Atmosphere, carbon dioxide and water vapour help?

  • #1
atmosphere, carbon dioxide and water vapour help??

Homework Statement





Homework Equations





The Attempt at a Solution



(c)
One cubic metre (1 m3) of atmosphere at sea level contains 3.80 × 10(power of)2 ppm of CO2 and 5.00 × 10(power of)3 ppm of water vapour.
If there is a total of 2.6 × 10(power of)25 molecules in 1 m3 of air, calculate how many molecules of CO2 and water vapour there are in 1 m3 of air. Give your answers to the appropriate number of significant figures.

my workings-CO2 - 380 ppm water vapour - 5000 ppm

7.063% CO2, 92.463% water vapour

2.6 x 10(power of)25 / 100 = 2.6 x 10(power of)23
2.6 x 10(power of)23 x 7.063

CO2 1.8368 x 10(power of)24
water vapour 2.416336 x 10 (power of)25

As you can see what i have managed to do is completely loose the plot and dont even understand what i have done myself any help would be great cheers.
 

Answers and Replies

  • #2
Redbelly98
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Welcome to Physics Forums.

Homework Statement





Homework Equations





The Attempt at a Solution



(c)
One cubic metre (1 m3) of atmosphere at sea level contains 3.80 × 10(power of)2 ppm of CO2 and 5.00 × 10(power of)3 ppm of water vapour.
If there is a total of 2.6 × 10(power of)25 molecules in 1 m3 of air, calculate how many molecules of CO2 and water vapour there are in 1 m3 of air. Give your answers to the appropriate number of significant figures.

my workings-CO2 - 380 ppm water vapour - 5000 ppm

7.063% CO2, 92.463% water vapour
Well, it doesn't really work that way. Besides the CO2 and water vapour, there are other molecules as well.

Here is a better way to think about it:
If you have 1 million molecules total, then 380 of them will be CO2. That is what "ppm" means: parts per million.
So if 380 out of 1 million molecules are CO2, what percentage would that be?
 
  • #3


hey 0.038% and 0.500%? if this is right how does this help me with the next part of the question?

cheers emz
 
  • #4


Or would the answer simply be 2.80x10(p-o)2 ppm + 5.00x10(p-o)3 ppm = 5380 ppm

Appropriate number of significant figures - 5400 ppm of carbon dioxide and water vapour? or 5380 ppm of carbon dioxide and water vapour?

Could it be that simple I was just way way way over complicating things?

cheers emz
 
  • #5
Redbelly98
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hey 0.038% and 0.500%? if this is right how does this help me with the next part of the question?
You're on the right track here.

So if you have "a total of 2.6 × 1025 molecules in 1 m3 of air" (quoted from your original post), and 0.038% of those molecules are water vapour molecules, how many water vapour molecules are there?

Or would the answer simply be 2.80x10(p-o)2 ppm + 5.00x10(p-o)3 ppm = 5380 ppm
Well, no. They are asking how many total molecules, of each type, are there in 1 m3 of air. Giving an answer in ppm is really not an answer to that question.
 
  • #6


9.88x10(p-o)21 water vapour and 1.3x10(p-o)23 carbon dioxide?

i really dont know

emz
 
  • #7
Redbelly98
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Your numbers are correct. One minor detail: the problem said to use the appropriate number of significant figures. That would change one of your answers.
 
  • #8


so the final answer would be 9.88x10(p-o)21 of water vapour and 1.30x10(p-o)23 of carbon dioxide?

cheers emz
 
  • #9


sorry 9.88x10(p-o)21 carbon dioxide and 1.30x10(p-o)23 water vapour?
 
  • #10
Redbelly98
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What is your reasoning? How many significant figures are there in all the numbers involved in the calculation?
 
  • #11


2 and 3? i really dont know my head is fried!

9.9x10(p-o)21 carbon dioxide and 1.3x10(p-o)23 water vapour?

emz
 
  • #12
Redbelly98
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Don't guess!

Here are the numbers, how many sig figs in each?

3.80 × 102 ppm of CO2
5.00 × 103 ppm of water vapour.
2.6 × 1025 total molecules

Good luck to you.
 
  • #13


3, 3 and 2 so they have 2 significant figures as you round to the least precise one?!

thanks emz
 
  • #15


thank you very very much

emz
 
  • #16


Emz: you do realise that you are not allowed to publish TMA questions and answers don't you?

I advise you talk to your Open University tutor in future.

EW
OU S104 2011B
 

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