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Two containers are at the same temperature. The first
containsgas with pressure p1,molecular
mass m1,and
root-mean-square speed vrms1.The
second contains gas with pressure 4p1,molecular
mass m2,and
average speed vavg2=
5vrms1.Find
the mass ratio m1/m2.

Answer :

Answer:

From Maxwell-Boltzmann distribution of molecules and their speeds,

average speed, Vavg = √([tex]\frac{8RT}{πM}[/tex])

and Root mean speed, Vrms = √([tex]\frac{3RT}{M}[/tex])

Given Vavg2 = 5Vrms1

Vavg2 = √([tex]\frac{8RT}{πM2}[/tex]) ................ eqn 1

Vrms1 = √([tex]\frac{3RT}{M1}[/tex]) --------------eqn 2

To find m1/m2, we divide eqn 1 by eqn 2,

we have, Vavg2/Vrms1 = √([tex]\frac{8m1}{3πm2}[/tex])

since Vavg2 = 5Vrms1, it becomes, 5 = √([tex]\frac{8m1}{3πm2}[/tex])

Therfore, m1/m2 = 25 * 3π/8 = 29.45

the mass ratio m1/m2 = 29.45.

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