Question

N molecules, each of mass m, of gas a and 2 n molecules, each of mass 2 m, of gas b are contained in the same vessel which is maintained at a temperature t. The mean square velocity of molecules of b type is denoted by $v_2$ and the mean square velocity of a type is denoted by $v_1$, then $\frac{v_1}{v_2}$ is

Answer 1

Answer:

The ratio of $v_1 to v_2$ is √2 :1 .

Explanation:

The formula used for root mean square speed is:

$\nu_{rms}=\sqrt{\frac{3kN_AT}{M}}$

where,

$\nu_{rms}$ = root mean square speed

k = Boltzmann’s constant = $1.38\times 10^{-23}J/K$

T = temperature

M = atomic mass of the gas

$N_A$ = Avogadro’s number = $6.02\times 10^{23}mol^{-1}$

Now put all the given values in the above root mean square speed formula, we get:

For gas-a :

$v_1=\sqrt{\frac{3kN_AT}{m}}$...[1]

For gas-b :

$v_2=\sqrt{\frac{3kN_AT}{2m}}$...[2]

[1] ÷ [2]

$\frac{v_1}{v_2}=\frac{\sqrt{\frac{3kN_AT}{m}}}{\sqrt{\frac{3kN_AT}{2m}}}$

$\frac{v_1}{v_2}=\frac{\sqrt{2}}{1}$

The ratio of $v_1 to v_2$ is √2 :1 .