Consider the low-speed flight of the Space Shuttle as it is nearing a landing. If the air
pressure and temperature of the shuttle are 1.2 atm and 300 K, respectively, what is the
air density? Note: 1 atm = 101325 N/m^2. Hint: Use ideal gas law!
To find the air density, we can use the ideal gas law, which states:
PV = nRT
Where:
P = pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = ideal gas constant
T = temperature of the gas
In this case, we know the pressure (P), temperature (T), and we need to find the air density. The number of moles (n) and the volume (V) are not given, but since we're only interested in finding the air density, we can assume that the volume of the gas is constant.
Therefore, we can rearrange the ideal gas law equation to solve for density (ρ):
PV = nRT
ρ = (n/V) = (P/RT)
To substitute values into this equation, we need to convert the pressure to pascals (Pa) and the temperature to Kelvin (K):
1 atm = 101325 N/m^2 (Pa)
1.2 atm = 1.2 * 101325 N/m^2 = 121590 N/m^2 (Pa)
Since the units of R are given in N·m/(mol·K), we don't need to convert it.
Now we can substitute the values into the equation:
ρ = (P/RT) = (121590 N/m^2) / (8.314 J/(mol·K) * 300 K)
Simplifying the equation further:
ρ = (121590 N/m^2) / (2494.2 J/(mol·K))
Finally, to convert the units from moles to kilograms, we need to use the molar mass of air, which is approximately 28.97 g/mol or 0.02897 kg/mol.
ρ = [(121590 N/m^2) / (2494.2 J/(0.02897 kg)]) = 1.18 kg/m^3
Therefore, the air density is approximately 1.18 kg/m^3.