Is the CO amount I found correct?

  • Thread starter chemman218
  • Start date
  • #1
14
0

Homework Statement


An automobile traveling 50 Km/hr emits 0.1% CO from the exhaust. If the exhaust rate is 80 m^3/min, what is the CO emission in grams per kilometer?

Homework Equations


PV=nrt

The Attempt at a Solution


flow rate volume of CO =(.8m^3/min)8(1min/.833km)=.960m^3/km*(1000L/1m^3)=960 L/km

PV=nRT
n=((1atm)(960L/km))/((.08206 L*atm/mol*k)*(298K))=39.25 moles/km of CO
39.25moles/km * 28g/mole of CO = 1099.65 g/km
 

Answers and Replies

  • #2
20,877
4,549
Yes, assuming that those 80 m^3/min are specifically referenced to 1 atm and 298K.

Chet
 
  • #3
14
0
Yes, it is at STP. I was trying to search the internet for reference on CO levels from automobiles to see if the number I got was correct but could not find anything. Thank you for your help!
 
  • #4
20,877
4,549
Yes, it is at STP. I was trying to search the internet for reference on CO levels from automobiles to see if the number I got was correct but could not find anything. Thank you for your help!
Isn't STP 1 atm and 0 C?

Chet
 
  • #5
14
0
Yes you are correct, but our teacher told us to use 298K.
 
  • #6
gneill
Mentor
20,913
2,862

The Attempt at a Solution


flow rate volume of CO =(.8m^3/min)8(1min/.833km)=.960m^3/km*(1000L/1m^3)=960 L/km
Shouldn't that be 0.08 m3/min if it's 0.1% of 80 m3/min?
 
  • #7
14
0
Thank you for the catch. I used the wrong percent to multiply. I used .01 instead of .001. Thank you so much for all your help!
 

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