The element Kr has van der Waals constants a = 2.325 atm•L2/mol2 and b = 0.0396 L/mol. Using both the ideal gas law and van der Waals’s equation, calculate the pressure expected for 30 mol of Kr gas in a 6.00-L container at 25 °C.
1) Using ideal gas law
Given:
V = 6.0 L
n = 30 mol
T = 25.0 oC
= (25.0+273) K
= 298 K
use:
P * V = n*R*T
P * 6 L = 30 mol* 0.08206 atm.L/mol.K * 298 K
P = 122 atm
Answer: 122 atm
2) suing van der Waals equation
Given:
V = 6.0 L
n = 30.0 mol
R = 0.08206 atm.L/mol.K
T = 298.0 K
a = 2.325 atm.L^2/mol^2
b = 0.0396 L/mol
use:
(P+an^2/V^2)*(V-nb) = n*R*T
(P + 2.325*30.0^2/6.0^2)*(6.0-30.0*0.0396) = 30.0*0.08206*298.0
(P + 58.125)*(4.812) = 733.6164
P + 58.125 = 152.4556
P = 94.3 atm
Answer: 94.3 atm
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