# Graham s Law of Diffusion and Effusion in C# Decoder Data Matrix ECC200 in C# Graham s Law of Diffusion and Effusion

Graham s Law of Diffusion and Effusion
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Graham s law defines the relationship of the speed of gas diffusion (mixing of gases due to their kinetic energy) or effusion (movement of a gas through a tiny opening) and the gases molecular mass The lighter the gas, the faster is its rate of effusion Normally this is set up as the comparison of the effusion rates of two gases, and the specific mathematical relationship is: r1 M2 = r2 M1 where r1 and r2 are the rates of effusion/diffusion of gases 1 and 2 respectively, and M2 and M1 are the molecular masses of gases 2 and 1 respectively Note that this is an inverse relationship For example, suppose you wanted to calculate the ratio of effusion rates for hydrogen and nitrogen gases Remember that both are diatomic, so the molecular mass of H2 is 2016 g/mol and the molecular mass of N2 would be 2802 g/mol Substituting into the Graham s law equation: rH MN 2 2 = rN MH
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r H2/r N2 = (2802 g/mol/2016 g/mol)
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= (13899)1/2 = 3728
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Hydrogen gas would effuse through a pinhole 3728 times as fast as nitrogen gas The answer is reasonable, since the lower the molecular mass, the faster the gas is moving Sometimes we measure the effusion rates of a known gas and an unknown gas, and use Graham s law to calculate the molecular mass of the unknown gas
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Gas Stoichiometry
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The gas law relationships can be used in reaction stoichiometry problems For example, suppose you have a mixture of KClO3 and NaCl, and you want to determine how many
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Gases 111
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grams of KClO3 are present You take the mixture and heat it The KCIO3 decomposes according to the equation: 2 KClO3(s) 2 KCl(s) + 3 O2(g) The oxygen gas that is formed is collected by displacement of water It occupies a volume of 542 mL at 27 C The atmospheric pressure is 7550 torr The vapor pressure of water at 27 C is 267 torr First, you need to determine the pressure of just the oxygen gas It was collected over water, so the total pressure of 7550 torr is the sum of the partial pressures of the oxygen and the water vapor: PTotal = PO + PH O (Dalton s law)
The partial pressure of water vapor at 27 C is 267 torr, so the partial pressure of the oxygen can be calculated by: PO = PTotal PH O = 7550 torr 267 torr = 7283 torr
At this point you have 542 mL of oxygen gas at 7283 torr and 300 K (27 C + 273) From this data you can use the ideal gas equation to calculate the number of moles of oxygen gas produced: PV = nRT PV /RT = n You will need to convert the pressure from torr to atm: (7283 torr) (1 atm/7600 torr) = 09583 atm and express the volume in liters: 542 mL = 0542 L Now you can substitute the quantities into the ideal gas equation: (09583 atm)(0542 L)/(00821 L atm/K mol)(300 K) = n 002110 mol O2 = n Now you can use the reaction stoichiometry to convert from moles O2 to moles KClO3 and then to grams KClO3: 2 mol KClO 12255 g KClO 3 3 (002110 mol O2 ) = 1723 g KClO3 3 mol O2 1 mol KClO3
Non-Ideal Gases
We have been considering ideal gases, that is, gases that obey the postulates of the Kinetic Molecular Theory But remember a couple of those postulates were on shaky ground The volume of the gas molecules was negligible, and there were no attractive forces between the gas particles Many times approximations are fine and the ideal gas equation works well But it would be nice to have a more accurate model for doing extremely precise work or when a gas exhibits a relatively large attractive force In 1873, Johannes van der Waals introduced a modification of the ideal gas equation that attempted to take into account the volume and attractive forces of real gases by introducing two constants a and b into the ideal gas equation Van der Waals realized that the actual volume of the gas is less than the ideal gas because gas molecules have a finite volume He also realized that the more moles of gas present, the greater the real volume He compensated for the volume of the gas particles mathematically with: corrected volume = V nb