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 equation for rms speed, which is given by:
v(rms) = sqrt((3kT) / m)
Where:
- v(rms) is the rms speed of the nitrogen atom
- k is the Boltzmann constant (1.38 x 10^-23 J/K)
- T is the temperature in Kelvin (K)
- m is the mass of the nitrogen atom (2.324 x 10^-26 kg)
(a) To find the temperature at which the rms speed of a nitrogen atom equals the escape speed from Earth (1.12 x 10^4 m/s):
We can rearrange the equation and solve for T:
T = (m * v(rms)^2) / (3k)
Substituting the given values:
T = (2.324 x 10^-26 kg * (1.12 x 10^4 m/s)^2) / (3 * (1.38 x 10^-23 J/K))
Now we can calculate this expression:
T = (2.61837 x 10^-18 J) / (4.14 x 10^-23 J/K)
T ≈ 6.315 K
Therefore, at a temperature of approximately 6.315 Kelvin, the rms speed of a nitrogen atom equals the escape speed from Earth.
(b) To find the temperature at which the rms speed of a nitrogen atom equals the escape speed from the Moon (2.37 x 10^3 m/s):
Using the same equation and substituting the given values:
T = (2.324 x 10^-26 kg * (2.37 x 10^3 m/s)^2) / (3 * (1.38 x 10^-23 J/K))
Now we can calculate this expression:
T = (1.38022 x 10^-20 J) / (4.14 x 10^-23 J/K)
T ≈ 333.6 K
Therefore, at a temperature of approximately 333.6 Kelvin, the rms speed of a nitrogen atom equals the escape speed from the Moon.