Calculate the volume occupied by 8.8 g of `CO_(2)` at `31.1^(@)C` and 1 bar pressure. R = 0.083 bar L `K^(-1)mol^(-1)`.

Explanation

According to ideal gas equations,
pV = nRT
`therefore V=(nRT)/(p)` but mole `= n=("Weight"_((g)))/("Molar mass (M)")`
`therefore V=(gRT)/(Mp)`
where, g = 8.8 gram
R = 0.083 bar L `K^(-1)mol^(-1)`
`T=31.1^(@)C=(31.1+273)K=304.1K`
M = molar mass of `CO_(2)=(12+32)=44g mol^(-1)`
p = 1 bar
`V=((8.8g)(0.083" bar L K"^(-1) mol^(-1))(304.1 K))/((44g mol^(-1))("1 bar"))`
`= 5.04806 L ~~ 5.048 L`