lecturenotes-May3

lecturenotes-May3 - Final Review: 130 Nicholson, 6 - 7 pm...

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Final Review : 130 Nicholson, 6 - 7 pm Friday (May 6 th )
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v rms = 3 pV nM 1/2 = 3 nRT nM v rms = 3 RT M p = nM v 2 ( ) ave 3 V pressure The RMS velocity depends on: Molar mass & Temperature Ideal Gas Law pV = nRT pV = Nk B T Boltzmann Constant k B = 1.38 × 10 -23 J/K R = k B N A gas constant R = 8.315 J/(mol K) = 0.0821 (L atm)/(mol K) = 1.99 calories/(mol K) n = number of moles N = number of particles ( v 2 ) avg = v rms
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KE = 3 2 k B T The KE of all ideal gas molecules depends only on the temperature (not mass!) E int, monotonic = N 3 2 k B T ( ) = 3 2 nRT The internal energy of an ideal gas depends only on the temperature Δ E int, monotonic = 3 2 nR Δ T ( ) KE = 1 2 m v rms ( ) 2 = 1 2 m 3 RT M M = mN A k = R N A B
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W by = pdV = area under p V graph 3)Cyclical process (closed cycle) Δ E int,closed cycle =0 a) net area in p-V curve is Q Δ E int = 0 Q = W Δ E int = W [ ] adiabatic Adiabatic expansion/contraction - NO TRANSFER OF ENERGY AS HEAT Q = 0 Δ E int = Q Constant-volume processes (isochoric)- NO WORK IS DONE W = 0 W by = pdV V i V f = V i = 0 Q Δ V = 0 = nC V Δ T W by = pdV V i V f = V i = p Δ V Q Δ P = 0 = nC P Δ T Constant-pressure processes Δ E int = Q p Δ V C V = C P R
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Chapter 20: Entropy and the Second law of thermodynamics 0 th law Thermal Equilibrium: A = B & B = C then A = C Q = nC Δ T Q 0 as T 0 1 st law Conservation of energy: E int = Q - W by = Q + W on Change in Internal energy = heat added minus work done by 2 nd law HEAT FLOWS NATURALLY FROM HOT OBJECT TO A COLD OBJECT Heat will NOT fow spontaneously ±rom cold to hot Today:
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Hall of fame Boltzmann constant k B =1.38 × 10 -23 J/K Ludwig Boltzmann (1844-1906)
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Irreversible Processes
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How to understand this: Entropy How to describe a system: P, T, V, E int , and n Entropy S, like T,V, P, E int , and n is state variable What is a process? expansion, compression, temperature rise, add mass Reversible vs irreversible {processes can be done infnitely slowly to ensure thermal equilibrium} [ Note: since T > 0, iF Q is positive (negative) the Δ S is positive (negative) ] system How to defne entropy? Easier to defne Change of entropy during a process. 0 ≤ Δ S total where Q is energy transFerred to or From a system during a process Temp in Kelvin Units: [J/K] Δ S part = S f S i = dQ T i f
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