At what temperature does the average speed of an oxygen molecule equal that of an airplane moving at 560 mph?
To find the temperature at which the average speed of an oxygen molecule equals that of an airplane moving at 560 mph, we can use the concept of the kinetic theory of gases. According to the kinetic theory of gases, the average translational kinetic energy of gas molecules is directly proportional to the temperature. Therefore, we can equate the translational kinetic energy of an oxygen molecule to that of an airplane and solve for the temperature.
First, let's calculate the average translational kinetic energy of an oxygen molecule and an airplane.
1. For an oxygen molecule:
The translational kinetic energy (KE) of a molecule is given by the equation: KE = (1/2) * m * v^2, where m is the mass of the molecule and v is the velocity/speed.
The mass of an oxygen molecule (O2) is approximately 5.31 x 10^-26 kg.
The average speed (v) of an oxygen molecule can be calculated using the formula: v = sqrt((3 * k * T) / m), where k is the Boltzmann constant (1.38 x 10^-23 J/K) and T is the temperature in Kelvin.
2. For an airplane:
The translational kinetic energy (KE) of an airplane can be calculated using the equation: KE = (1/2) * m * v^2, where m is the mass of the airplane and v is the velocity/speed.
Now, let's set the kinetic energies of the oxygen molecule and the airplane equal to each other and solve for the temperature (T).
(1/2) * m * v^2 (O2) = (1/2) * m * v^2 (airplane)
Using the formulas mentioned earlier, we have:
(1/2) * m * (sqrt((3 * k * T) / m))^2 (O2) = (1/2) * m * (560 mph)^2 (airplane)
Simplifying and canceling out terms:
(3 * k * T) / m (O2) = (560 mph)^2 (airplane)
Now, we can substitute the values into the equation and solve for T:
(3 * (1.38 x 10^-23 J/K) * T) / (5.31 x 10^-26 kg) = (560 mph)^2
Let's convert mph to m/s and solve for T:
(3 * (1.38 x 10^-23 J/K) * T) / (5.31 x 10^-26 kg) = (560 mph * (0.44704 m/s/mph))^2
(3 * (1.38 x 10^-23 J/K) * T) / (5.31 x 10^-26 kg) = (560 * 0.44704)^2 m^2/s^2
(3 * (1.38 x 10^-23 J/K) * T) / (5.31 x 10^-26 kg) = (560 * 0.44704)^2 m^2/s^2
T = ((560 * 0.44704)^2 * (5.31 x 10^-26 kg)) / (3 * (1.38 x 10^-23 J/K))
Evaluating this expression, we can determine the temperature (T) at which the average speed of an oxygen molecule equals that of an airplane moving at 560 mph.
Please note that the numerical value will depend on the accuracy of the input variables used in the calculations.