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

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.