Question

At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C ? (atomic mass of Ar = 39.9 u, of He = 4.0 u).

Answer 1

Given that the temperature of Helium gas is −20 °C, atomic mass of argon is 39.9 u and of helium is 4.0 u.

Let v He be the root mean square velocities of helium, then

v Ar = 3R T Ar M Ar (1)

Here, R is the universal gas constant, T Ar is the temperature of argon gas and M Ar is the atomic mass of argon.

Let v He be the root mean square velocities of Helium, then

v He = 3R T He M He (2)

Here, T He is the temperature of helium gas and M He is the atomic mass of helium.

The root mean square velocity of both the gases is equal, hence

v Ar = v He 3R T Ar M Ar = 3R T He M He T Ar M Ar = T He M He T Ar M Ar = T He M He T Ar = T He M He × M Ar

Substitute the values in the above equation.

T Ar = ( 273+( −20 ) ) 4 ×39.9 = 253 4 ×39.9 =2523.67 K

Hence, at 2523.67 K temperature, the root mean square speed of an atom of argon gas cylinder is equal to that of helium gas atom at −20 °C.