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Chem 402 Lecture 5: L5 1:27:17
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Washington University in St. Louis

University College - Chemistry

University College - Chemistry Chem 402

Barnes Alexander

Spring

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25 January 2017
L5: When Terms Drop Out, Boltzman Statistics, and QM vs. Classical
I. Thermodynamic Variable Derivations from Q
A. Derivation of the Ideal Gas Law + Insights!
1. ππ΄ = βπππ β πππ + οππ, ( ) ππ΄ = βπ, A = -kTlnQ, Q = q /N!
ππ π,π
πππ(π /π!) πππ(π )βπππ!)
2. π = ππ( ππ ) = ππ( ππ )
π,π π,π
a. After separating the ln term, the N power goes in front of the ln
π 2ππππ 3/2
3. π = πππ ππ (ln π β lnβ‘(π!))π,π, then sub in π‘ππππ π,π) = π β ( β2 )
a. kT is important because it represents the energy of molecules hitting walls
b. The N can drop out in front because pressure scales with the number of molecules
hitting the wall
3
π 2ππππ 2
4. π = πππ ππ (ln(π β ( β2 ) ) β ln π! )
π,π
π 2ππππ 3 πlnβ‘(π) 1
= πππ (ln((π) + lnβ‘( ) ) β lnβ‘(π!)) = πππ( ) = πππ( )
ππ β2 π,π ππ π,π π
a. After you sub in transV, T), separate the ln term
b. (lnN!) drops out because it doesnβt matter how many distinguishable particles you
have with respect to volume
β’ It doesnβt matter if they are distinguishable for pressure
β’ Pressure is just molecules hitting walls, so this term goes to zero
2ππππ
c. The ππ( )3/2 term will drop out because it goes to zero with respect to V
β2
β’ Also, kT and N were already accounted for once, so this cancels
d. Then we are left with P = kT (d/dV of lnV), which is just = 1/V
β’ Makes sense since as volume decreases pressure must increase (physical
insight)
1 π
1
5. π = πππ( ) π π( )π( )πο¨ PVπ= nRNT
π
a. For n moles, N =nN And k = R/N = A = R/(N/n), gives us the ideal gas law!
B. Calculating U = from the Translational Degree of Freedom
1. π’ =β‘ πΈ β‘= ππ 2 π (πππ) = ππ 2 π (ππ(q /N!) ) = ππ 2 π (ππ(q ) β
ππ π,π ππ π,π ππ
lnβ‘(N!) = ππ 2 π (πππ(q) β lnβ‘(N!))
π,π ππ π,π
a. Sub in U and then Q = q /N! and separate out the ln
3
2 π 2ππππ 2 2 π 3/2
2. = ππ ππ (πππ(π β ( β2 ) ) β lnβ‘(N!)) = ππ ππ(πππ(π ) +
π,π
3
(πππ( 2πππ ) β V) β lnβ‘(N!)) = ππ 2 π ( ππππ) = πππ 2π (πππ)
β 2 ππ 2 π,π 2 ππ π,π
π,π
a. Sub in qtransthen separate out T
3
2πππ 2
b. The (πππ( β2 ) β V) term cancels to zero because this is a constant and will not
change with respect to T c. lnN! Also goes to zero since it doesnβt matter if the particles are distinguishable
when trying to calculate energy
d. The (3/2) on T then comes down (ln rules) and can be pulled out since it is a
constant, N is also constant here and can be pulled out
3. 3 πππ 2 π (πππ ) = πππ 2 1ο¨ πΈ β‘= πππ = ππ
π 3
2 ππ π,π 2 π 2 2
a. First derivative of lnT is (1/T)
b. For n moles (N = nN ) and k = R/N , gives us the for a monoatomic ideal
A A
gas
Μ
Μ
Μ
Μ
3
c. In molar quantity: πΈ β‘= π
π 2
C. Specific Heat Capacity at Constant Volume (C ) v
1. Since C pf water = 4.184 J/Β°Cβ’g, essentially energy over temp
2. Molar C (vor a monoatomic gas in the electronic ground state)
π( π
π)
a. πΆ = ( ) ππ = ( 2 ) = π
3

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