At what temperature would the rms speed of a nitrogen atom equal the following speeds? (Note: The mass of a nitrogen atom is 2.324 10-26 kg.)
(a) the escape speed from Earth, 1.12 104 m/s
_____K
(b) the escape speed from the Moon, 2.37 103 m/s
_____K
To find the temperature at which the root mean square (rms) speed of a nitrogen atom equals a given speed, we can use the formula for rms velocity, which is related to temperature through the kinetic theory of gases.
The formula for rms velocity is given by:
vrms = √(3kT/m)
Where:
vrms is the root mean square speed,
k is the Boltzmann constant (1.38 x 10^-23 J/K),
T is the temperature in Kelvin,
m is the mass of the nitrogen atom (2.324 x 10^-26 kg).
We can rearrange the formula to solve for T:
T = (vrms^2 * m) / (3k)
Let's calculate the temperature using this formula for each given speed:
(a) Escape speed from Earth: vrms = 1.12 x 10^4 m/s
Substituting the values into the formula, we have:
T = ((1.12 x 10^4)^2 * 2.324 x 10^-26) / (3 * 1.38 x 10^-23)
T ≈ 6.25 K
Therefore, at a temperature of approximately 6.25 Kelvin, the rms speed of a nitrogen atom will be equal to the escape speed from Earth.
(b) Escape speed from the Moon: vrms = 2.37 x 10^3 m/s
Using the same formula:
T = ((2.37 x 10^3)^2 * 2.324 x 10^-26) / (3 * 1.38 x 10^-23)
T ≈ 0.07 K
Hence, at a temperature of approximately 0.07 Kelvin, the rms speed of a nitrogen atom will be equal to the escape speed from the Moon.
To find the temperature at which the root mean square (rms) speed of a nitrogen atom is equal to a given speed, we can use the formula for the rms speed, given by:
v_rms = sqrt((3 * k * T) / m)
where:
v_rms is the root mean square speed,
k is the Boltzmann constant (1.38 * 10^-23 J/K),
T is the temperature in Kelvin, and
m is the mass of the nitrogen atom.
Let's calculate the temperature for each case:
(a) Escape speed from Earth (v = 1.12 * 10^4 m/s):
First, we need to find the rms speed.
v_rms = 1.12 * 10^4 m/s
Now, let's rearrange the formula to solve for T:
T = (m * v_rms^2) / (3 * k)
Plugging in the values:
T = (2.324 * 10^-26 kg * (1.12 * 10^4 m/s)^2) / (3 * 1.38 * 10^-23 J/K)
Calculating:
T ≈ 152.09 K
Therefore, at a temperature of approximately 152.09 Kelvin, the rms speed of a nitrogen atom would be equal to the escape speed from Earth.
(b) Escape speed from the Moon (v = 2.37 * 10^3 m/s):
Using the same formula, let's find the temperature:
T = (m * v_rms^2) / (3 * k)
T = (2.324 * 10^-26 kg * (2.37 * 10^3 m/s)^2) / (3 * 1.38 * 10^-23 J/K)
Calculating:
T ≈ 77.29 K
Therefore, at a temperature of approximately 77.29 Kelvin, the rms speed of a nitrogen atom would be equal to the escape speed from the Moon.