Two gases A and B are held at the same temperature. One of the gases is diatomic, and the other is not. It is found that the ratio of the rms speeds of the two gases is:
V(a)/V(b) = 0.7143
Find the most likely composition of A and B. That is, what are the two gases?
For the purpose of this exercise, assume the gases are pure, and are composed of the most common isotope
V(a)/V(b) = 0.7143
Find the most likely composition of A and B. That is, what are the two gases?
For the purpose of this exercise, assume the gases are pure, and are composed of the most common isotope
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I think the two gases are Hydrogen and Helium.
The ratio of the velocities squared will be inversely proportional to the masses of the molecules/atoms. You can use the rms speed or kinetic energy; either way works.
Set V(a) / [√3kT/m(a)] = V(b) / [√3kT/m(b)]
Rearrange to get V(a)/V(b) = √3kT/m(a) / [√3kT/m(b)] = 0.7143
Use just this part √3kT/m(a) / [√3kT/m(b)] = 0.7143
Cancel the constants and square both sides.
m(b)/m(a) = 0.5102
So, m(a) is about twice as much as m(b). If m(a) is helium, it's about twice the mass as diatomic hydrogen. The ratio works, hydrogen is diatomic, helium is not. I'm pretty sure this is the right answer.
The ratio of the velocities squared will be inversely proportional to the masses of the molecules/atoms. You can use the rms speed or kinetic energy; either way works.
Set V(a) / [√3kT/m(a)] = V(b) / [√3kT/m(b)]
Rearrange to get V(a)/V(b) = √3kT/m(a) / [√3kT/m(b)] = 0.7143
Use just this part √3kT/m(a) / [√3kT/m(b)] = 0.7143
Cancel the constants and square both sides.
m(b)/m(a) = 0.5102
So, m(a) is about twice as much as m(b). If m(a) is helium, it's about twice the mass as diatomic hydrogen. The ratio works, hydrogen is diatomic, helium is not. I'm pretty sure this is the right answer.