The van der Waals constants for a substance are `a=300.003 kPa dm^(6)mol^(-2)` and `b=40.8 cm^(3) mol^(-1)`. Find the critical constants of this substance.

Explanation

Given
`a=300.0kPa dm^(6)mol^(-2)`
`b=40.0 cm^(3)mol^(-1)=40xx10^(-3)dm^(3)mol^(-1)`
(`i`) `V_(c)=3b=3xx40.0 cm^(3)mol^(-1)=120.0 cm^(3)`
(`ii`) `P_(c)=(8a)/(27b^(2))`
`=(8xx300)/(27xx(40xx10^(-3))^(2))=(2400xx10^(6))/(43200)`
`=0.055xx10^(6)kPa`
(`iii`) `T_(c)=(8a)/(27Rb)`
`=(8xx300)/(27xx8.314xx40xx10^(-3))=(2400)/(8979.12)`
`=0.2672xx10^(3)=267 K`
Hence, `V_(c)=120.0 cm^(3)`, `P_(c)=0.055xx10^(6) kPa`,
`T_(c)=267 K`