R Of Helium

  

The tables below have been prepared from the professional units conversion program Uconeer by Katmar Software. These tables contain 188 values for the Universal Gas Constant in the most likely combinations of units. These include the most commonly used values for the universal gas constant when working in metric units, SI units, English engineering units and US customary units.

Almost every known helium reserve on the planet was discovered by accident, and the helium was merely a byproduct of natural gas harvesting. The United States has been the largest producer of. Helium is the second most abundant element in the universe, but here on earth, it's rather rare. Most people guess that we extract helium from the air, but actually we dig it out of the ground. Helium can be found in certain parts of the world, notably in Texas, as a. Now, we consider the Helium atom andwill see that due to the attendant 3-bodyproblemforwhichwecannot determineaclosed-form, rst-principles analytic solution, we will have to nd recourse in approximate methods. The Helium atom has 2 electrons with coordinates r1 and r2 as well as a single nucleus with coordinate R. The nucleus carries a Z. The Helium 1 creating a featherstick for a fire. His company’s philosophy is simple, “Every blade is made to use with functionality, design, and precision.” Form follows function with R&N Blades and performance is paramount.

However, the dedicated Gas Constant calculator in Uconeer calculates the value of 'R' based on over 900 combinations of units, so if you don't find what you want here please go to the Uconeer main page to read the program description and to download the program. Uconeer will even allow you to add your own units, so if you do not find the units you need in the tables below, or built into Uconeer, you can add them in for yourself.

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Gas Constant Values based on Energy Units

If you are using kg mol as the unit for quantity of matter then multiply the appropriate value in the g mol column by 1000.

g mol
Kelvin
lb mol
Kelvin
g mol
Rankine
lb mol
Rankine
Btu 0.0078806 3.57458 0.00437811 1.98588
cal 1.98588 900.779 1.10327 500.433
Chu 0.00437811 1.98588 0.00243228 1.10327
erg (dyn.cm) 8.31447e+7 3.77138e+10 4.61915e+7 2.09521e+10
ft.lbf 6.13244 2781.63 3.40691 1545.35
hp.h 3.09719e-6 0.00140486 1.72066e-6 0.000780479
J 8.31447 3771.38 4.61915 2095.21
kcal 0.00198588 0.900779 0.00110327 0.500433
kgf.m 0.847840 384.574 0.471022 213.652
kJ 0.00831447 3.77138 0.00461915 2.09521
kW.h 2.30958e-6 0.00104761 1.2831e-6 0.000582003
N.m 8.31447 3771.38 4.61915 2095.21
Copyright (C) Katmar Software 2001-2020

Gas Constant Values based on Pressure and Volume Units

This table is based on g moles as the unit for quantity of matter and Kelvin as the unit for temperature. If you are using kg mol as the unit for quantity of matter then multiply the value given below by 1000.

cm^3 liter m^3 ft^3 inch^3
atm 82.0575 0.0820575 8.20575e-5 0.00289783 5.00745
bar 83.1447 0.0831447 8.31447e-5 0.00293623 5.0738
foot H2O 2781.63 2.78163 0.00278163 0.0982323 169.745
inch H2O 33379.5 33.3795 0.0333795 1.17879 2036.94
inch Hg 2455.27 2.45527 0.00245527 0.086707 149.83
kgf/cm^2 84.7840 0.0847840 8.47840e-5 0.00299412 5.17384
kPa 8314.47 8.31447 0.00831447 0.293623 507.38
lbf/ft^2 173651 173.651 0.173651 6.13244 10596.9
mbar 83144.7 83.1447 0.0831447 2.93623 5073.8
meter H2O 847.840 0.847840 0.00084784 0.0299412 51.7384
mm H2O 847840 847.840 0.847840 29.9412 51738.4
mm Hg 62363.8 62.3638 0.0623638 2.20236 3805.68
Pa 8.31447e+6 8314.47 8.31447 293.623 507380
PSI 1205.91 1.20591 0.00120591 0.0425864 73.5893
Copyright (C) Katmar Software 2001-2020

Gas Constant Values based on Pressure and Volume Units

This table is based on lb moles as the unit for quantity of matter and Rankine as the unit for temperature.

