Calculate Resulting Pressure of Mixed Gases in 12L Vessel

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SUMMARY

The resulting pressure of mixed gases in a 12L vessel can be calculated using the ideal gas law and principles of gas mixtures. For the given quantities: 4.0 L of Neon (Ne) at 2.0 atm, 2.0 L of Helium (He) at 3.0 atm, and 2.0 L of Argon (Ar) at 5.0 atm, the final pressure is determined to be 2.0 atm. Two methods are recommended for this calculation: one involves calculating the number of moles of each gas and applying the ideal gas law, while the other method involves converting all gas volumes to standard conditions before compressing to the desired volume.

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Homework Statement


What is the resulting pressure when the quantities of gases listed below
are mixed and placed in a 12 L vessel at constant temperature?
4.0 L Ne measured @ 2.0 atm
2.0 L He measured @ 3.0 atm
2.0 L Ar measured @ 5.0 atm

Homework Equations


P1V1=P2V2

The Attempt at a Solution


the answer is 2.0 atm.and i have no clue how they figured that out.
because i tried to do it by the above equation. and that definitely did not work out.and then i tried adding up the initial volumes:8 L and the initial temperatures:10 atm.and that didn't work either.so basically;i need helpp.
and hopefully.before tuesday morning.because that's when my ap chem test is. which is why you see me here posting all these questions.

thanks a lot physicsforum people.you guys seriously save my life these days :redface:
 
Last edited:
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One method: calculate the number of mole of each component, add up, use the perfect gas law for the desired volume.

Second method: convert all components as volumes for 1 atm, add up the volumes still at 1 atm, compress to the desired volume. (this method is the same as the previous one, but moles are replaced by "standard volume").
 

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