Topic: mole% to %(v/v) (Read 2974 times)

maya_ver · « **on:** December 08, 2014, 05:02:19 PM »

The gas mixture contains of H2, N2, O2, CO, CO2, CH4, C3H8, C3H6 and C4H10.
1 m3 of gas mixture (at 101.325 Pa and 288,15 K) contains 10 mole% of CO.

How to calculate % (v/v) of CO in the gas mixture?

Borek · « **Reply #1 on:** December 08, 2014, 05:30:32 PM »

Have you heard about Avogadro's law?

maya_ver · « **Reply #2 on:** December 08, 2014, 05:52:08 PM »

Yes of course and I used it for calculation before I asked

As I got 0,2% (v/v) and my colleague got 10% I decided to ask an expert

Borek · « **Reply #3 on:** December 09, 2014, 02:43:45 AM »

How do you get 0.2% v/v from 10% mole using Avogadro's law?

maya_ver · « **Reply #4 on:** December 09, 2014, 06:22:57 AM »

Sorry, I wrote 0,2 instead of 0,1. But I see now it was 0,1 m3 not 0,1%.
V = (1 m3 / 100 mole%) * 10 mole% = 0,1 m3 CO
So I think my colleague was right when he got 10% v/v

Borek · « **Reply #5 on:** December 09, 2014, 07:42:01 AM »

Direct application of Avogadro's law tells you that molar fractions and volume fractions for gases are identical. So yes - 10% v/v is an obvious answer.

maya_ver · « **Reply #6 on:** December 09, 2014, 08:53:16 AM »

Thank you very much for your help

Topic: mole% to %(v/v) (Read 2974 times)

maya_ver

mole% to %(v/v)

Borek

Re: mole% to %(v/v)

maya_ver

Re: mole% to %(v/v)

Borek

Re: mole% to %(v/v)

maya_ver

Re: mole% to %(v/v)

Borek

Re: mole% to %(v/v)

maya_ver

Re: mole% to %(v/v)

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