Question

At what temperature would the root mean square speed of oxygen molecules be

Answer 1

Ves = Vrms

11.2 x 10^3 = √(3KT/m)

T = [(11.2 x 10^3)^2 m] / 3k

Putting value of m and k

T = 8.360 x 10^4 k

Answer 2

here is the answer

Taking,

Mass of Oxygen molecule, M = 2.76x10^-26 kg

Boltzmann constant, k= 1.38x10^-23 J/K

Escape velocity, v= 11.2x10^3 m/s

Rms velocity = (3kT/M)^1/2

Rms velocity = (3 x 1.38x10^-23 x T/2.76 x 10^-26)^1/2 = 11.2x10^3

Squaring both sides,

3 x 1.38x10^-23 J/K x T/2.76 x 10^-26 kg = 1.25x10^8 m/s

T=1.25x10^8 m/s x 2.76x10^-26 kg /3 x 1.38x10^-23 J/K

Solving the above equation you’d get,

T = 8.3 x 10^4 K

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