Calculate the pressure exerted by `22g` of `CO_(2)` in `0.5 dm^(3)` at `300 K` using (`a`) the ideal gas law and (`b`) the van der Waals equation. Given `a=300.0 kPa dm^(6) mol^(-2)` and `b=40.0 cm^(3) mol^(-1)`.

Explanation

Moles of `CO_(2)=(w)/(Molar weight)=(22 g)/(44 g mol^(-1))=0.5 mol`
`V=0.5 dm^(3)`, `T=300 K`, `a=300.0 kPa dm^(6)mol^(-2)`
`b=40.0 cm^(3)mol^(-1)=0.04dm^(3)mol^(-1)`
(`a`) From ideal gas law `P=(nRT)/(V)`, we have
`P=((0.5)(8.314)xx300)/(0.5)=2494.2 Pa=2.49xx10^(3)kPa`
(`b`) From the van der Waals equation, we get
`P=(nRT)/(V-nb)-(n^(2)a)/(V^(2))`
`:. P=((0.5)xx(8.314)xx300)/((0.5)-(0.5)(0.04))-((0.5)^(2)xx(300))/((0.5)^(2))`
`=2598.12 kPa-300kPa=2298.12 kPa`