If 1.00 mol of argon is placed in a 0.500-l container at 27.0 degree 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/mol. express your answer to two significant figures and include the appropriate units.

2.0 atm is the difference between the ideal pressure and  the real pressure.

Explanation:

If 1.00 mole of argon is placed in a 0.500-L container at 27.0 °C

Moles of argon = n = 1.00 mol

Volume of the container,V  = 0.500 L

Ideal pressure of the gas = P

Temperature of the gas,T = 27 °C = 300.15 K[/tex]

Using ideal gas equation:

Vander wall's of equation of gases:

The real pressure of the gas=

For argon:

b=0.03219 L/mol.

Difference :

2.0 atm is the difference between the ideal pressure and  the real pressure.

< > according to newton's third law of motion, forces always act in equal but opposite pairs. another way of saying this is for every action, there is an equal but opposite reaction.< >

b. accuracy of the data matters alot more than its number of digits, organization or amount of detail. quantitative observation is more accepted than the qualitative observation.