Three Uses Of Helium


cm^3 liter m^3 ft^3 inch^3
atm 20678.1 20.6781 0.0206781 0.730241 1261.86
bar 20952.1 20.9521 0.0209521 0.739917 1278.58
foot H2O 700958 700.958 0.700958 24.7541 42775.1
inch H2O 8.41150e+6 8411.50 8.41150 297.049 513301
inch Hg 618717 618.717 0.618717 21.8498 37756.5
kgf/cm^2 21365.2 21.3652 0.0213652 0.754505 1303.79
kPa 2.09521e+6 2095.21 2.09521 73.9917 127858
lbf/ft^2 4.37594e+7 43759.4 43.7594 1545.35 2.67036e+6
mbar 2.09521e+7 20952.1 20.9521 739.917 1.27858e+6
meter H2O 213652 213.652 0.213652 7.54505 13037.9
mm H2O 2.13652e+8 213652 213.652 7545.05 1.30379e+7
mm Hg 1.57154e+7 15715.4 15.7154 554.985 959014
Pa 2.09521e+9 2.09521e+6 2095.21 73991.7 1.27858e+8
PSI 303885 303.885 0.303885 10.7316 18544.2
Copyright (C) Katmar Software 2001-2020


Multi-electron Hamiltonians

The second element in the periodic table provides our first example of a quantum-mechanical problem which cannot be solved exactly. Nevertheless, as we will show, approximation methods applied to helium can give accurate solutions in perfect agreement with experimental results. In this sense, it can be concluded that quantum mechanics is correct for atoms more complicated than hydrogen. By contrast, the Bohr theory failed miserably in attempts to apply it beyond the hydrogen atom.

Figure (PageIndex{1}) shows a schematic representation of a helium atom with two electrons whose coordinates are given by the vectors (r_1) and (r_2). The electrons are separated by a distance (r_{12} = r_1-r_2 ). The origin of the coordinate system is fixed at the nucleus. As with the hydrogen atom, the nuclei for multi-electron atoms are so much heavier than an electron that the nucleus is assumed to be the center of mass. Fixing the origin of the coordinate system at the nucleus allows us to exclude translational motion of the center of mass from our quantum mechanical treatment.

The Hamiltonian operator for the hydrogen atom serves as a reference point for writing the Hamiltonian operator for atoms with more than one electron. Start with the same general form we used for the hydrogen atom Hamiltonian

[hat {H} = hat {T} + hat {V} label {6.7.1}]

Include a kinetic energy term for each electron and a potential energy term for the attraction of each negatively charged electron for the positively charged nucleus and a potential energy term for the mutual repulsion of each pair of negatively charged electrons. The He atom Hamiltonian is

[ hat {H} = -dfrac {hbar ^2}{2m_e} (nabla ^2_1 + nabla ^2_2) + V_1 (r_1) + V_2 (r_2) + V_{12} (r_{12}) label {6.7.2}]

where

[ V_1(r_1) = -dfrac {2e^2}{4 pi epsilon _0 r_1} label {6.7.3}]

[ V_2(r_2) = -dfrac {2e^2}{4 pi epsilon _0 r_2} label {6.7.4}]

[ V_{12}(r_{12}) = dfrac {e^2}{4 pi epsilon _0 r_{12}} label {6.7.5}]

The two-electron Hamiltonian in Equation (ref{6.7.2}) can be extended to any atom or ion by replacing the He nuclear charge of +2 with a general charge (Z); e.g.

[V_1(r_1) = -dfrac {Ze^2}{4 pi epsilon _0 r_1} label {6.7.6}]

and including terms for the additional electrons. The subsequent multi-electron atom with (n) electron is

R Of Helium

[hat {H} = underbrace{-dfrac {hbar ^2}{2m_e} sum_i^n nabla ^2_i}_{text{Kinetic Energy}} + underbrace{sum_i^n V_i (r_i)}_{text{Coulombic Attraction}} + underbrace{ sum_{i ne j}^{n,n} V_{ij} (r_{ij})}_{text{electron-electron Repulsion}} label {6.7.7}]

This multi-electron Hamiltonian is qualitatively similar to the 2-electron Hamiltonian (Equation ref{6.7.1}) with each electron having its own kinetic energy and nuclear potential energy terms (Equations (ref{6.7.3}) and (ref{6.7.4})). The other big difference between single electron systems and multi-electron systems is the presence of the (V_{ij}(r_{ij})) terms which contain (1/r_{ij}), where (r_{ij}) is the distance between electrons (i) and (j). These terms account for the electron-electron repulsion that we expect between like-charged particles.

Exercise (PageIndex{1}): Multi-electron atom Hamiltonians

For the generalized multi-electron atom Hamiltonian (Equation (ref{6.7.7})):

  1. Explain the origin of each of the three summations.
  2. What do these summations over (i.e., what is the origin of the summing index)?
  3. Write expressions for (V_i(r_i)) and (V_{ij}(r_{ij})).

