The density of air is determined by the closeness of air molecules. It is influenced by factors such as temperature, pressure, and humidity. When the temperature is around 25 degrees Celsius and standard atmospheric pressure is present, the density of dry air is approximately 1.225 kilograms per cubic meter.
If specific values for temperature, pressure, and humidity are available, a formula known as the ideal gas law can be utilized. By making adjustments, a more accurate value of air density can be obtained.
Air density can be affected by different temperatures, pressures, and humidity levels. When it gets hotter, the air becomes less dense. Similarly, when pressure increases, air density is also increased. A slight impact on air density can also be observed due to the presence of water vapor in the air, which is known as humidity.
The mass of air contained in a given volume of the Earth’s atmosphere, known as air density (ρ), diminishes as altitude increases. Moreover, air density is also influenced by temperature and humidity alterations. According to the International Standard Atmosphere (ISA), air density at sea level and 15°C is approximately 1.225 kg/m3.
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Formula for Density of Air
Density
ρ = m/V
where,
ρ = density [kg/m3] or [lb/ft3]
m = mass [kg] or [lb]
V = volume [m3] or [ft3]
Formula for Specific Weight of Air
Specific Weight
γ = (m x g)/V = ρ x g
where,
γ = specific weight [N/m3] or [lbf/ft3]
m = mass [g] or [lb]
g = acceleration due to gravity [m/s2]
V = volume [cm3] or [ft3]
ρ = density [g/cm3] or [lb/ft3]
Density of air at standard atmospheric temperature – Imperial Units [PDF]
| Temperature (°F) | Density (slugs/ft3) |
| -40 | 0.002939 |
| -20 | 0.002805 |
| 0 | 0.002683 |
| 10 | 0.002626 |
| 20 | 0.002571 |
| 30 | 0.002519 |
| 40 | 0.002469 |
| 50 | 0.002420 |
| 60 | 0.002373 |
| 70 | 0.002329 |
| 80 | 0.002286 |
| 90 | 0.002244 |
| 100 | 0.002204 |
| 120 | 0.002128 |
| 140 | 0.002057 |
| 160 | 0.001990 |
| 180 | 0.001928 |
| 200 | 0.001870 |
| 250 | 0.001756 |
| 300 | 0.001624 |
| 400 | 0.001435 |
| 500 | 0.001285 |
| 750 | 0.001020 |
| 1000 | 0.000845 |
| 1500 | 0.000629 |
Specific Weight of air at standard atmospheric temperature – Imperial Units [PDF]
| Temperature (°F) | Specific Weight (lb/ft3) |
| -40 | 0.09456 |
| -20 | 0.09026 |
| 0 | 0.08633 |
| 10 | 0.08449 |
| 20 | 0.08273 |
| 30 | 0.08104 |
| 40 | 0.07942 |
| 50 | 0.07786 |
| 60 | 0.07636 |
| 70 | 0.07492 |
| 80 | 0.07353 |
| 90 | 0.07219 |
| 100 | 0.07090 |
| 120 | 0.06846 |
| 140 | 0.06617 |
| 160 | 0.06404 |
| 180 | 0.06204 |
| 200 | 0.06016 |
| 250 | 0.05598 |
| 300 | 0.05224 |
| 400 | 0.04616 |
| 500 | 0.04135 |
| 750 | 0.03280 |
| 1000 | 0.02717 |
| 1500 | 0.02024 |
Density of air at standard atmospheric temperature – SI Units [PDF]
| Temperature (°C) | Density (kg/m3) |
| -40 | 1.514 |
| -20 | 1.395 |
| 0 | 1.293 |
| 10 | 1.247 |
| 20 | 1.204 |
| 30 | 1.165 |
| 40 | 1.127 |
| 50 | 1.109 |
| 60 | 1.060 |
| 70 | 1.029 |
| 80 | 0.9996 |
| 90 | 0.9721 |
| 100 | 0.9461 |
| 200 | 0.7461 |
| 300 | 0.6159 |
| 400 | 0.5243 |
| 500 | 0.4565 |
| 1000 | 0.2772 |
Specific Weight of air at standard atmospheric temperature – SI Units [PDF]
| Temperature (°C) | Specific Weight (N/m3) |
| -40 | 14.85 |
| -20 | 13.68. |
| 0 | 12.67 |
| 10 | 12.23 |
| 20 | 11.81 |
| 30 | 11.43 |
| 40 | 11.04 |
| 50 | 10.88 |
| 60 | 10.40 |
| 70 | 10.09 |
| 80 | 9.803 |
| 90 | 9.533 |
| 100 | 9.278 |
| 200 | 7.317 |
| 300 | 6.040 |
| 400 | 5.142 |
| 500 | 4.477 |
| 1000 | 22.719 |
Example – Lifting Force of a Hot Air Balloon
An air balloon with volume 10 m3 is heated to 100 oC. The temperature of the surrounding air is 20 oC. The change in gravity force (weight) of the air volume is the potential lifting force of the balloon.
The lifting force can be calculated as,
FL = dm ag
= V dρ ag
= 10 x (1.205 – 0.946) x 9.81 = 25.4 N
where,
FL= lifting force – change in gravity force [N]
ag = acceleration of gravity = 9.81 [m/s2]
dm = V dρ = change of mass in the balloon [kg]
dρ = change in density due to temperature difference [kg/m3]
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FAQ’s
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Conclusion
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