Calculating Moles and Molecules of Oxygen in a Gas Cylinder

[nafo man]

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

 

[Borek]

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.

 

[nafo man]

i am not sure,where did i go wrong?

 

[Borek]

For example: you have pressure in Pascals, but you try to use R with pressure in atm.

 

[nafo man]

if i am using atm and liters, then the value of R to use is 0.082L atm/mol K ?

 

[Borek]

Yes.

http://en.wikipedia.org/wiki/Gas_constant

 

[nafo man]

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)

 

[nafo man]

1200 / 2493 = 0.481347774

 

[nafo man]

am i on right track?

 

[nafo man]

the value of R=0.082 is in litre
n=1200/24.6 =48.78
I hope this is the right answer

 

[Borek]

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.