Ch 20/#1 Today’s To Do List

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  1. Chem 300 - Ch 20/#1 Today’s To Do List • Start Chapter 20: • Spontaneous Processes • Entropy

  2. Spontaneous Processes • Mechanical Processes • Falling Ball - • Decrease in Potential Energy

  3. Spontaneous Processes • Spontaneous reactions: • 2H2 + O2 --> 2H2O H < 0 • H2O(s) --> H2O(l) at +5 oC H >0 • What determines them? • More than U or H

  4. 1st Law not sufficient • Even H = 0 processes can be spontaneous

  5. Spontaneous flow into Vacuum

  6. Spontaneous Mixing

  7. --->> Increased Disorder? No!! • Spontaneity associated with • lowered energy (U or H) • increased energy dispersal • or some combination of these • Find State function to measure dispersal of energy

  8. Effect of q on T, V of IDG • dqrev = dU - dw = Cv(T)dT + PdV = Cv(T)dT + nRTdV/V • dqrev is not a state function • thus right side can’t be state function • but Cv(T)dT is fcn just of T so is state fcn • 2nd term is a work term & not state fcn

  9. Continued... • dqrev = Cv(T)dT + nRTdV/V • Divide all by T: • dqrev/T = Cv(T)dT/T + nRdV/V • Now both terms are integrable • Right side is exact differential • dqrev/T now is an exact differential (the derivative of a state function)

  10. Entropy • dS = dqrev/T • S = entropy • is a state function • path independent

  11. Calc. Q & S

  12. Summary • Path A (rev. isothermal expan) • qrev = nRT1ln(V2/V1) • S = nRln(V2/V1) • Path BC (adiabatic + const V) • qrev = Cv(T)dT (from T2 to T1) • dS = dqrev/T = Cv(T)dT/T • S = Cv(T)dT/T = Cv(T)ln(T1/T2) • = nRln(V2/V1) [Refer to Ch. 19, lecture 2, slide #7]

  13. More Summary • Path DE (const P + const V)(pp.823-4) • The Const P path (T1 T3 & V1V2) • dU = CVdT = dqrev - P1dV (P1 = nRT/V) • dqrev = CVdT + P1dV = CVdT + nRTdV/V • The Const V path (dV = 0, T3 T1) • dU = CVdT = dqrev - 0 • Total: the two CVdT terms cancel: S = nR dV/V = nRln(V2/V1) • Thus all 3 paths have same S

  14. Next Time • 2nd Law • Entropy Changes • Carnot Cycle • Efficiency of Heat Engines

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