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Topic: Mole-Volume Ratio Relationship in Gases  (Read 7868 times)

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Offline Vicstar

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Mole-Volume Ratio Relationship in Gases
« on: July 21, 2007, 10:44:58 PM »
Is there such a relationship in gases?

Because if there isn't, how can you answer this question:

The output of a diesel generator, which contains a mixture of CO2 and H2O gases, is collected in a
sealed vessel at 447 K. As the mixture cools to 298 K, the pressure falls from 3.0 atm to 0.80 atm.
What is the ratio of CO2 to H2O in the mixture?

Offline Yggdrasil

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Offline lrn24gve

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Re: Mole-Volume Ratio Relationship in Gases
« Reply #2 on: July 26, 2007, 09:41:03 PM »
You don't need to know a mole-volume relationship of gases in order to figure this out. By rearranging the ideal gas law, pv=nrt, you'll see that n = pv/rt. First of all, you want to find out the total moles when you started out, which would be:

moles of CO2 + moles of H2O = PV/RT

You know that the boiling point of water is 373 K, so by 298 K, H2O isn't a gas anymore and is a liquid now. Because there is only one gas left, you want to find the number of moles of that one gass, therefore giving you:

moles of CO2 = PV/RT

However, you're not done yet. You know that the ratio would be moles of CO2 / moles of H2O. In order to get that, you can divide the two formulas on the top.

(moles of CO2 + moles of H2O) / moles of CO2= (P1V/RT1) / (P2V/RT2)

You always know that R will be constant and through Gay-Lussac's law, you know that volume is constant. Therefore, simplifying your problem to:

(moles of CO2 + moles of H2O) / moles of CO2= (P1/T1) / (P2/T2)

and through some algebraic simplifications...

moles of H2O / moles of CO2 = P1T2 / P2T1

Just plug in the numbers for pressure and temperature and you'll get the ratio.

Offline AWK

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Re: Mole-Volume Ratio Relationship in Gases
« Reply #3 on: July 27, 2007, 01:48:12 AM »
You should take into account also a condensation of H2O
AWK

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