It turns out that the van der Waals constant b
equals four times the total volume actually occupied by the molecules of a mole of gas. Calculate the fraction of the volume in a container actually occupied by Ar
atoms (b
= 0.0322 L/mol
):
at 150 atm
pressure and 0 ∘C
. (Assume for simplicity that the ideal-gas equation still holds.)
The van der Waals equation is given by:
(P + a(n/V)^2)(V - nb) = nRT
Where:
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
a = van der Waals constant
b = van der Waals constant
Given:
P = 150 atm
T = 0°C = 273 K
a = 0 (for simplicity, assume ideal gas behavior for Ar)
b = 0.0322 L/mol
To calculate the fraction of volume actually occupied by Ar atoms, we need to solve for V_actual and V_total:
V_actual = V - nb
V_total = V
Since b = 4(V_actual), we have:
V_actual = b/4
V_total = V
Substitute b/4 for V_actual in the van der Waals equation:
(150 + 0)(V - 0.0322) = nRT
150V - 4.83 = nRT
Since V_total = V, we have:
150V = nRT
Divide the two equations:
150V - 4.83 = 150V
-4.83 = 0
This means that the volume occupied by the Ar atoms is negligible compared to the total volume in the container. Therefore, the fraction of the volume in the container actually occupied by Ar atoms is approximately 0.