Calculate the weight of methane in a `9 L` cylinder at `16 atm` and `27^(@)C` temperature. (`R=0.08 L atm K^(-1)mol^(-1)`)

Explanation

The ideal gas equation is
`PV=nRT`
Given `P=16 atm`, `R=0.08`, `V=9 L`, `T=273+27^(@)C=300 K`
Putting these value in the above equation, we get
`16xx9=nxx0.08xx300`
` :.` Moles `(n)=(16xx9)/(0.08xx300)=6`
We know
Moles `(n)=(Weight)/(Molecular weight)`
Weight of methane`= `Moles `xx `Molecular weight of methane
`=6xx16 (CH_(4)=12+4=16)`
`=96 g`