Question

Consider 20.0 moles of CO2 in a 1.0 liter container at 300.0 K. What is the pressure predicted by the van der Waals equation? The ideal gas law constant is 0.08206 [L•atm] / [mol•K]. For CO2, the pressure correction constant is 3.658 L2•atm / mol 2, and the volume correction constant is 0.04286 L / mol.

Answer 1

Pressure predicted by van der waals equation is 1984.7 atm

Step-by-step explanation:

Van der waal gas equation-

`(P+\tfrac{n^(2)a}{V^(2)})(V-nb)=nRT`

Where, P is pressure of gas , n is number of moles of gas, a is pressure correction constant, V is volume of gas, b is volume correction constant , R is gas constant and T is temperature in kelvin scale.

So, plug-in all the given values in the above equation-

`[P+((20.0mol)^(2)* (3.658L^(2).atm.mol^(-2)))/((1.0L)^(2))][1.0L-(20.0mol*0.04286 L.mol^(-1) )]=(20.0mol)* (0.08206L.atm.mol^(-1).K^(-1))* (300.0K)`

or,
`[P+((20.0mol)^(2)* (3.658L^(2).atm.mol^(-2)))/((1.0L)^(2))]`=
`3447.90atm`

or, P = 1984.7 atm

So, Pressure predicted by van der waals equation is 1984.7 atm