How do you calculate the volume occupied by 64.0 grams of #CH_4# at 127°C under a pressure of 1535 torr?

1 Answer
Mar 2, 2017

Answer:

For a start let's specify the pressure as #(1535*mm*Hg)/(760*mm*Hg*atm^-1)=2.02*atm#.

Explanation:

#mm*Hg# are used for low pressures, or for pressures round about #1*atm#. You do NOT use a mercury column for pressures over #1*atm#. Why not? Because you risk contaminating your laboratory with elemental mercury, where it will inhabit every crack and every hole - this is major clean-up job which contract cleaners won't touch.

So now we can use the Ideal Gas Equation, #PV=nRT#, and thus, #V=(nRT)/P=((64.0*g)/(16.04*g*mol^-1)xx0.0821*(L*atm)/(K*mol)xx400*K)/(2.02*atm)#

#~=65-70*L#