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Water’s Phase Diagram Source: P.W. Atkins, Physical Chemistry, 2 ed., 1978, p.193.
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The Final Monday, 1:15–3:15 PM, here About 50% old and 50% new stuff 300 points (double other tests) MC, essay, worked problems Calculators permitted, probably unneeded
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Thermodynamic Processes molecular models § 19.5–19.7
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Named Path Types No heat transfer Q = 0 U = –W Examples: Rapid expansion or compression (piston engine) Large system (atmospheric parcels) Adiabatic
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Named Path Types Constant volume W = 0 U = Q Examples: Rigid containers Engine spark Bomb calorimetry Isochoric
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Named Path Types Constant pressure W = p V Examples: Open-air processes Most lab chemistry Isobaric
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Named Path Types Constant temperature Examples: Slow processes (thermal equilibrium) Thermostatted systems If an ideal gas: U = 0, so Q = W constant pV Isothermal
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Named Path Types Expansion at zero pressure W = 0 Q = 0 U = 0 Absolutely irreversible If Ideal Gas: T = 0 No work done on individual molecules Free Expansion
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Free Expansion of Real Gases T < 0 expand against mutual attraction of molecules U = 0 anyway Potential energy gain from separation of molecules, so they slow down
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CPS Question Water ice melting at 0 °C is an example of a(n) process. (Add correct answers together and enter the sum.) 1.adiabatic 2.isochoric 4.isothermal 8.isobaric 16.free expansion
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CPS Question A hot-air balloon expanding as it rises is an example of a(n) process. (Add correct answers together and enter the sum.) 1.adiabatic 2.isochoric 4.isothermal 8.isobaric 16.free expansion
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CPS Question The crew of Soyuz-11 died during re-entry shortly after a pressure seal failed at an altitude of 168 km. This disaster was an example of a(n) process. (Add correct answers together and enter the sum.) 1.adiabatic 2.isochoric 4.isothermal 8.isobaric 16.free expansion Source: Wikimedia Commons
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Ideal Gases U = f(T) and nothing else! Monatomic ideal gas U = 3/2 NkT Diatomic ideal gas U = 5/2 NkT etc. U = nC v T No intermolecular potentials
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Constant-Volume Heating dU = dK + pdV K tr = 3/2 NkT dK tr = 3/2 NkdT dV= 0 nC v = dU/dT = 3/2 Nk = 3/2 nR C v = 3/2 R C v of a monatomic ideal gas
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CPS Question To raise the temperature of a mole of ideal from T 1 to T 2 at constant pressure requires the same temperature increase at constant volume. A.less heat than B.the same amount of heat as C.more heat than D.The processes cannot be compared.
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Constant-Pressure Heating Some internal energy becomes work Source: Young and Freedman, Fig. 19.4a
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Constant-Pressure Heating dU = dK + pdV K tr = 3/2 NkT V = NkT/p nC v = dU/dT = dK/dT + dV/dT = 3/2 Nk + pNk/p = 5/2 Nk = 5/2 nR C v = 5/2 R C p of a monatomic ideal gas
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Any Ideal Gas C p = C v + R C v is energy to increase molecular K 1/2 kT/molecule = 1/2 RT/mole per mode R is work to expand against constant p p V = p (NkT/p) = Nk T = nR T/mole No complication from intermolecular interactions
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Heat Capacity Ratio = C p /C v > 1 always Useful for analyzing adiabatic processes (§19.8)
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Group Work 1.Qualitatively sketch a pV plot for each described process A B. a)System is heated at constant pressure until volume doubles, then cooled at constant volume to the initial temperature. b)System is heated at constant volume until its absolute temperature doubles, allowed to expand at constant temperature to twice its volume, then cooled at constant volume to the initial temperature. c)System is allowed to expand into a vacuum (free expansion) to twice its volume. d)Volume is gradually doubled while maintaining a constant temperature.
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Group Work 2.Give the formulas for W of each process. 3.Give the formulas for Q of each process.
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Example Problem 19.38 A cylinder contains 0.1 moles of an ideal monatomic gas initially at pressure 1.0 10 5 Pa and volume 2.5 10 –3 m 3. a)Find the initial temperature of the gas. b)If the gas is allowed to expand to twice its initial volume, find the final temperature and pressure if the expansion is i.isothermal ii.isobaric iii.adiabatic
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Otto Cycle Source: Young and Freedman, Fig. 20.5
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