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Refractive Index by Pressure/Temperature 
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#1
Jul2314, 02:23 PM

P: 24

The refractive index of H_{2} = 1.0001594
under the following conditions : P = 101325 Pa T = 273.15 K I cannot find any source for the refractive index of gases under any other conditions 3 Questions : a. What is the pressure gradient for the refractive index? b. What is the thermal gradient for the refractive index? c. Is there a formula for calculating the refractive index based on the Individual Gas Constant? Thanks! 


#2
Jul2414, 01:17 AM

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P: 3,633

Sounds like a thread apt for the homework forum.



#3
Jul2414, 06:12 AM

P: 24

It's an engineering problem  in need of a solution
The refractive index should be directly proportional to pressure where half the pressure = half the refractive index In liquids, I believe the refractive index thermal gradient is around 0.000045 / K where the refractive index (and density) increases as temperature decreases, but I don't think this gradient applies to gases. I found the equation for refractive index below : 


#4
Jul2414, 04:01 PM

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Refractive Index by Pressure/Temperature
Do you know a relationship between concentration and p and T for dilute gasses? 


#5
Jul2514, 08:20 AM

P: 24

calculated through the ideal gas equation. [itex]PV = nRT[/itex] For example, we take the refractive index of H_{2} at 101325 Pa and 273.15 K which = 1.0001594 The individual gas constant for H_{2} = 4124 J / kg K Therefore, using the ideal gas equation, 1 m² of H_{2} at 101325 Pa and 273.15 K has a concentration of 0.089949 mol multiplying the molar concentration by the Avogadro constant  6.02214129 x 10^{23} we have a density of 5416861 x 10^{22} H_{2} molecules / m³ with the temperature remaining constant, if we reduce the pressure by 2.0265 x 10^{22} at 5 x 10^{18} Pa and 273.15 K the refractive index of H_{2} then = 4.9354 x 10^{23} again, using the ideal gas equation, 1 m² of H_{2} at 5 x 10^{18} Pa and 273.15 K has a concentration of 4.43864225 x 10^{24} mol multiplying the molar concentration by the Avogadro constant we now have a density of H_{2} = 2 (.673) molecules / m³ So, we now have an H_{2} concentration of 4.43864225 x 10^{24} mol / m³ and a refractive index of 4.9354 x 10^{23} so, if we multiply 2.673 by the temperature difference between 273.15 K and 2.7 K (101.166..) we have a molar concentration of 4.483455 x 10^{22} and a density of 270 H_{2} molecules per m³ the refractive index increases as temperature decreases how then do we calculate the refractive index of H_{2} at 2.7 K ? . 


#6
Jul2514, 09:27 AM

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P: 3,633

Huh, you said nothing about being interested in 2.7 K and extremely low pressures. What's this, interstellar vacuum?
Maybe you could explain in more detail what you are really trying to do. 


#7
Jul2514, 09:32 AM

P: 24

for example 1% of the 273.15 K baseline temperature = 2.7 K or we can take 10% of 273.15 K, 27 K as an example. The melting point of H_{2} is typically listed as 13.95 K however the concentration at this melting point is not specified. The melting point of H_{2} increases with concentration. Melting and boiling points are a function of the intermolecular forces (van der Waals forces). For molecular H_{2} the only intermolecular forces are London dispersion forces, and London dispersion forces depend on the polarizability of the molecule. The greater the number of electrons per volume the greater the strength of the London dispersion forces. So, at an extremely low molar concentration of 4.483455 x 10^{22} mol / m³ the melting point should be below 2.7 K In fact below a critical concentration, H_{2} as a solid should not even be possible. However, to simplify the problem, we can take 27 K as a temperature and a molar concentration of 4.49 x 10^{23} mol / m³ is there an equation for calculating the refractive index based on concentration alone? . 


#8
Jul2614, 10:14 AM

P: 24

actually, the real question is..
does anyone here know the equation for calculating refractive index based on molar comcentration? . 


#9
Jul2614, 02:54 PM

P: 24

So, I finally got my answer from a chemist :
The temperature is irrelevant to a specific quantitative concentration. Concentration itself is a coefficient of temperature and pressure, however once the concentration is defined  that's all we need to know. we can pack up the tents and go home. The refractive index of H_{2} at a molar concentration of 4.483455 x 10^{22} mol / m³ = 4.9354 x 10^{23} . 


#10
Jul2714, 08:04 AM

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P: 3,633

No, the refractive index will be very nearly 1. If the index 1 specifies the standard state and the index 2 the actual state you are interested in then:
##(n_21)/(n_11)=(p_2T_1)/(p_1T_2)## 


#11
Jul2714, 12:21 PM

P: 24

Let's check.. p_{1} = 101325 p_{2} = 5 x 10^{18} T_{1} = 273.15 T_{2} = 2.7 (p_{2}T_{1}) = 0.00000000000000136575 (p_{1}T_{2}) = 273577.5 (p_{2}T_{1}) / (p_{1}T_{2}) = 4.992186857471831565 x 10^{21} it doesn't quite add up to 1 but the result is 101.15060293941385835182181384441 x the previous result and if we multiply T_{2} by this factor we have 273.1066 which would suggest that we are simply multiplying the previous result by (T_{1}/T_{2}) . 


#12
Jul2814, 01:23 AM

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#13
Jul2814, 06:23 AM

P: 24

Of course, I was only responding to your previous reply I assumed you meant that the refractive index should be very nearly 1 in order to achieve a target refractive index, we first reduce the pressure at a constant temperature until we reach the target, then we reduce the temperature to T_{2}, which increases the refractive index so we compensate by reducing the pressure again  using the equation remaining at T_{2}, and we end up with the target refractive index and temperature. simple Your equation is exactly what I needed.. Thanks! 


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