Question

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

Answer 1

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`