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.

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