A mixture of two gasses A and B is kept in a vessel which is maintained at a temperature T . In the vessel there are N molecules each of mass m , of the gas A and 2N molecules each of mass 2m of the gas B . The mean square velocity of the molecules of the gas B is v2 and the mean square velocity of the molecules of the gas A along the x direction is w2 . The value of (w2/v2) is

Question

A mixture of two gasses A and B is kept in a vessel which is maintained at a temperature T . In the vessel there are N molecules each of mass m , of the gas A and 2N molecules each of mass 2m of the gas B . The mean square velocity of the molecules of the gas B is v2 and the mean square velocity of the molecules of the gas A along the x direction is w2 . The value of (w2/v2) is (A)  text2 (B)  text1 (C) (1/3) (D) (2/3)

Tardigrade Admin · Accepted Answer

Mean kinetic energy of the two types of molecules should be equal. The mean square velocity of A type molecules

= ω^{2} + ω^{2} + ω^{2} = 3 ω^{2}

Therefore,

\frac{1}{2} m (3 (ω)^{2}) = \frac{1}{2} (2 m) v^{2}

This gives

\frac{ω ^{2}}{v ^{2}} = \frac{2}{3}

$2$

$1$

$\frac{1}{3}$

$\frac{2}{3}$

Mean kinetic energy of the two types of molecules should be equal. The mean square velocity of A type molecules $= ω^{2} + ω^{2} + ω^{2} = 3 ω^{2}$
Therefore, $\frac{1}{2} m (3 (ω)^{2}) = \frac{1}{2} (2 m) v^{2}$
This gives $\frac{ω ^{2}}{v ^{2}} = \frac{2}{3}$

2

1

31​

32​

Mean kinetic energy of the two types of molecules should be equal. The mean square velocity of A type molecules =ω2+ω2+ω2=3ω2 Therefore, 21​m(3(ω)2)=21​(2m)v2 This gives v2ω2​=32​

$2$

$1$

$\frac{1}{3}$

$\frac{2}{3}$

Mean kinetic energy of the two types of molecules should be equal. The mean square velocity of A type molecules $= ω^{2} + ω^{2} + ω^{2} = 3 ω^{2}$
Therefore, $\frac{1}{2} m (3 (ω)^{2}) = \frac{1}{2} (2 m) v^{2}$
This gives $\frac{ω ^{2}}{v ^{2}} = \frac{2}{3}$