Question

A flask contains argon and chlorine in the ratio 2:1 by mass. the temperature of the mixture is 270C. Obtain the ratio of A) Average K. E. per molecule B) Root mean square speed vmax of the molecules of the two gases. Give :Atomic mass of argon =39.9u'. Molecular mass of chlorine =70.9u.

Answer 1

hope u understand the calculations yourself

Answer 2

(A)Ratio of average KE per molecule is 1:1

(B) Root mean square speed vmax of the molecules of Ar to Cl is 1.33:1

The Kinetic Energy for any gas is given by (3/2)Kb where KbT  is Boltzmann's constant and T is the temperature.

(A)  Given that the temperature of the mixture is constant, so the Average K.E. per molecule is the same for both argon and chlorine.

So, Ratio of average KE per molecule is 1:1

(B) We know, K.E. = (1/2)mv², where v is the rms velocity of the gas.

mv² = 2 KE

⇒ v² = 2KE/m

where m is the molecular mass of the gas.

For Ar, m = 39.9 u

v'² = 2KE/39.9

For Cl, m = 70.9 u

v''² = 2KE/70.9

KE is the same, as shown above.

So,

v'²/v''² =  70.9/39.9 = 1.78/1

Taking square root on both sides,

v'/v'' = 1.33/1

So, the ratio of rms velocity of Argon to Chlorine is 1.33:1