To solve this problem, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
1. First, let's convert the given temperature of 30°C to Kelvin:
T = 30°C + 273.15 = 303.15 K
2. We can use the ideal gas law to find the number of moles of air in the tank:
PV = nRT
n = PV / RT
Given:
P = 4.00 × 10^5 Pa
V = 20.0 L = 0.02 m^3
R = 8.314 J/(mol·K)
T = 303.15 K
n = (4.00 × 10^5 Pa × 0.02 m^3) / (8.314 J/(mol·K) × 303.15 K)
n ≈ 32.560 mol
3. To find the mass of air, we can use the molar mass of air (M = 28.8 g/mol):
m = n × M
m = 32.560 mol × 28.8 g/mol
m ≈ 938.752 g
m ≈ 0.939 kg (rounded to three decimal places)
4. Now, let's determine the volume of the air at normal atmospheric pressure (1 atm) and 0°C (273.15 K):
Use the combined gas law:
P₁V₁ / T₁ = P₂V₂ / T₂
Given:
P₁ = 4.00 × 10^5 Pa
V₁ = 0.02 m^3
T₁ = 303.15 K
P₂ = 1 atm ≈ 101325 Pa
V₂ = ?
T₂ = 273.15 K
Rearranging the equation:
V₂ = (P₁V₁T₂) / (T₁P₂)
V₂ = (4.00 × 10^5 Pa × 0.02 m^3 × 273.15 K) / (303.15 K × 101325 Pa)
V₂ ≈ 0.087 m^3
V₂ ≈ 87.402 L (rounded to three decimal places)
Therefore, the mass of air in the tank is approximately 0.939 kg, and it would occupy a volume of approximately 87.402 L at normal atmospheric pressure and 0°C.