- #36

TSny

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See your post #5. You can get T from E.Sir but how do I solve the third part of the question? What can I say about theequilibrium temperature?

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- Thread starter tanaygupta2000
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- #36

TSny

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See your post #5. You can get T from E.Sir but how do I solve the third part of the question? What can I say about theequilibrium temperature?

- #37

tanaygupta2000

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We deduced, Nε/2 < E < NεSee your post #5. You can get T from E.

Since E = Nε/(1+e

=> Nε/2 < Nε/(1+e

=> 1/2 < 1/(1+e

=> 2 > 1+e

=> 1 > e

=> 0 > ε/kT > -∞

Is this correct ?

- #38

TSny

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- #39

tanaygupta2000

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Using E = Nε/(1+e

Substituting E = Nε/8 + 5Nε/8 = 3Nε/4

=> T = (ε/k) /ln(1/3) = (ε/k) /(-1.098) =

Hence the system as a whole is having NEGATIVE TEMPERATURE.

- #40

TSny

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Yes. That looks right if the total energy is 3Nε/4.Using E = Nε/(1+e^{(ε/kT)}), I am getting T = (ε/k) /ln(Nε/E - 1)

Substituting E = Nε/8 + 5Nε/8 = 3Nε/4

=> T = (ε/k) /ln(1/3) = (ε/k) /(-1.098) =-0.91ε/k

Hence the system as a whole is having NEGATIVE TEMPERATURE.

However, as I mentioned earlier, I don't believe it's possible for system B to have an initial energy of 5Nε/8. The maximum possible energy of B (or A) is (N/2)ε, since B has only N/2 particles.

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