To determine the pressure exerted by 1.00 mol of oxygen gas at 50 K in a volume of 0.0224 m³, we can use both the ideal gas law and the van der Waals equation. Let's break this down step by step.
Using the Ideal Gas Law
The ideal gas law is expressed as:
PV = nRT
Where:
- P = pressure (in Pascals)
- V = volume (in m³)
- n = number of moles of gas
- R = ideal gas constant (8.314 J/(mol·K))
- T = temperature (in Kelvin)
Plugging in the values:
- n = 1.00 mol
- V = 0.0224 m³
- R = 8.314 J/(mol·K)
- T = 50 K
Now, substituting these values into the ideal gas law:
P = (nRT) / V
P = (1.00 mol × 8.314 J/(mol·K) × 50 K) / 0.0224 m³
P = (415.7 J) / 0.0224 m³
P ≈ 18500.0 Pa or 18.5 kPa
Applying the Van der Waals Equation
The van der Waals equation accounts for the volume occupied by gas molecules and the intermolecular forces. It is given by:
(P + a(n/V)²)(V - nb) = nRT
Where:
- a = van der Waals constant for attraction (0.138 J·m³/mol²)
- b = van der Waals constant for volume (3.18 × 10⁻⁵ m³/mol)
First, we need to calculate the values of n/V and nb:
n/V = 1.00 mol / 0.0224 m³ ≈ 44.64 mol/m³
nb = 1.00 mol × 3.18 × 10⁻⁵ m³/mol ≈ 3.18 × 10⁻⁵ m³
Now, substituting these values into the van der Waals equation:
(P + 0.138(44.64)²)(0.0224 - 3.18 × 10⁻⁵) = 1.00 × 8.314 × 50
Calculating the term a(n/V)²:
0.138 × (44.64)² ≈ 0.138 × 1997.63 ≈ 275.4
Now, substituting back into the equation:
(P + 275.4)(0.0224 - 0.0000318) = 415.7
(P + 275.4)(0.0223682) = 415.7
Now, solving for P:
P + 275.4 = 415.7 / 0.0223682
P + 275.4 ≈ 18500.0
P ≈ 18500.0 - 275.4
P ≈ 18224.6 Pa or 18.2 kPa
Summary of Results
To summarize, the pressures calculated are:
- Ideal Gas Law Pressure: 18.5 kPa
- Van der Waals Pressure: 18.2 kPa
These results show how the van der Waals equation slightly adjusts the pressure downward compared to the ideal gas law, reflecting the real behavior of gases under conditions where intermolecular forces and molecular volume become significant.
