I got $$S = -Nk \frac {μB}{kT} tanh (\frac {μB}{kT}) + ln (2cosh(\frac{μB}{kT}))$$. Letting T→∞ gives ##S = -Nk (\frac{μB}{kT})^2 + ln 2##. Something doesn't seem right. I didn't even get to use the expression for U at T→∞ to get this entropy even though the question wants me to use it?
Hmm, I'm still not able to get the expression for entropy at infinite T as given in the question. Here's my attempt so far:
$$S = - \frac {∂A}{∂T} = - \frac {∂}{∂T} (-kT ln(Z_N)$$ where ##Z_N = (2 cosh(\frac {μB}{kT})^N##. When T→∞, cosh(x)→1 so ##Z_N = 2^N##. Therefore $$S = k ln(2^N)$$
But...
Okay, I got tanh(x) ≈ x from the Taylor expansion, so I managed to obtain the expression ##U = - \frac {N (μB)^2}{kT}## for the T→∞ limit. Is this correct?
I'm still confused about the entropy part. Am I supposed to calculate the helmholtz energy first?
Homework Statement
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I'm stuck on part (b) and (c) of the following question
Homework Equations
The Attempt at a Solution
The partition function was ##Z_N = 2 cosh(μBβ)## where ##β = \frac {1}{kT}##. From there I used ##U = - \frac {∂}{∂β} ln (Z_n)## to get ##U = -NμB tanh( \frac...
Homework Statement
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A "molecular zipper" has two rows of molecules, and each row has a large number of monomers. A monomer from one row is weakly linked to a monomer in the other row. The zipper can unzip from one end by breaking the bond between pairs of monomers. A bond can be broken...
Thanks, I managed to obtain the follow expression for the CoM of the pyramid with the cavity:
$$CoM = \frac {B^2 H^2 - b^2 h^2} {4(B^2 H - b^2 h)}$$
Is this correct?
Okay, I think the equation makes sense as you put it. Just to confirm, it is H/4 - h/4 in this case, right? I think I'm just having a hard time understanding it intuitively because from the diagram it looks like the pyramid is heavier at the top in hollow case.
I'm assuming yes, or it is the volume that is important? I'm struggling to understand why it would shift down and not up if it is hollow from the bottom?