The Universal and Individual Gas Constants in fluid mechanics and thermodynamics. Individual gas constant is given for the most common gases.

1. 5 Common Uses Of Helium
2. Molar Mass Of Helium Gas
3. Molar Mass Of Helium In Kg
4. Molar Mass Of Helium Molecule

The specific heat (= specific heat capacity) at constant pressure and constant volume processes, and the ratio of specific heats and individual gas constants - R - for some commonly used 'ideal gases', are in the table below (approximate values at 68 o F (20 o C) and 14.7 psia (1 atm)). Definitions of molecular mass, molecular weight, molar mass and molar weight. Molecular mass (molecular weight) is the mass of one molecule of a substance and is expressed in the unified atomic mass units (u). (1 u is equal to 1/12 the mass of one atom of carbon-12) Molar mass (molar weight) is the mass of one mole of a substance and is.

The Universal and Individual Gas Constants are known from the Ideal Gas Law.

The Individual Gas Constant - R

The Individual Gas Constant depends on the particular gas and is related to the molecular weight of the gas. The value is independent of temperature. The induvidual gas constant, R, for a gas can be calculated from the universal gas constant, R_u (given in several units below), and the gas molecular weight, M_gas:

R = R_u/M_gas [1]

In the imperial system the most common units for the individual gas constant are ft lb/slug ^oR. In the SI system the most common units are J/kg K.

Unit conversion: 1 J/kg K = 5.97994 ft lb/slug °R, and 1 ft lb/slug °R = 0.167226 J/kg K.

The Individual Gas Constant for gases:

For full table - rotate the screen!

The Universal Gas Constant - R_u

The Universal Gas Constant - R_u - appears in the ideal gas law and can be expressed as the product between the Individual Gas Constant - R - for the particular gas - and the Molecular Weight - M_gas - for the gas, and is the same for all ideal or perfect gases:

5 Common Uses Of Helium

R_u = M_gas R [2]

The Universal Constant defined in Terms of the Boltzmann's Constant

The universal gas constant can be defined in terms of Boltzmann's constant k as:

R_u = k N_A [3]

where
k = Boltzmann's constant = 1.381 x 10^-23 [J/K]
N_A = Avogadro Number = 6.022 x 10²³ [1/mol]

Molar Mass Of Helium Gas

The Molecular weight of a Gas Mixture

The average molecular weight of a gas mixture is equal to the sum of the mole fractions of each gas multiplied by the molecular weight of that particular gas:

M_mixture = Σx_i*M_i = (x₁*M₁ + ......+ x_n*M_n) [4]

where

x_i = mole fractions of each gas
M_i = the molar mass of each gas

The Universal Gas Constant - R_u - in alternative Units

• atm.cm³/(mol.K) : 82.057338
• atm.ft³/(lbmol.K) : 1.31443
• atm.ft³/(lbmol.^oR) : 0.73024
• atm.l/(mol.K) : 0.082057338
• bar.cm³/(mol.K) : 83.144598
• bar.l/(mol.K) : 0.083144598
• Btu/(lbmol.^oR) : 1.9872036
• cal/(mol.K) : 1.9859
• erg/(mol.K) : 83144598
• hp.h/(lbmol.^oR) : 0.0007805
• inHg.ft³/(lbmol.^oR) : 21.85
• J/(mol.K) : 8.3144598
  
• kJ/(kmol.K) : 8.3144598
  
• J/(kmol.K) : 8314.472
• (kgf/cm²).l/(mol.K) : 0.084784
• kPa.cm³/(mol.K) : 8314.4598
• kWh/(lbmol.^oR) : 0.000582
• lbf.ft/(lbmol.^oR) : 1545.349
• mmHg.ft³/(lbmol.K) : 999
• mmHg.ft³/(lbmol.^oR) : 555
• mmHg.l/(mol.K) : 62.363577
• Pa.m³/(mol.K) : 8.3144598
• psf.ft³/(lbmol.^oR) : 1545.3465
• psi.ft³/(lbmol.^oR) : 10.73
• Torr.cm³/(mol.K) : 62364

See also:
- More material properties
- The Ideal Gas Law - Gases are highly compressible with changes in density directly related to changes in temperature and pressure.
- A Mixture of Gases - Properties of mixtures of gases.
- More about temperature

