How do you calculate the volume occupied by 64.0 grams of CH_4CH4 at 127°C under a pressure of 1535 torr?

1 Answer
anor277
Mar 2, 2017

For a start let's specify the pressure as (1535*mm*Hg)/(760*mm*Hg*atm^-1)=2.02*atm1535⋅mm⋅Hg760⋅mm⋅Hg⋅atm−1=2.02⋅atm.

Explanation:

mm*Hgmm⋅Hg are used for low pressures, or for pressures round about 1*atm1⋅atm. You do NOT use a mercury column for pressures over 1*atm1⋅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=nRTPV=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)V=nRTP=64.0⋅g16.04⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×400⋅K2.02⋅atm

~=65-70*L≅65−70⋅L

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