Lecture12Ch161Mar4

# Lecture12Ch161Mar4 - Summary of last lecture Paramagnetism...

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Summary of last lecture Paramagnetism Maxwell Distribution Classical CPF

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Key Concepts and Lessons: a) Maxwell Distribution of velocities, mean velocity b) Classical Harmonic Oscillator c) Quantum Harmonic Oscillator d) Heat capacity of solids-Einstein model Reading: Reif, Ch6
Maxwell Distribution We can now focus on any particle (not necessarily in an ideal gas but in a classical system) and ask what is the probability distribution of its velocity. Consider, say, particle ‘’1’’ out of N in our system: P p 1 , p 2 ....... p N ( ) = dq 1 dq 2 ... dq N e 1 2 mkT p i 2 i = 1 N e 1 kT U q 1 , q 2 ... q N ( ) Z = dq 1 dq 2 ... dq N e 1 2 mkT p i 2 i = 1 N e 1 kT U q 1 , q 2 ... q N ( ) dq 1 dq 2 ... dq N dp 1 dp 2 ... dp N e 1 2 mkT p i 2 i = 1 N e 1 kT U q 1 , q 2 ... q N ( ) = = e 1 2 mkT p i 2 i = 1 N dp 1 dp 2 ... dp N e 1 2 mkT p i 2 i = 1 N = 1 2 π mkT i = 1 N 3/2 e p i 2 2 mkT and we get famous Maxwell distribution for velocities of each individual particle (hence tilde sign below): P v x , v y , v z ( ) dv x dv y dv z = m 2 kT 3/2 e m 2 kT v x 2 + v y 2 + v z 2 ( ) dv x dv y dv z which gives us the probability that velocity of the particle lies in the range v x , v x + dv x ; v x , v y + dv y ; v x , v z + dv z ( ) ( note change of normalization as we switched from p to v)

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Maxwell Distribution, Cont Distribution of speeds. Now consider a system of N molecules and ask what fraction of them have absolute value of speed between v and v +dv. This is given by the relation: This distribution is not symmetric! f ( v ) f ( v ) dv = 4 π P v x , v y , v z ( ) v 2 dv = 4 m 2 kT 3 v 2 e mv 2 2 kT dv we used a relation 4 v 2 dv for an element of volume in 3-dimensional v-space
Maxwell Distribution- mean speed We can evaluate mean speed of a particle as: v = vf ( v ) dv = 4 π m 2 kT 0 3/2 e mv 2 2 kT 0 v 3 dv = 4 m 2 kT 3/2 1 2 m 2 kT 2 ; v = 8 kT m On the other hand means SQUARED speed (related to kinetic energy): 1 2 mv 2 = E k = 4 m 2 kT 3/2 v 4 e mv 2 2 kT dv = 0 4 m 2 kT 3/2

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## This note was uploaded on 05/04/2010 for the course CHEM 161 taught by Professor Shaklovich during the Spring '10 term at Harvard.

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Lecture12Ch161Mar4 - Summary of last lecture Paramagnetism...

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