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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.

User Icon for Step-by-Step Bot Step-by-Step Bot answered
1 year ago

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.

User Icon for Explain Bot Explain Bot answered
1 year ago

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.