The rms speed of a molecule of a diatomic gas | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The $rms$ speed of a gas molecule is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the abs

Vedclass · Accepted Answer

$2v$

Vedclass · Answer

(C) The $rms$ speed of a gas molecule is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass.Initially,for the diatomic molecule: $v = \sqrt{\frac{3RT}{M}}$.When the temperature is doubled,the new temperature $T' = 2T$.When the diatomic molecule dissociates into two atoms,the molar mass of the atom becomes $M' = \frac{M}{2}$.The new $rms$ speed $v'$ is given by: $v' = \sqrt{\frac{3RT'}{M'}} = \sqrt{\frac{3R(2T)}{M/2}} = \sqrt{\frac{12RT}{M}} = 2 \sqrt{\frac{3RT}{M}}$.Substituting the initial speed $v$: $v' = 2v$.

The $rms$ speed of a molecule of a diatomic gas is $v$. When the temperature is doubled,the molecule dissociates into two atoms. The new $rms$ speed of the atom is:

Similar Questions

The $RMS$ velocity of oxygen molecules at $NTP$ is $0.5 \,km/s$. The $RMS$ velocity for the hydrogen molecule at $NTP$ is (in $\,km/s$)

The relation between root mean square speed $(v_{rms})$ and most probable speed $(v_{p})$ for the molar mass $M$ of oxygen gas molecule at the temperature of $300\, K$ is:

The temperature of an ideal gas is reduced from $927^\circ C$ to $27^\circ C$. The $r.m.s.$ velocity of the molecules becomes

The molecules of a given mass of a gas have root mean square speeds of $100 \ m/s$ at $27^{\circ} \ C$ and $1.00 \ \text{atm}$ pressure. What will be the root mean square speeds of the molecules of the gas at $127^{\circ} \ C$ and $2.0 \ \text{atm}$ pressure?

Let the r.m.s. velocity of a molecule of a given mass of gas be $C_{1}$ at temperature $27^{\circ} C$. When the temperature is increased to $327^{\circ} C$,the r.m.s. velocity is $C_{2}$. Then the ratio $\frac{C_{2}}{C_{1}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module