Thermodynamics - Entropy&Heat dispersion

In summary, at temperatures near absolute zero, the heat capacity is not a constant and you have to take into account that the entropy at temperatures near absolute zero is not a constant.
  • #1
AbedeuS
133
0
Hey, just a small mathematical enquiry about why temperature tends to average out in, say a box of gases. Let's just use the standard "Partitioned vessel" view of it, with One side having 1 mole of Helium at..say 398k and the other side with 1 mole of helium at 298K. The heat capacity is (rounded up) 21 J/Mol K and I'm assuming it's the same over the temperature range.

Now to calculate the final temperature when both gases are mixed I'm just going to use [tex]q=nC_{v}\deltaT[/tex] to find the total thermal energy then calculate for T:

[tex]q_{298K} = 1*21*298 = 6258J[/tex]
[tex]q_{398K} = 1*21*398 = 8358J[/tex]

So:
[tex]T= \frac{q_{298K}+q_{398K}}{2*21} = 348K[/tex]

Now with the final temperature, and discounting any residual T = 0 crystal entropy the entropy of that mixture is calculated by:

[tex]\int^{398}_{0} \frac{C_{v}}{T} dT = C_{p}ln(\frac{398}{0})[/tex]
Stuck here, ln(0) is not possible I did have a grand plan to try and prove it to myself but how do you deal with this part in the integral?
 
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  • #2
solve as an improper integral?

maybe, I haven't taken calc II in about 3 years.
 
  • #3
This isn't a homework or coursework question by the way, it's just me enquiring (Directed at whomever moved it)
 
  • #4
AbedeuS said:
Now with the final temperature, and discounting any residual T = 0 crystal entropy the entropy of that mixture is calculated by:

[tex]\int^{398}_{0} \frac{C_{v}}{T} dT = C_{p}ln(\frac{398}{0})[/tex]
Stuck here, ln(0) is not possible I did have a grand plan to try and prove it to myself but how do you deal with this part in the integral?
You take into account that the heat capacity at temperatures near absolute zero is not a constant. In the Debye theory of solids, for example, it is
proportional to the third power of the temperature, so C/T goes as the
square of T. Then you can do the integral.
 
  • #5
So its:
[tex]C_p ln(398) - 0^2[/tex]?
 
  • #6
No, I don't think so. My suggestion was just to break the range of the integration into several sections: 0 to T1, T2, ..., 398. and I only addressed my remarks to the range near zero (the part of your post in bold print). Many (solid) materials follow an empirical law of the type

[tex] C= aT^3 + \gamma T [/tex]

over a small range in the neighbourhood of absolute zero of temperature. (i.e. the range 0 to T1. In this range, you can do the integral, thus avoiding the problem of log(0).
The values of the constants in the equation above as well as the temperature T1 will depend upon the material you are considering. For most materials, you will also have to divide the range from T1 to 398 into several smaller ranges as well because 1) the above equation loses its validity and 2) there may be phase transitions which will also contribute (i.e. the heat capacity is not a continuous function over the whole temperature range). For example, for oxygen, the ranges are

0 to 14 K contributing 0.54 cal/deg
14 to 23.66 K contributing 1.5
phase change contributing 0.948
23.66 to 43.76 contributing 4.661
phase change contributing 4.058
43.76 to 54.39 contributing 2.397
fusion contributing 1.954
54.39 to 90.13 (liquid phase) contributing 6.462
Vaporization contributing 18.07
---------------------------------------------
Total: Entropy of gas (T=90.13) is 40.59

For higher temperatures, you can use cp=7R/2, up to about T=298 K; at
temperatures above this, you need another empirical equation.

As you see, the calculation of entropy is not exactly trivial. For helium, it might be
less complicated but I don't really know. A final remark: I have only addressed myself to the question you asked about the integral in neighbourhood of T=0, not to the rest of your post. Maybe you don't need to consider the absolute entropy but only the
change in entropy from the boiling point to the higher temperature. In that case, you could simply reformulate the problem to avoid the lower temperature range(s)!
Would that work in your case?
 

1. What is entropy in thermodynamics?

Entropy is a measure of the disorder or randomness in a system. In thermodynamics, it is a measure of the amount of energy that is unavailable to do work. It is often referred to as the "arrow of time" as it shows the direction in which a system will naturally tend to move.

2. How does entropy relate to heat dispersion?

Entropy and heat dispersion are closely related as heat dispersion is one of the main factors that increases entropy in a system. As heat flows from a hotter object to a cooler object, the overall entropy of the system increases. This is because heat dispersion leads to a more disordered and random arrangement of particles within the system.

3. Can entropy be reversed?

The second law of thermodynamics states that the total entropy of a closed system will always increase over time. While it is possible to decrease the entropy of a specific part of a system, the total entropy of the system will still increase. Therefore, entropy cannot be completely reversed in a closed system.

4. How does entropy affect energy efficiency?

Entropy plays a crucial role in determining the efficiency of energy conversion processes. The greater the entropy, the less efficient the process will be. This is because some energy will inevitably be lost as heat and increase the overall entropy of the system. Therefore, reducing entropy is important in improving energy efficiency.

5. What is the relationship between entropy and probability?

Entropy and probability are inversely related. As the probability of a particular arrangement of particles increases, the entropy decreases. This is because a more ordered and less random arrangement has a lower probability of occurring. Conversely, a more random and disordered arrangement has a higher probability and therefore a higher entropy.

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