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Starting with the definition of rate of effusion and Graham’s finding relating rate and molar mass, show how to derive the Graham’s law equation, relating the relative rates of effusion for two gases to their molecular masses.

Therefore, the relative rates of effusion will be inversely proportional to the square root of their molecular masses.

01:45

Aadit S.

Chemistry 101

Chapter 9

Gases

Carleton College

University of Central Florida

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So we're continuing to look at the fusion, and we know that the rate of effusion of gas A is in vastly proportional to the one over the route off the mass molecular weight. So if we had two masters so the rate off effusion of a bad by the rate of effusion off B that is proportional to Group B spies by route and a so the kinetic energy of gasses depend only on the temperature, so of two glasses are at the same temperature. This means that they have the same kinetic energy so kinetic energy. It's 1/2 m v squared where k a kinetic energy A is equal to the kinetic energy R B therefore, 1/2 i m a b b b A squared is equal to 1/2 m b b b squared. So we have B A squad. Bye bye, B b squared people to, um, be divided by m A. On dso last step Here we have is a over B B equal to root be by the by route A. Therefore, the relative rates of effusion will be inversely proportional to the square root of the molecular masses.

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