Answer:
Explanation:
From the given information:
The density of O₂ gas = [tex]d_{ideal} = \dfrac{P\times M}{RT}[/tex]
here:
P = pressure of the O₂ gas = 310 bar
= [tex]310 \ bar \times \dfrac{0.987 \ atm}{1 \ bar}[/tex]
= 305.97 atm
The temperature T = 415 K
The rate R = 0.0821 L.atm/mol.K
molar mass of O₂ gas = 32 g/mol
∴
[tex]d_{ideal} = \dfrac{305.97 \ \times 32}{0.0821 \times 415}[/tex]
[tex]d_{ideal}[/tex] = 287.37 g/L
To find the density using the Van der Waal equation
Recall that:
the Van der Waal constant for O₂ is:
a = 1.382 bar. L²/mol² &
b = 0.0319 L/mol
The initial step is to determine the volume = Vm
The Van der Waal equation can be represented as:
[tex]P =\dfrac{RT}{V-b}-\dfrac{a}{V^2}[/tex]
where;
R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol
Replacing our values into the above equation, we have:
[tex]310 =\dfrac{0.08314\times 415}{V-0.0319}-\dfrac{1.382}{V^2}[/tex]
[tex]310 =\dfrac{34.5031}{V-0.0319}-\dfrac{1.382}{V^2}[/tex]
[tex]310V^3 -44.389V^2+1.382V-0.044=0[/tex]
After solving;
V = 0.1152 L
∴
[tex]d_{Van \ der \ Waal} = \dfrac{32}{0.1152}[/tex]
[tex]d_{Van \ der \ Waal}[/tex] = 277.77 g/L
We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.