Related Topics

• Fluid Mechanics - The study of fluids - liquids and gases. Involves velocity, pressure, density and temperature as functions of space and time
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• Air Psychrometrics - The study of moist and humid air - psychrometric charts, Mollier diagrams, air-condition temperatures and absolute and relative humidity and moisture content
• Material Properties - Material properties for gases, fluids and solids - densities, specific heats, viscosities and more

Related Documents

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Graham's Law of Effusion
Examples and Problems only

Example #1: 8.278 x 10¯⁴ mol of an unidentified gaseous substance effuses through a tiny hole in 86.9 s Under identical conditions, 1.740 x 10¯⁴ mol of argon gas takes 81.3 s to effuse.

a) What is the molar mass of the unidentified substance (in g/mol)?
b) What is the molecular formula of the substance?
c) Under identical conditions, how many moles of ethene (C₂H₄) gas would effuse in 91.0 s?

Example #2: It takes 354 seconds for 1.00 mL of Xe to effuse through a small hole. Under the same conditions, how long will it take for 1.00 mL of nitrogen to effuse?

Example #3: What is the rate of effusion for a gas that has a molar mass twice that of a gas that effuses at a rate of 4.2 mol/min?

Example #4: It takes 110. seconds for a sample of carbon dioxide to effuse through a porous plug and 275 seconds for the same volume of an unknown gas to effuse under the same conditions. What is the molar mass of the unknown gas (in g/mol)?

Example #5: What is the molar mass of a compound that takes 2.65 times as long to effuse through a porous plug as it did for the same amount of XeF₂ at the same temperature and pressure?

Example #6: If a gas effuses 4.25 times faster than iodine gas (I₂), what is its molar mass?

A. 59.7 g/mol
B. 163 g/mol
C. 123 g/mol
D. 158 g/mol

Example #7: Calculate the density of a gas at STP, if a given volume of the gas effuses through an apparatus in 6.60 min and the same volume of nitrogen, at the same temperature andpressure, effuses through this apparatus in 8.50 minutes.

Example #8: If 0.0949 moles of NH₃ effuses in 881 seconds, how many seconds would it take for the same number of moles of B₂H₆ to effuse?

Example #9: The rate of diffusion of an unknown gas was determined to be 2.92 times greater than that of NH₃. What is the approximate molar mass of the unknown gas?

Example #10: If a molecule of C₂H₆ diffuses a distance of 0.978 m from a point source, calculate the distance (m) that a molecule of CH₄ would diffuse under the same conditions for the same period of time.

Example #11: A sample of oxygen gas (O₂) effuses into a vacuum 1 times faster than an unknown gas. O₂ has a molecular weight of about 32.00 g mol¯¹. What is the molecular weight of this unknown gas (in g mol¯¹)?

Example #12: Argon effuses from a container at a rate of 0.0260 L/s. How long will it take for 3.00 L of I₂ to effuse from the same container under identical conditions?

Example #13: Calculate the density of a gas at STP, if a given volume of the gas effuses through an apparatus in 6.60 min and the same volume of oxygen at the same temperature and pressure, effuses through this apparatus in 8.50 minutes.

Bonus Example #1: The rate of effusion of an unknown gas at 480 K is 1.6 times the rate of effusion of SO₂ gas at 300 K. Calculate the molecular weight of the unknown gas.

Bonus Example #2: Heavy water, D₂O (molar mass = 20.0276 g mol¯¹), can be separated from ordinary water, H₂O (molar mass = 18.0152 g mol¯¹), as a result of the difference in the relative rates of diffusion of the molecules in the gas phase. Calculate the relative rates of diffusion for H₂O when compared to D₂O.

Problem #1: If equal amounts of helium and argon are placed in a porous container and allowed to escape, which gas will escape faster and how much faster?

Problem #2: What is the molecular weight of a gas which diffuses 1/50 as fast as hydrogen?

Problem #3: Two porous containers are filled with hydrogen and neon respectively. Under identical conditions, 2/3 of the hydrogen escapes in 6 hours. How long will it take for half the neon to escape?

Problem #4: If the density of hydrogen is 0.090 g/L and its rate of effusion is 5.93 times that of chlorine, what is the density of chlorine?

Problem #5: How much faster does hydrogen escape through a porous container than sulfur dioxide?

Problem #6: Compare the rate of diffusion of carbon dioxide (CO₂) & ozone (O₃) at the same temperature.

Problem #7: 2.278 x 10¯⁴ mol of an unidentified gaseous substance effuses through a tiny hole in 95.70 s. Under identical conditions, 1.738 x 10¯⁴ mol of argon gas takes 81.60 s to effuse. What is the molar mass of the unidentified substance?

