- #1

tahaha

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## Homework Statement

A system with 100,000 molecules has 2 energy levels, A and B. At first, the 2 energy levels are populated equally. After a reversible process, energy level A is populated by 65% of the molecules and the system is at 293K.

a) What is the difference in energy between the two levels?

b) How much heat was added or removed from the system?

c) What is the change in entropy?

## Homework Equations

Boltzmann distribution

Probabilistic definition of entropy

## The Attempt at a Solution

I computed the Boltzmann factor for the two energy levels and got the correct answer for (a) = 2.5*10^-21 J.

Then, when I looked at (b) and (c), I thought they wouldn't ask for the change in entropy after asking for the heat, if the change in entropy was required for the calculations of the heat. So I pondered, and got this:

Since in the process, 50%-(100%-65%)=15% of the molecules were transferred from B to A, so the energy change of the process would be= 0.15(100,000)(-ΔUab) = 0.15(100,000)(-2.5*10^-21 J) = -3.75*10^-17 J.

But then I was stuck, because I only had the energy change, and just couldn't find the heat without knowing whether there was work and how much.

And then the solution did what exactly I thought they wouldn't do:

1. Find ΔS using the probabilistic definition:

Initial S= -(100,000)(kb)[2(0.5ln0.5)]

Final S= -(100,000)(kb)[0.65ln0.65 + 0.35ln0.35)]

ΔS= -4570kb

2. ΔS= q/T (reversible)

∴ q= TΔS = -1.85*10^-17 J

Then for (c), they put "see above."

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My questions/confusions are:

1). Does my attempt (multiplying the energy change for one molecule by the number of molecules) really get me to the energy change?

2). I'm confused about how they use ΔS= q/T to get the heat here. The problem doesn't say the process is isothermal?

According to the Boltzmann factor:

Nb/Na = exp(-ΔUab/kbT)

The process changes Nb/Na, and ΔUab has to be constant (does it?), T has to change.

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I hope the question isn't too long! Thanks for any help!