If 1.00 mol
of argon is placed in a 0.500- L
container at 26.0 ∘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.
To calculate the difference between the ideal pressure and the real pressure for argon in the given conditions, we can use the van der Waals equation:
(P + a(n/V)^2)(V - nb) = nRT
Where:
P = real pressure
a = 1.345 L^2•atm/mol^2
n = 1.00 mol
V = 0.500 L
b = 0.03219 L/mol
R = 0.08206 L•atm/(mol•K) (ideal gas constant)
T = 26.0°C = 299.0 K
First, let's calculate the ideal pressure using the ideal gas law:
PV = nRT
P = (nRT)/V
P = (1.00 mol * 0.08206 L•atm/(mol•K) * 299.0 K) / 0.500 L
P = 49.1 atm
Now, let's calculate the real pressure using the van der Waals equation:
(P + a(n/V)^2)(V - nb) = nRT
(P + (1.345 (1.00/0.500)^2) )((0.500) - (0.03219 * 1.00)) = (1.00 mol * 0.08206 L•atm/(mol•K) * 299.0 K)
(P + 10.76)(0.4678) = 24.45
(P + 10.76) = 52.26
P = 41.5 atm
The difference between the ideal pressure and real pressure is:
|41.5 atm - 49.1 atm| = 7.6 atm
Therefore, the difference between the ideal pressure and the real pressure is 7.6 atm.