Problem #8: A compound composed of carbon, hydrogen, and chlorine diffuses through a pinhole 0.411 times as fast as neon. Select the correct molecular formula for the compound:

(a) CHCl₃
(b) CH₂Cl₂
(c) C₂H₂Cl₂
(d) C₂H₃Cl

Problem #9: Which pair of gases contains one which effuses at twice the rate of the other in the pair?

(a) He and Ne
(b) Ne and CO₂
(c) He and CH₄
(d) CO₂ and HCl
(e) CH₄ and HCl

Problem #10: If a molecule of CH₄ diffuses a distance of 0.530 m from a point source, calculate the distance (in meters) that a molecule of N₂ would diffuse under the same conditions for the same period of time.

Bonus Problem #1: Calculate the density of a gas at STP, if a given volume of the gas effuses through an apparatus in 6.60 min and the same volume of nitrogen at the same temperature and pressure, effuses through this apparatus in 8.50 minutes.

Bonus Problem #2: At 25.0 °C and 380.0 mmHg, the density of sulfur dioxide is 1.31 g/L. The rate of effusion of sulfur dioxide through an orifice is 4.48 mL/s. (a) What is the density of a sample of gas that effuses through an identical orifice at the rate of 6.78 mL/s under the same conditions? (b) What is the molar mass of the gas?

Problem #11: What is the rate of effusion for a gas that has a molar mass twice that of a gas that effuses at a rate of 3.62 mol/min?

Problem #12: Calculate the rate of effusion of NO₂ compared to SO₂ at the same temperature and pressure.

Problem #13: Assume you have a sample of hydrogen gas containing H₂, HD, and D₂ that you want to separate into pure components. What are the various ratios of relative rates of effusion?

Molar Mass Of Helium In Kg

Problem #14: A 3.00 L sample of helium was placed in container fitted with a porous membrane. Half of the helium effused through the membrane in 25 hours. A 3.00 L sample of oxygen was placed in an identical container. How many hours will it take for half of the oxygen to effuse though the membrane?

Problem #15: At a certain temperature, hydrogen molecules move at an average velocity of 1.84 x 10³ m/s. Estimate the molar mass of a gas whose molecules have an average velocity of 311 m/s.

Problem #16: An unknown gas effuses 1.66 times more rapidly than CO₂. What is the molar mass of the unknown gas.

Problem #17: A sample of hydrogen gas effuse through a porous container 8.91 times faster than an unknown gas. Estimate the molar mass of the unknown gas.

Problem #18: N₂ is contaminated with a noble gas.The contaminant effuses at 1.87x N₂. What is the noble gas?

Problem #19: In an effusion experiment, it was determined that nitrogen gas, N₂, effused at a rate 1.812 times faster than an unknown gas. What is the molar mass of the unknown gas?

Problem #20: Why are the rates of diffusion of nitrogen gas and carbon monoxide almost identical at the same temperature?

Problem #21: In running a diffusion experiment, ammonia is found to diffuse 30.0 cm during the same amount of time hydrogen chloride moves 20.0 cm. Calculate the percentage deviation from Graham's Law.

Problem #22: A sample of Br₂(g) take 10.0 min to effuse through a membrane. How long would it take the same number of moles of Ar(g) to effuse through the same membrane?

Problem #23: At a particular pressure and temperature, it takes just 8.256 min for a 4.893 L sample of Ne to effuse through a porous membrane. How long would it take for the same volume of I₂ to effuse under the same conditions?

Problem #24a: How much faster does U²³⁵F₆ effuse than U²³⁸F₆?

Problem #24b: Calculate the ratio of effusion rates for U²³⁸F₆ and U²³⁵F₆. Express your answer using five significant figures and as the following ratio:

Molar Mass Of Helium Molecule

rate U²³⁸F₆ / rate U²³⁵F₆

Problem #25: O₃ effuses 0.8165 times as fast as O₂. What % of the molecules effusing first would be O₂?

Bonus Problem #1: HCl and NH₃ diffuse through a tube and a white disc of NH₄Cl is formed. Where in the tube?

Bonus Problem #2: One way of separating oxygen isotopes is by gaseous diffusion of carbon monoxide. The gaseous diffusion process behaves like an effusion process. Calculate the relative rates of effusion of:

¹²C¹⁶O
¹²C¹⁷O
¹²C¹⁸O