The v_{rms} of gas molecules in a container i | vedclass

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

Question

(C) The root mean square velocity $(v_{rms})$ of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.Here,$R$ is the universal gas

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(C) The root mean square velocity $(v_{rms})$ of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.Here,$R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.Since the temperature $(T)$ remains constant and the nature of the gas $(M)$ does not change,the $v_{rms}$ depends only on the temperature.Even if the amount of gas in the container decreases due to leakage,the temperature remains constant.Therefore,the $v_{rms}$ of the remaining gas molecules remains unchanged at $400 \ ms^{-1}$.

The $v_{rms}$ of gas molecules in a container is $400 \ ms^{-1}$. If half of the gas leaks out of the container at a constant temperature,the $v_{rms}$ of the remaining gas molecules will be . . . . . . $ms^{-1}$.

Similar Questions

If the temperature of an ideal gas is increased from $27^{\circ}C$ to $927^{\circ}C$,what happens to its $r.m.s.$ velocity?

The temperature at which oxygen molecules will have the same r.m.s. speed as helium molecules at $57^{\circ} C$ is (molecular masses of oxygen and helium are $32$ and $4$ respectively.) (in $K$)

The temperature of an ideal gas is increased from $100 \ K$ to $400 \ K$. If '$x$' is the root mean square velocity of its molecules at $100 \ K$,what is the new root mean square velocity?

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

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