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Topic: Partial Pressure  (Read 6803 times)

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Offline Korokian

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Partial Pressure
« on: January 28, 2007, 01:02:34 AM »
N2O4(g) ---> 2NO2(g)
constant temperature
starting pressure of N2O4(g) is 0.100 atm
after some time the pressure increases from 0.100 atm to 0.148 atm.
partial pressure of N2O4?

Offline Borek

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Re: Partial Pressure
« Reply #1 on: January 28, 2007, 05:13:36 AM »
Over 80 posts already and you still don't know you have to try by yourself first?
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Offline Korokian

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Re: Partial Pressure
« Reply #2 on: January 28, 2007, 12:27:22 PM »
i do try it first.. don't you know that i post because i need help? or are you just going to keep criticizing me?
i don't know how to start, why else would i post here....

okay let me state the obvious borek

total pressure= p1 + p2
PV=nRT
........you going to help or just keep saying try by yourself first

Offline Borek

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Re: Partial Pressure
« Reply #3 on: January 28, 2007, 12:46:15 PM »
total pressure= p1 + p2
PV=nRT

Can you calculate initial n(N2O4) from these equations and data given in the question?
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Offline Korokian

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Re: Partial Pressure
« Reply #4 on: January 28, 2007, 01:12:59 PM »
you know that 1 mol N2O4(g) makes 2 mol NO2(g)
but you don't have the volume or temperature
temperature is constant but you still don't have volume

Offline Korokian

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Re: Partial Pressure
« Reply #5 on: January 28, 2007, 01:41:03 PM »
i did .148/ 3
and got the partial pressure of N2O4 to be .049333 atm

Offline Borek

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Re: Partial Pressure
« Reply #6 on: January 28, 2007, 03:21:15 PM »
No, you are on the wrong track.

Try to assume you start with 1 mole of N2O4 - that will give you initial n and V.

If so T & R are just constants so you don't have to think about them.
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Offline Korokian

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Re: Partial Pressure
« Reply #7 on: January 28, 2007, 05:20:27 PM »
so initial mol is 1 mol
initial V would be 10 L

final V is 10 L
final P is .148 atm
final mol is 1.48 mol?

and 1.48 / 3 ?

Offline Borek

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Re: Partial Pressure
« Reply #8 on: January 28, 2007, 05:28:02 PM »
so initial mol is 1 mol
initial V would be 10 L

Huh? 1 mol of gas at STP is 22.4L, you have 10 times lower pressure, so the volume must be 10 times that.
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Offline Korokian

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Re: Partial Pressure
« Reply #9 on: January 29, 2007, 12:27:36 PM »
so then initial volume is 224 L

i really don't see where this is going

Offline Borek

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Re: Partial Pressure
« Reply #10 on: January 29, 2007, 01:29:41 PM »
What will be amount of BOTH gases after n moles of N2O4 decomposes?

Try with 0.1 mole if n doesn't work for you.
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Offline Korokian

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Re: Partial Pressure
« Reply #11 on: January 29, 2007, 01:59:37 PM »
.1-x + 2x mols

Offline Borek

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Re: Partial Pressure
« Reply #12 on: January 29, 2007, 02:37:18 PM »
OK, now that you know total number of moles you can calculate both total pressure and partial pressure of N2O4. Rest should be easy ;)
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