A tank attached to an air compressor contains 20.0 litre of air at a temperature of 30¡ãc and guage pressure of 4.00¡Á10^5 pa. What is the mass of air, and what volume would it occupy at normal atmospheric pressure and 0¡ãc. Air is a mixture of gases, consisting of about 78% nitrogen and 21% oxygen, with small percentage of other gases. The average molecular mass is M= 28.8g/mol.
Pls, I need help I keep getting mass to be 0.117 instead of 0.115 and affects the volume 0.941 instead of 0.874.
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.
To find the mass of air in the tank, we can use the ideal gas law equation: 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.
First, we need to convert the given pressure from Pascals to atmospheres (atm):
1 atm = 101325 Pa
So, 4.00 x 10^5 Pa = (4.00 x 10^5 Pa) / (101325 Pa/atm) = 3.95 atm (approximately)
Next, we need to convert the given volume from liters to moles:
1 mole of gas occupies 22.4 liters at standard temperature and pressure (STP).
So, 20.0 liters = (20.0 liters) / (22.4 liters/mole) = 0.893 moles (approximately)
Now, we can calculate the mass of air in the tank using the molecular mass and the number of moles:
Mass = n x M
Mass = 0.893 moles x 28.8 g/mol = 25.7 g (approximately)
So, the mass of air in the tank is approximately 25.7 g.
To find the volume the air would occupy at normal atmospheric pressure (1 atm) and 0°C, we can use the combined gas law equation: (P1 x V1) / T1 = (P2 x V2) / T2, where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature.
Given:
P1 = 3.95 atm (from earlier calculation)
V1 = 20.0 liters
T1 = 30°C = 273 + 30 = 303 K
P2 = 1 atm
T2 = 0°C = 273 K
Substituting the values into the equation:
(3.95 atm x 20.0 liters) / 303 K = (1 atm x V2) / 273 K
Solving for V2 (the final volume):
V2 = (3.95 atm x 20.0 liters x 273 K) / (303 K)
V2 = 17.70 liters (approximately)
So, the volume the air would occupy at normal atmospheric pressure and 0°C is approximately 17.70 liters.
Therefore, the mass of air in the tank is approximately 25.7 g and the volume it would occupy at normal atmospheric pressure and 0°C is approximately 17.70 liters.