Question

What is the pressure of 64 grams of O2 in a 1L container at a temperature of 10°C?

Answer 1

So PV=nRT
P=?

We know that:
-V=1L
-R=8.314kPa•L/mol•K
-T=10°C (we need to converse °C in K so 10+273) =283K
For n, we know that we have 64g of O2 and were looking for the quantity of mole in these 64g. 1 mole of O2 is 15.999g so 64g would be 4.0mol ((64g • 1mol)/15.999g) so n=4.0

P•1L= 4.0mol • 8.314kPa•L/mol•K • 283K
P • 1L = 9411
P = 9411/1L
P = 9411kPa