molecular thermodynamics 1. This problem is intended to show the structure of th
ID: 3041112 • Letter: M
Question
molecular thermodynamics
1. This problem is intended to show the structure of the partition function for a system consisting of weakly interacting distinguishable particles. It starts with an example. Consider a system that consists of two distinguishable particles, a and b. Suppose that there are two energy states that are available to each particle with energies 1 and 2. Particles a and b can each have either of these energies. Since the particles are distinguishable, there are no restrictions on the number of particles in the states. In this case, the possible assignments of particles to states are listed in the table: Particle States of the two-particle system 4 The energies of the states of the two-particle system are the sums of the energies of particles a and b. For example, the energy of state l of the two-particle system is E,- + . Then the partition function for the case of two distinguishable particles with two available states is a) Show that the partition function above factorizes into the form b) Now consider the case with two distinguishable particles and three energy states that are available to each particle, , , and . Find the partition function in this case, either by preparing a table that corresponds to the table above or directly. c) Show that the partition function from part b factorizes intoExplanation / Answer
Ans: By means of weakly interacting particles is they are allowed to exchange the energy among themselves. However, no amount of energy is stored within any interaction potential. Hence total energy can be considered as a sum of the individual energies.
So, if we have a N particle system, particle N being present in energy level Kn with energy EKn.
Then total energy is given by the relation
E = EK1 + EK2 + EK3 + -------
In this problem, we have two distinguishable particles with. Hence probability for the occurrence will be simple 2! = 4.
A)
So, Q = exp[(1+1)/kT] + exp[(1+2)/kT] + exp[(2+1)/kT] + exp[(2+2)/kT]
Q = exp(1/kT)[exp(1)/kT + exp(2)/kT] +exp(2/kT)[ exp(1)/kT + exp((2)/kT] on taking out common factor from exponential terms.
Q = [exp(1/kT) + exp(2/kT)] [exp(1)/kT + exp(2)/kT] Proof
B)
States
a
b
1
1
1
2
1
2
3
1
3
4
2
1
5
2
2
6
2
3
7
3
1
8
3
2
9
3
3
Hence, we will have 9 possible configurations for both the particles.
3! = 9
And corresponding particle function will
Q = exp[(1+1)/kT] + exp[(1+2)/kT] + exp[(1+3)/kT] + exp[(2+1)/kT] + exp[(2+2)/kT] + exp[(2+3)/kT] + exp[(3+1)/kT] + exp[(3+2)/kT] + exp[(3+3)/kT]
c.)
On taking out the common exp factor from RHS of above equation
We will have
Q = exp(1/kT)[exp(1)/kT + exp(2)/kT + exp(3)/kT ] + exp(2/kT)[ exp(1)/kT + exp((2)/kT + exp(3)/kT] + exp(3/kT)[ exp(1)/kT + exp((2)/kT + exp(3)/kT]
On taking out the common multiplier exp(1)/kT + exp((2)/kT + exp(3)/kT from every part,
We will have the
Q = [exp(1/kT) + exp(2/kT) + exp(3/kT)] [exp(1)/kT + exp(2)/kT + exp(3)/kT] Proof
States
a
b
1
1
1
2
1
2
3
1
3
4
2
1
5
2
2
6
2
3
7
3
1
8
3
2
9
3
3
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.