N particles are distributed amongst three levels having energies 0, kT and 2kT. If
the total equilibrium energy of the system is approximately 425kT, what is the value
of N?
Thnx
stat mech question
Re: stat mech question
Let me try:
Total Energy = Sum [ (Number of particles in a state i) * (energy of state i) ]
425kT = N exp(0)*0 + N exp(1)*kT + N exp(2)*2kT
= NkT [ exp(1) + 2 exp(2) ]
= .503 NkT
N = 845
Total Energy = Sum [ (Number of particles in a state i) * (energy of state i) ]
425kT = N exp(0)*0 + N exp(1)*kT + N exp(2)*2kT
= NkT [ exp(1) + 2 exp(2) ]
= .503 NkT
N = 845
Re: stat mech question
er.. i think each of the 3 levels does not have N particles  its more like n1+n2+n3 = Nmatonski wrote:Let me try:
Total Energy = Sum [ (Number of particles in a state i) * (energy of state i) ]
425kT = N exp(0)*0 + N exp(1)*kT + N exp(2)*2kT
= NkT [ exp(1) + 2 exp(2) ]
= .503 NkT
N = 845
And the choices in the answer are
1001
335
425
390
181
Re: stat mech question
n1*E1 + n2*E2 + n3*E3 = 425kT
where n1+n2+n3 = N
P(n1)/P(n2) = n1/n2 = exp(0*kt/kT)/exp(kt/kT) = e
Similarly, n1/n3 = e^2
Hence 0*exp(0)kT + (n2*exp(1))kT + (n3*exp(2))*2kT = 425kT, which has n1 = 665. Then we use the above relative n relations to get n2 = 244, n3 = 90.
These numbers have N = 999 (numerical error suggests it should be the first option)
where n1+n2+n3 = N
P(n1)/P(n2) = n1/n2 = exp(0*kt/kT)/exp(kt/kT) = e
Similarly, n1/n3 = e^2
Hence 0*exp(0)kT + (n2*exp(1))kT + (n3*exp(2))*2kT = 425kT, which has n1 = 665. Then we use the above relative n relations to get n2 = 244, n3 = 90.
These numbers have N = 999 (numerical error suggests it should be the first option)
Last edited by WakkaDojo on Mon Aug 24, 2009 7:54 pm, edited 1 time in total.
Re: stat mech question
My thought was that N x the probability being in state i = ni. My mistake was that exp(E/kT) is not the probability of being in a state because it's not normalized. Oh well. Shows that I need to review.rohit wrote:er.. i think each of the 3 levels does not have N particles  its more like n1+n2+n3 = N
And the choices in the answer are
Re: stat mech question
thats it! thnxWakkaDojo wrote:n1*E1 + n2*E2 + n3*E3 = 425kT
where n1+n2+n3 = N
P(n1)/P(n2) = n1/n2 = exp(0kT)/exp(kT) = e
Similarly, n1/n3 = e^2
Hence 0*exp(0)kT + (n1*exp(1))kT + (n2*exp(2))*2kT = 425kT, which has n1 = 665. Then we use the above relative n relations to get n2 = 244, n3 = 90.
These numbers have N = 999 (numerical error suggests it should be the first option)

 Posts: 378
 Joined: Wed Mar 26, 2008 9:14 am
Re: stat mech question
WakkaDojo wrote: Hence 0*exp(0)kT + (n1*exp(1))kT + (n2*exp(2))*2kT = 425kT, which has n1 = 665. Then we use the above relative n relations to get n2 = 244, n3 = 90.
shouldn't this be 0*exp(0)kT + (n2*exp(1))kT + (n3*exp(2))*2kT = 425kT ??
and i am still not getting n1=665
Re: stat mech question
Yes, sorry for the typo (at the time I was quite used to indexing things n0, n1, ... instead of n1, n2, ... since I was doing a lot of computational work). But my calculations are still correct:blackcat007 wrote:WakkaDojo wrote: Hence 0*exp(0)kT + (n1*exp(1))kT + (n2*exp(2))*2kT = 425kT, which has n1 = 665. Then we use the above relative n relations to get n2 = 244, n3 = 90.
shouldn't this be 0*exp(0)kT + (n2*exp(1))kT + (n3*exp(2))*2kT = 425kT ??
and i am still not getting n1=665
n1/n2 = e > n2 = n1 / e
n1/n2 = e^2 > n3 = n1 / e^2
So we have
n1 (e^{1} + 2*e^{2}) = 425, and we get n1 = 665. Then just plug in n2 = n1 / e, and n3 = n2 / e.