Plot Heat Capacity vs Temperature for a 2 state system microcananonical ensemble

[binbagsss]

Homework Statement

I have $C= NK_B (\frac{\epsilon}{K_B T})^{2}e^{\frac{\epsilon}{K_B T}}\frac{1}{(e^{\frac{\epsilon}{K_BT}}+1)^2}$

and need to sketch $C$ vs. $T$

Homework Equations

See above

The Attempt at a Solution

I have $C= NK_B (\frac{\epsilon}{K_B T})^{2}e^{\frac{\epsilon}{K_B T}}\frac{1}{(e^{\frac{\epsilon}{K_BT}}+1)^2}$

Considering asymptotic limits I have:

$C \to e^{-\frac{\epsilon}{K_{B}T}}$ as $T \to 0$
$C \to \frac{1}{T^{2}}$ as $T \to \infty$

The solution is attached.

So from these limits I get the shape at these ends, and deduce there is a maximum to allow me to sketch the rest of it.

I am unsure how to deduce this maximum?

Differentiating gives quite a mess and it seems that it should be obvious to conclude the maximum is at $\epsilon / K_{B}$, or at least a better method to find this point? (My knowledge of graph sketching is quite poor).

Many thanks in advance.

 

[binbagsss]

Homework Statement

I have $C= NK_B (\frac{\epsilon}{K_B T})^{2}e^{\frac{\epsilon}{K_B T}}\frac{1}{(e^{\frac{\epsilon}{K_BT}}+1)^2}$

and need to sketch $C$ vs. $T$

Homework Equations

See above

The Attempt at a Solution

I have $C= NK_B (\frac{\epsilon}{K_B T})^{2}e^{\frac{\epsilon}{K_B T}}\frac{1}{(e^{\frac{\epsilon}{K_BT}}+1)^2}$

Considering asymptotic limits I have:

$C \to e^{-\frac{\epsilon}{K_{B}T}}$ as $T \to 0$
$C \to \frac{1}{T^{2}}$ as $T \to \infty$

The solution is attached.

So from these limits I get the shape at these ends, and deduce there is a maximum to allow me to sketch the rest of it.

I am unsure how to deduce this maximum?

Differentiating gives quite a mess and it seems that it should be obvious to conclude the maximum is at $\epsilon / K_{B}$, or at least a better method to find this point? (My knowledge of graph sketching is quite poor).

Many thanks in advance.

bump

 

[vela]

The only suggestion I'd have is to let $u=\frac{\epsilon}{k_\text{B}T}$ and find the extremum of
$$\frac{u^2 e^u}{(1+e^u)^2}.$$ It shouldn't be that messy.

 

[binbagsss]

The only suggestion I'd have is to let $u=\frac{\epsilon}{k_\text{B}T}$ and find the extremum of
$$\frac{u^2 e^u}{(1+e^u)^2}.$$ It shouldn't be that messy.

I have:

$2e^u+2+u-e^u u =0$ , unsure of where to go now...

 

[vela]

You'd have to solve that numerically. To get a qualitative idea of where the root lies, you can rewrite that equation as
$$e^u = \frac{u+2}{u-2}.$$ Plot graphs of the two sides of the equations and see where they intersect.