At an altitude of 50.00 km, the average atmospheric temperature is essentially 0 degrees C. What is the average number of air molecules per cubic centimeter of air at this altitude?
To determine the average number of air molecules per cubic centimeter of air at an altitude of 50.00 km, we can use the ideal gas law equation. The ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature.
Since we want to find the number of air molecules per cubic centimeter, we can assume the volume is 1 cubic centimeter (V = 1 cm^3). The pressure at this altitude can be approximated as the atmospheric pressure, which is around 0.001 atm. The temperature is given as 0 degrees Celsius, which needs to be converted to Kelvin. Adding 273.15 to 0 gives 273.15 K.
The gas constant, R, is approximately 0.0821 L·atm/(mol·K). However, to match our units, we need to convert L to cm^3. There are 1000 cm^3 in one liter, so the gas constant can be written as 0.0821 cm^3·atm/(mol·K).
Now, we can rearrange the ideal gas law equation to solve for the number of moles of air molecules (n): n = PV/(RT).
Plugging in the known values:
P = 0.001 atm
V = 1 cm^3
R = 0.0821 cm^3·atm/(mol·K)
T = 273.15 K
n = (0.001 atm * 1 cm^3)/((0.0821 cm^3·atm/(mol·K)) * 273.15 K)
Simplifying the equation, we find n ≈ 5.17 x 10^-23 moles of air molecules per cubic centimeter.
To convert moles to number of molecules, we need to multiply by Avogadro's number, which is approximately 6.022 x 10^23 molecules per mole.
So, the average number of air molecules per cubic centimeter at an altitude of 50.00 km is approximately (5.17 x 10^-23 moles) * (6.022 x 10^23 molecules/mole) = 3.11 molecules of air per cubic centimeter.