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.
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.
1 answer