October 17, 2019, 02:09:47 PM
Forum Rules: Read This Before Posting Topic: Heat capacity in systems with different energy-levels (Boltzmann)  (Read 182 times)

0 Members and 1 Guest are viewing this topic. Bobby Blothunger

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« on: October 01, 2019, 12:38:35 PM »
Hello!

I am struggeling to wrap my head around why the heat capacity is different for systems with different energy-levels.

https://imgur.com/a/yUvi5ZT

I'll try to explain the graph a little bit.
CVA: distance between energy-level: 100 K (equidistant)
CVB: distance between energy-level: 20 K   (equidistant)
CVC: distance between main energy-levels: 200 K, each with 5 sublevels of 20 K

Cv is defined as Cv=(δU/δT) with V and N being constant.

What I find difficult to understand is how to explain why the graphs look different and why when
T ∞ Cv R . How do i relate entropy to this equation? Is it possible to use 1/T = (δS/δU) with V and N being constant.

What I believe is the reason:
CVA: Since the distance between the energy-levels are large it takes more energy to push up and "use" the next level, and therefore the slope is also flatter.
CVB: Since the distance between the energy-levels are small it takes less energy for the system to use them. Moreover more energy-levels also mean that the entropy will be greater (ΔS aswell), which explains the greater slope in comparison to CVA.
CVA: This one I don't really understand, why is Cv flat and negative between 10-35 K?

I'm sorry if it hard to understand my question, english is not my first language.
Studying 5 year program M.Sc. in biotechnology in Sweden, currently doing my second year. mjc123

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« Reply #1 on: October 02, 2019, 09:42:32 AM »
Are you familiar with the Boltzmann distribution (the title of your post suggests you are)? This tells us how different energy levels are populated at a given temperature. If there are N0 molecules in the ground state, then the number NE in a state of energy E is given by
NE = N0e-E/kT
Thus the higher the energy E, the less populated the state is.
Consider the system with energy levels 100k apart (write small k for Boltzmann's constant, so it doesn't look like a temperature of 100 K). At a low temperature, say 10 or 20 K, where E>>kT, there are very few molecules in the excited states. If you increase the temperature by 1 K, "very few" becomes "very few plus a tiny bit more". The increase of energy due to this "tiny bit more" going to a higher level is very small, so the heat capacity (energy input to raise the temperature by 1 K) is also small. As the temperature rises, and the population of the higher levels increases, the heat capacity also increases.
Now consider the system with levels 20k apart. The higher levels are easier to reach, and at 10 or 20 K, they are already significantly populated, and more molecules are promoted for a given increase in temperature, so the heat capacity is higher than for system A at the same temperature.
System C is different, in that you have a few closely-spaced levels, then a big gap, then a few more closely-spaced levels, and so on. At low temperatures we can assume that the 200k level and above are essentially unpopulated, and what we have is a system of 5 levels 20k apart. To begin with, it behaves like system B, but there comes a point when, in system B, levels 100, 120, 140 etc. would start to be populated, but in system C these don't exist. For a finite system, at high temperature the levels would tend to become equally populated, with little change with increasing temperature, so the heat capacity would tend towards zero. That's why you see the peak in Cv and the decrease at intermediate temperatures (note, Cv is not negative, the slope of the curve is negative). Of course, the system is not really finite, and as T increases further, the higher levels (>200k) start to be populated and contribute to the heat capacity, so it rises again. Bobby Blothunger

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« Reply #2 on: October 03, 2019, 06:58:14 AM »
Thank you mjc123 for your informative answer.

I now understand the reasoning behind the behavior of system A and B and I think that I am beginning to understand system C as well. Although I still have a hard time understanding why the slope of the curve is negative, is it possible to relate it to a fundamental equation with δS instead of δU? And are you saying that if the system had been finite with just the 5 sublevels Cv 0 when T ∞? Why is that the case? Would it not just stagnate, since δU and δT never becomes negative. δU juste becomes smaller and smaller the higer the tempature.

To summerise: how is it that the slope of the curve becomes negative when neither δU nor δT becomes negative?

"Cv = δU/δT"

Edit: missed a questionmark
Studying 5 year program M.Sc. in biotechnology in Sweden, currently doing my second year. mjc123

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« Reply #3 on: October 03, 2019, 07:53:14 AM »
Do you understand slopes and derivatives? What becomes negative is not Cv, but dCv/dT = d2U/dT2. The energy continuously increases with temperature, but the rate of increase (the heat capacity) starts off low, then increases, goes through a maximum, decreases for a while, then starts increasing again.

Yes, I am saying that for a finite system Cv 0 as T ∞. As you say, "δU juste becomes smaller and smaller the higer the tempature". In other words dU/dT tends to zero as T increases.