If 1.00 mol of argon is placed in a 0.500-L container at 22.0 ∘C , what is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation)? For argon, a=1.345(L2⋅atm)/mol2 and b=0.03219L/mo

The answer to your question is ΔP = 1.57 atm

Explanation:

Data

moles = 1

volume = 0.5 L

Temperature = 22°C  or 295°K

Pressure = ?

a) Ideal Gas

                      PV = nRT          R = 0.082 atm L / mol°K

-Solve for P

                     P = nRT/V

                     P = (1)(0.082)(295)/ 0.5

                     P = 23.944/ 0.5

                    P = 47.9 atm

b) Van der Waals

                     (P + a/v²)(v - b) = RT

- Substitution

                     (P + 1.345/0.5²)(0.5 - 0.03219) = (0.082)(295)

- Simplification

                     (P + 5.38)(0.46781) = 24.19

                     P + 5.38 = 24.19/0.46781

                     P + 5.38 = 51.71

-Solve for P

                      P = 51.71 - 5.38

                      P = 46.33 atm

c) The difference between both Pressures is

        ΔP = 47.9 - 46.33

             = 1.57 atm