Exercise (PageIndex{2}): Boron Atom

Gas Constant Of Helium

Boron is the fifth element of the periodic table (Z=5) and is located in Group 13.

  1. Write the multi-electron Hamiltonian for a (ce{^{11}B}) atom.
  2. Would it be any different for a (ce{^{11}B^{+}}) ion?
  3. Would it be any different for a (ce{^{10}B}) atom?
Answer

a.

[hat {H}_{ce{B}}(vec r_1,vec r_2,vec r_3,vec r_4,vec r_5) = -dfrac {hbar ^2}{2m_e} sum_i^{5} nabla ^2_i + sum_i^{5} dfrac {-5e^2}{4 pi epsilon _0 r_i} + sum_{i ne j}^{5,5} dfrac {e^2}{4 pi epsilon _0 r_{ij}} nonumber]

which expands to 20 terms

[ begin{align*} hat {H}_B(vec r_1,vec r_2,vec r_3,vec r_4,vec r_5) = &-dfrac {hbar ^2}{2m_e} nabla ^2_1 -dfrac {hbar ^2}{2m_e} nabla ^2_2 -dfrac {hbar ^2}{2m_e} nabla ^2_3 -dfrac {hbar ^2}{2m_e} nabla ^2_4 -dfrac {hbar ^2}{2m_e} nabla ^2_5 [4pt] &- dfrac {-5e^2}{4 pi epsilon _0 r_1} - dfrac {-5e^2}{4 pi epsilon _0 r_2} - dfrac {-5e^2}{4 pi epsilon _0 r_3} - dfrac {-5e^2}{4 pi epsilon _0 r_4} - dfrac {-5e^2}{4 pi epsilon _0 r_5} [4pt] &+ dfrac {e^2}{4 pi epsilon _0 r_{12}} + dfrac {e^2}{4 pi epsilon _0 r_{13}} + dfrac {e^2}{4 pi epsilon _0 r_{14}} +dfrac {e^2}{4 pi epsilon _0 r_{15}} + dfrac {e^2}{4 pi epsilon _0 r_{23}} +dfrac {e^2}{4 pi epsilon _0 r_{24}} + dfrac {e^2}{4 pi epsilon _0 r_{25}} + dfrac {e^2}{4 pi epsilon _0 r_{34}} + dfrac {e^2}{4 pi epsilon _0 r_{35}} + dfrac {e^2}{4 pi epsilon _0 r_{45}} end{align*}]

b. Yes, (ce{^{11}B^{+}}) has one less electron than (ce{^{11}B}). Its Hamiltonian is

About avogadro. [hat {H}_{ce{B^+}}(vec r_1,vec r_2,vec r_3,vec r_4 ) = -dfrac {hbar ^2}{2m_e} sum_i^{4} nabla ^2_i + sum_i^{4} dfrac {-5e^2}{4 pi epsilon _0 r_i} + sum_{i ne j}^{4,4} dfrac {e^2}{4 pi epsilon _0 r_{ij}} nonumber]

or expanded to 14 terms

[ begin{align*} hat {H}_B(vec r_1,vec r_2,vec r_3,vec r_4) = &-dfrac {hbar ^2}{2m_e} nabla ^2_1 -dfrac {hbar ^2}{2m_e} nabla ^2_2 -dfrac {hbar ^2}{2m_e} nabla ^2_3 -dfrac {hbar ^2}{2m_e} nabla ^2_4 [4pt] &- dfrac {-5e^2}{4 pi epsilon _0 r_1} - dfrac {-5e^2}{4 pi epsilon _0 r_2} - dfrac {-5e^2}{4 pi epsilon _0 r_3} - dfrac {-5e^2}{4 pi epsilon _0 r_4} [4pt] &+ dfrac {e^2}{4 pi epsilon _0 r_{12}} + dfrac {e^2}{4 pi epsilon _0 r_{13}} + dfrac {e^2}{4 pi epsilon _0 r_{14}} + dfrac {e^2}{4 pi epsilon _0 r_{23}} +dfrac {e^2}{4 pi epsilon _0 r_{24}} + dfrac {e^2}{4 pi epsilon _0 r_{34}} end{align*}]

c. No effect. Changing the number of neutrons in the nucleus does not affect kinetic nor potential energies of the electrons. The Hamilitonian for (ce{^{10}B}) is identical to (ce{^{11}B}). This is technically correct for this level of discussion, but as we we will see in later, if we expand the Hamiltonian with hyperfine structure the number of neutrons can play a role.