Number of moles and molecules

  • Thread starter nafo man
  • Start date
  • #1
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Homework Statement


A gas cylinder contains of 0.8 × 10^-3 m^3 volume oxygen.the temperature of the oxygen is 320k,and the pressure of the gas is 1.5 × 10^6 Pa.how to calculate:
number of moles and molecules of Oxygen?
mass of Oxygen if its molar mass is 32.0 × 10^-3 kg ?
The mass of a single molecule of gas?



Homework Equations



PV = nRT → n=PV/RT

The Attempt at a Solution


PV = nRT → n=PV/RT
p=1.5 × 10^6 Pa
T=320 k
V= 0.8 × 10^-3 m^3 = 0.8 litre.
Is the value of the R is 0.082L atm/mol K ?
so n= 1.5 x10^6 x 0.8/0.082 x 320

n=5 716.46341
is this the right way for part one?
thanks
 

Answers and Replies

  • #2
Borek
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You are on the right track, but R value is wrong. Please check wikipedia page on gas constant and select R value that fits all units you use. Alternatively, convert all data to units of your R constant.
 
  • #3
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i am not sure,where did i go wrong?
 
  • #4
Borek
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28,635
3,107
For example: you have pressure in Pascals, but you try to use R with pressure in atm.
 
  • #5
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if i am using atm and liters, then the value of R to use is 0.082L atm/mol K ?
 
  • #7
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n=PV/RT
P=1.5x10^6
V=0.8x10^-3
T=300 k
R=8.31
n=(1.5x10^6 x 0.8x10^-3)/8.31x300)
 
  • #8
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1200 / 2493 = 0.481347774
 
  • #9
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am i on right track?
 
  • #10
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the value of R=0.082 is in litre
n=1200/24.6 =48.78
I hope this is the right answer
 
  • #11
Borek
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It is not about hope, it is about calculations.

Please don't ignore units - you do it all the time:

n=PV/RT
P=1.5x10^6
V=0.8x10^-3
T=300 k
R=8.31
n=(1.5x10^6 x 0.8x10^-3)/8.31x300)
so you are not sure what units your answer has.

List all the units in your calculations - see what cancels out, what is left, then you will know if what you did is OK.

Even if I will tell you if the number you listed is right, you will still have no idea why.
 

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