There are N ( < 100 ) molecules of a gas having velocities 1 , 2 , 3 , … , N km / s respectively. Then

v r m s = N 1 2 + 2 2 + ⋯ + N 2 = 6 ( N + 1 ) ( 2 N + 1 ) and v a vg = N 1 + 2 + ⋯ + N = 2 N + 1 . So v a vg v r m s = 6 2 N + 1 2 N + 1 .

Start Free Mock LoginAlready registered? Login here

Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoKTGMedium2 minQB

PhysicsMediumsingle choice

There are $N (< 100)$ molecules of a gas having velocities $1, 2, 3, \dots, N km / s$ respectively. Then

Options:

Answer:

Solution:

$v_{r m s} = \frac{1 ^{2} + 2 ^{2} + \dots + N ^{2}}{N} = \frac{( N + 1 ) ( 2 N + 1 )}{6}$ and $v_{a vg} = \frac{1 + 2 + \dots + N}{N} = \frac{N + 1}{2}$ . So $\frac{v _{r m s}}{v _{a vg}} = \frac{2}{6} \frac{2 N + 1}{N + 1}$ .

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:KTG

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...