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a tank contains 2 m3 of air at -93oc and a gage pressure of 1.4 mpa. determine the mass of air, in kg. the local atmospheric pressure is 1 atm.

User Konny
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Final answer:

The mass of the air in the tank can be determined using the ideal gas law equation, which relates the pressure, volume, and temperature of a gas to its mass. In this case, the mass of the air is approximately 383.6 kg.

Step-by-step explanation:

The mass of the air can be determined using the ideal gas law equation, which relates the pressure, volume, and temperature of a gas to its mass. The ideal gas law equation is:

pV = mRT

Where:

  • p is the pressure of the gas
  • V is the volume of the gas
  • m is the mass of the gas
  • R is the ideal gas constant
  • T is the temperature of the gas in Kelvin

In this question, the volume is given as 2 m³, the pressure is given as 1.4 MPa, and the temperature is given as -93°C. To convert the temperature to Kelvin, we add 273.15:

T = -93 + 273.15 = 180.15 K

Plugging these values into the equation:

1.4 MPa * 2 m³ = m * R * 180.15 K

Now, we need to convert the pressure from MPa to Pa and the volume from m³ to L:

1.4 MPa = 1.4 * 10^6 Pa

2 m³ = 2 * 10³ L

Substituting these values into the equation:

1.4 * 10^6 Pa * 2 * 10³ L = m * R * 180.15 K

Simplifying:

2.8 * 10^6 * 10³ = m * R * 180.15

Dividing both sides by the ideal gas constant R, which has a value of 8.314 J/(mol·K):

m = (2.8 * 10^6 * 10³) / (8.314 * 180.15) kg

Calculating this value:

m ≈ 383.6 kg

User Ekow
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