Chap10-Part4 - LASTCLASS(a 3RT(3(8.314 J K mol(298.1 K J 21...

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(a) Calculate the average kine2c energy of a gas molecule at 25 o C in units of Joules. molecule J x mol molecules x K mol K J N RT 21 23 10 2 . 6 ) / 10 02 . 6 ( 2 ) 1 . 298 )( / 314 . 8 )( 3 ( 2 3 = = = ε (b) Calculate the rms speed for H 2 and for O 2 at 25 o C in m/s. s m kg gm J s m kg mol gm K mol K J O u s m kg gm J s m kg mol gm K mol K J H u M RT u 500 ] 10 ][ / ][ / 32 ) 1 . 298 )( / 314 . 8 )( 3 ( [ ) ( 900 , 1 ] 10 ][ / ][ / 2 ) 1 . 298 )( / 314 . 8 )( 3 ( [ ) ( 3 3 2 2 2 3 2 2 2 = = = = = 2 LAST CLASS…….
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Graham’s Law of Effusion 1 2 2 1 2 1 2 1 2 1 / 3 / 3 M M r r M RT M RT u u r r = = = Effusion is the process in which gas molecules escape through a small hole into an evacuated space. Graham’s Law states that, for two gases at equal temperatures and pressures, the relaBve rates of effusion (r) are inversely proporBonal to the raBo of the square root of their molecular masses. The faster the molecules are moving, the more likely that they will encounter the hole and escape.
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The 2me required for 25‐mL samples of different gasses to diffuse through a pinhole into a vacuum. Diffusion ‐ The rate at which two gases mix. Effusion ‐ The rate at which a gas escapes through a pinhole into a vacuum.
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‐ Since KE avg is dependent only upon T, two different gases at the same temperature must have the same KE avg . The KineBc Molecular Theory &Graham's Law ½ m H V H 2 = ½ m O V O 2 m H V H 2 = m O V O 2 V H 2 V O 2 = m O m H V H V O = m O m H
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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?
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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?
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