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!
Take the hint.
Density = P/(RT)
R = 0.08205 liter*atm/(mole*K)
Use P = 1.2 atm. That will give you the density in mole/liter
Air has 28.9 grams per mole
To find the air density, we need to use the ideal gas law, which states that the product of pressure (P), volume (V), and the number of moles (n) of a gas is equal to the product of the gas constant (R) and the temperature (T) in Kelvin.
The ideal gas law equation is given by: PV = nRT
Here, we are given the following information:
Pressure (P) = 1.2 atm
Temperature (T) = 300 K
First, we need to convert the pressure from atm to Pascal (Pa) since the gas constant is given in SI units.
1 atm = 101325 N/m^2 (Pa)
Therefore, 1.2 atm = 1.2 * 101325 N/m^2 = 121590 N/m^2
Now we can rearrange the ideal gas law equation to solve for the number of moles (n) of the gas:
n = (PV) / (RT)
To calculate the gas constant (R), we can use the value R = 8.314 J/(mol K).
Now, substituting the given values into the equation, we have:
n = (121590 N/m^2 * V) / (8.314 J/(mol K) * 300 K)
Now we need to find the volume (V).
Since the space shuttle is a closed system, the volume (V) of the gas remains constant throughout its flight. Therefore, we can assume that V = constant.
So, we need to determine the value of V from the given information to find the air density.
Without knowing the volume of the space shuttle, we cannot calculate the air density using the given information alone. Additional information, such as the volume or mass of the gas, would be required to solve the problem completely.