If m' represents the mass of each molecule of | vedclass

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

Question

(A) The root mean square speed $(v_{rms})$ of a gas molecule is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$ where $R$ is the universal gas

Vedclass · Accepted Answer

$m^{-\frac{1}{2}} T^{\frac{1}{2}}$

Vedclass · Answer

(A) The root mean square speed $(v_{rms})$ 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. Since $M = m 	imes N_A$ (where $m$ is the mass of one molecule and $N_A$ is Avogadro's number),we can substitute this into the equation: $v_{rms} = \sqrt{\frac{3RT}{mN_A}}$ Since $3$,$R$,and $N_A$ are constants,we have: $v_{rms} \propto \sqrt{\frac{T}{m}}$ $v_{rms} \propto m^{-\frac{1}{2}} T^{\frac{1}{2}}$

If $m'$ represents the mass of each molecule of a gas and $T'$ its absolute temperature,then the root mean square speed of the gas molecule is proportional to

Similar Questions

$A$ gas is at a temperature of $0^{\circ}C$. To what temperature in $^{\circ}C$ must the gas be heated so that the $rms$ speed of its molecules becomes double?

What will be the effect on the root mean square velocity of oxygen molecules if the temperature is doubled and oxygen molecule dissociates into atomic oxygen?

The temperature of an ideal gas is increased from $100 \ K$ to $400 \ K$. If '$x$' is the r.m.s. velocity of its molecules at $100 \ K$,then at $400 \ K$,the r.m.s. velocity becomes:

The $r.m.s.$ velocity will be greater for

The speeds of $5$ molecules of a gas (in arbitrary units) are as follows: $2, 3, 4, 5, 6$. The root mean square speed for these molecules is

Vedclass Test Series

Exam Paper Generator

Online Exam Module