31m/s is a typical highway speed for a car.
At what temperature do the molecules of nitrogen gas have an rms speed of 31 m/s?
(Answer in K)
The subject is Physics, ending with an s.
The average kinetic energy of any molecule of a gas is (3/2)kT. Set that equal to (1/2)m V^2, and solve for T.
T = m V^2/(3 k)
m is the mass of a nitrogen molecule and k = 1.38*10^-23 joule/K
thanks
To determine the temperature at which the molecules of nitrogen gas have an rms (root mean square) speed of 31 m/s, we can use the equation of kinetic theory:
v_rms = √(3kT/m)
Where:
v_rms = root mean square speed of the molecules
k = Boltzmann's constant (1.38 × 10^(-23) J/K)
T = temperature in Kelvin
m = mass of a nitrogen molecule (approximately 2.8 x 10^(-26) kg)
Rearranging the equation, we can solve for T:
T = (v_rms^2 * m) / (3k)
Plugging in the given values:
v_rms = 31 m/s
m = 2.8 x 10^(-26) kg
k = 1.38 × 10^(-23) J/K
Now we can calculate the temperature:
T = (31^2 * 2.8 x 10^(-26)) / (3 * 1.38 × 10^(-23))
By performing the calculations, we find:
T ≈ 129 K
Therefore, the temperature at which the molecules of nitrogen gas have an rms speed of 31 m/s is approximately 129 Kelvin.