Calculate the pressure exerted by one mole of `CO_(2)` gas at `273 K` van der Waals constant `a=3.592 dm^(6) atm mol^(-2)`. Assume that the volume occupied by `CO_(2)` molecules is negligible.
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The van der Waals equation for one mole of a gas is
`(P+(a)/(V^(2)))(V-b)=RT`
It is given that the volume occupied by `CO_(2)` molecules is negligible. Hence, the equation of state becomes
`(P+(a)/(V^(2)))(V)=RT`
or `p=(RT)/(V)-(a)/(V^(2))`
Assuming `V_(m)=22.414 dm^(3) mol^(-1)`, we get
`p=((8.314 kPa dm^(3) K^(-1) mol^(-1))(273 K))/((22.414 dm^(3) mol^(-1))-(3.592 dm^(6)atm mol^(-2))/((22.414 dm^(3) mol^(-1))^(2))`
`=101.246 kPa-7.15xx10^(-3)atm`
`=101.246 kPa-(7.15xx10^(-3)atm)((101.325 kPa)/(1 atm))`
`=101.264 kPa-0.724 kPa`
`=100.601 kPa`