The root mean square speed of the molecules of a diatomic gas is v . When the temperature is doubled, the molecules dissociate into two atoms. The new root mean square speed of the atom is

Question

The root mean square speed of the molecules of a diatomic gas is v . When the temperature is doubled, the molecules dissociate into two atoms. The new root mean square speed of the atom is (A) √2 v (B) v (C) 2 v (D) 4 v

Tardigrade Admin · Accepted Answer

v rms =√(3 R T/M) . According to problem T will become T / 2 and M will become M / 2 so the value of v rms will increase by √4=2 times i.e., new root mean square velocity will be 2 v.

Q. The root mean square speed of the molecules of a diatomic gas is $v .$ When the temperature is doubled, the molecules dissociate into two atoms. The new root mean square speed of the atom is

$2 v$

v

2 v

4 v

$v_{r m s} = \frac{3 RT}{M} .$
According to problem $T$ will become
$T /2$ and $M$ will become $M /2$ so the value of $v_{r m s}$ will increase by $4 = 2$ times i.e., new root mean square velocity will be $2 v$ .

Q. The root mean square speed of the molecules of a diatomic gas is v. When the temperature is doubled, the molecules dissociate into two atoms. The new root mean square speed of the atom is

2​v

v

2 v

4 v

vrms​=M3RT​​. According to problem T will become T/2 and M will become M/2 so the value of vrms​ will increase by 4​=2 times i.e., new root mean square velocity will be 2v.

Q. The root mean square speed of the molecules of a diatomic gas is $v .$ When the temperature is doubled, the molecules dissociate into two atoms. The new root mean square speed of the atom is

$2 v$

$v_{r m s} = \frac{3 RT}{M} .$
According to problem $T$ will become
$T /2$ and $M$ will become $M /2$ so the value of $v_{r m s}$ will increase by $4 = 2$ times i.e., new root mean square velocity will be $2 v$ .