If 1.00 mol of argon is placed in a 0.500-L container at 30.0 ∘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/mol.
Express your answer to two significant figures and include the appropriate units.
Ideal gas equation-
PV = nRT
So,
Pideal = nRT/V = 1mol*0.0821Latm/molK*303K / 0.500L
Pideal = 49.75atm
Van der waals equation-
(P + n2a/V2)(V-nb) = nRT
P = [nRT/(V-nb)] - n2a/V2
P=[1mol*0.0821Latm/molK*303K/(0.5L-1mol*0.03219L/mol)
- 1mol2*1.345L2atm/mol2/0.5L*0.5L
Preal = 24.876Latm/0.46781L - (1.345L2atm/0.25L2)
Preal = 53.175L - 5.38atm
Preal = 47.79atm
Difference = Pideal - Preal = 49.75atm - 47.79atm
Difference = 1.96atm = 2.0atm
If 1.00 mol of argon is placed in a 0.500-L container at 30.0 ∘C , what...
If 1.00 mol of argon is placed in a 0.500-L container at 30.0 ∘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/mol.
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