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Grahams Law r a = m b r b m a

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Grahams Law r a = rate at which substance a travels r b = rate at which substance b travels m a = mass of substance a m b = mass of substance b

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Grahams Law Since rate is distance over time, this equation can also be rearranged as follows: t a = m a t b m b

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Grahams Law The practical effect of Grahams Law is that small molecules travel faster than larger molecules. This is true for both diffusion and effusion.

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Grahams Law Diffusion – molecules moving through the air Effusion – molecules moving through very small holes, such as the holes in a balloon

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Grahams Law What does this mean about air leaking out of a helium balloon?

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Grahams Law Helium, with a mass of 4 amu will leave the balloon faster than nitrogen and oxygen gases with masses of 28 and 32 amu.

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Grahams Law Solving problems with Grahams law is mostly just algebra.

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Grahams Law For example: A sample of an unknown gas flows through the wall of a porous cup in 39.9 min. An equal volume of gaseous hydrogen, measured at the same temperature and pressure, flows through in 9.75 min. What is the molar mass of the unknown gas?

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Grahams Law We need the equation using time. t a = m a t b m b

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Grahams Law What do we know? t a = 39.9 min t b = 9.75 min m a = x amu m b = 2 amu

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Grahams Law Plug into the equation: 39.9 = x 9.75 2

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Grahams Law Cross multiply and solve for x 39.9( 2) = 9.75( x) 56.43 = 9.75( x) 5.78 = ( x) 5.78 2 = x 33.49 = x

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