|Aug7-12, 10:56 PM||#1|
Entropy of liquid nitrogen.
I tried to find entropy of liquid nitrogen in various data books, using nitrogen's CAS number, MSDS, but I was able to get the entropy of nitrogen till only 100K, not below that.
Does anyone know what's the entropy of liquid nitrogen or that where can I find it?
|Aug8-12, 01:26 AM||#2|
I suggest the following approximation to extrapolate the value of the entropy below 100°K. The approximation won't be valid for temperatures far below 100° K, but it should be valid within a limited range of 100°K.
Hypothesize that the specific heat of liquid nitrogen is constant. Furthermore, hypothesize that assume that the liquid nitrogen is being cooled down from 100K at whatever pressure you got that value of entropy from.
The specific heat at constant pressure is defined as C_P where,
You can look up C_P for liquid nitrogen in tables. Since C_P is almost constant for liquids, it doesn't matter precisely.
T is the temperature, S is the entropy and P is the pressure.
If you assume that C_P is constant, then you can integrate this equation easily.
where ΔS is the change in entropy, T is the new temperature and,
Putting the two equations together,
You say you have S(100°K), and you can look up C_P. Thus, you can estimate the entropy at any temperature, i.e., S(T).
Here are some hints.
Note "ln" is the Neparian logarithm, or the "natural logarithm".
The difficult part may be finding the best value of C_P.
For a liquid, there is very little difference between C_P or C_V. The specific heat at constant pressure is almost the same as the specific heat at constant volume for a liquid. So if your table gives you C_V instead of C_P, just use it.
As I said, C_P is almost constant over a large temperature range for a liquid. Similarly, it will be constant over a limited pressure range for a liquid. Therefore, if you can only find C_P at 77°K, use it.
Final hint: There may be something wrong with your value of S(100°K).
Check very carefully whether that CAS number is really for 100°K or 77°K. The standard tables prefer values at 1 Atmosphere. Your question implies a larger value of pressure, which is possible. However, check.
You may have found the entropy of gaseous nitrogen at 100°K. This will not work.
Also note that all this is based on an approximation. One must be wary of approximations. However, this hypothesis is probably the best estimate possible given the limited information in your problem.
|Aug8-12, 02:03 AM||#3|
Thanks for your reply.
I will like to add few things.
A correction - I have the entropy of nitrogen gas at 100K. Not of liquid nitrogen at 100K.
The equation from which I derived the entropy recommends to use it in the temperature range of 100K-500K. It might be safe to extrapolate till 78K, but below this temperature there will be a sudden drop in entropy(due to change in phase, gas to liquid). Again after phase change there will be slow and smooth reduction in entropy with reduction in temperature.
I wanted to know the entropy of nitrogen at phase change point or below the phase change point(liquid nitrogen) at 1bar pressure.
|Aug8-12, 07:09 AM||#4|
Entropy of liquid nitrogen.
Well, finding entropy of liquid nitrogen is more difficult than I thought.
As a rough approximition, I took it 120J/moleK.
Following is the reason;
EN2(liquid) = (EBr2(liquid)/EBr2(gas))*EN2(gas)
EN2(liquid) = (152/245)*192 = 120
I guess, this should be close enough.
|Aug8-12, 01:18 PM||#5|
At 1 Atmosphere, the boiling point of nitrogen is 77°K. The nitrogen gas has to be cooled down to the boiling point, condense into a liquid, and then cooled down to the temperature that you want the entropy at.
1) S(gas, 77°K)=S(gas, 100°K)+C_P[diatomic ideal gas] ln(77°K/100°K)
2) S(liquid, 77°K)=S(gas, 77°K)-ΔH(vaporization, nitrogen)/77°K
3) S(liquid, T)=S(liquid, 77°K)+C_P[liquid nitrogen] ln(T/77°K)
Here, ΔH(vaporization, nitrogen) is the heat of vaporization for nitrogen, C_P[diatomic ideal gas] is the specific heat at constant pressure for a diatomic gas, C_P[liquid nitrogen] is the specific heat at constant pressure for liquid nitrogen and T is the temperature of interest. The heat of fusion is also called the enthalpy of fusion.
The three steps above should be valid down to the freezing point of nitrogen. Extending this method to temperatures just below the freezing point should be obvious. At temperatures far below the freezing point, the specific heat will stop being constant with temperature. However,
You will need to look up C_P[liquid nitrogen] and ΔH(vaporization, nitrogen) in a table. C_P[diatomic ideal gas] is the same for all diatomic ideal gases. Nitrogen is a diatomic gas. The units have to be consistent. Therefore, you may have to manipulate the values that you look up on a table.
C_P[liquid nitrogen]≈C_V[liquid nitrogen]
You already have S(gas, 100°K). I couldn't find it in the CRC manual, but you have it. Here is the rest of the information that you will need.
Handbook of Chemistry and Physics, 84th Edition, 2003-2004
David R. Linde, Editor in Chief
Properties of Cryogenic Liquids
T(Nitrogen vaporization)=77.35 °K
ΔH(vaporization, nitrogen)=25.3 J/g
C_P[Nitrogen gas]=1.34 J/(g °K)
C_P[Liquid nitrogen]=2.042 J/(g °K)
These values, together with the formulas above, determine the entropy density in units of J/(g °K) at any temperature T in °K.
|Aug8-12, 04:35 PM||#6|
I gave another post where I detail a better way to calculate the entropy. The main physical reason I used that method is because entropy is a perfect differential for reversible processes. Therefore, one can break up the entropy into three components. The proportionality that you just used is based on - what?-
The method that I suggested requires you looking up the heat of vaporization of nitrogen and the specific heat of liquid nitrogen. I think that would work far better than your proportionality.
|Aug8-12, 05:47 PM||#7|
Try the NIST database. http://webbook.nist.gov/chemistry/fluid/
|Aug8-12, 10:52 PM||#8|
Thanks Darwin and Q_Goest!
The information is there in the NIST site that Q_Goest mentioned.
Entropy of liquid nitrogen at 1bar pressure and 77.244K is 79.313J/moleK.
I was quite off the mark.
I will like to close this thread now.
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