At a pressure of 24 times 10^5 , text{dyne/cm | vedclass

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

Question

(C) The formula for $r.m.s.$ velocity is $v_{rms} = \sqrt{\frac{3P}{\rho}}$,where $\rho = \frac{m}{V}$.Substituting $\rho$,we get $v_{rms} = \sqrt{\fr

Vedclass · Accepted Answer

$600$

Vedclass · Answer

(C) The formula for $r.m.s.$ velocity is $v_{rms} = \sqrt{\frac{3P}{ho}}$,where $ho = \frac{m}{V}$.Substituting $ho$,we get $v_{rms} = \sqrt{\frac{3PV}{m}}$.Given: $P = 24 	imes 10^5 \, 	ext{dyne/cm}^2 = 24 	imes 10^4 \, 	ext{N/m}^2$,$V = 10 \, 	ext{litre} = 10 	imes 10^{-3} \, 	ext{m}^3$,$m = 20 \, 	ext{gm} = 20 	imes 10^{-3} \, 	ext{kg}$.$v_{rms} = \sqrt{\frac{3 	imes (24 	imes 10^4) 	imes (10 	imes 10^{-3})}{20 	imes 10^{-3}}} = \sqrt{\frac{3 	imes 24 	imes 10^4 	imes 10^{-2}}{20 	imes 10^{-3}}} = \sqrt{\frac{72 	imes 10^2}{20 	imes 10^{-3}}} = \sqrt{3.6 	imes 10^6} = 600 \, 	ext{m/s}$.

At a pressure of $24 \times 10^5 \, \text{dyne/cm}^2$,the volume of $O_2$ is $10 \, \text{litre}$ and its mass is $20 \, \text{gm}$. The $r.m.s.$ velocity will be ....... $m/s$.

Similar Questions

When the r.m.s. velocity of a gas is denoted by '$v$' at temperature '$T$',which of the following relations is true?

At what temperature is the $rms$ speed of hydrogen molecules equal to the $rms$ speed of oxygen molecules at $47^{\circ}C$ (in $K$)?

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:

Derive the equation of $v_{rms}$ in terms of molar mass.

If the pressure of a gas is doubled while keeping the temperature constant,what will happen to the root mean square $(RMS)$ speed of the gas molecules?

Vedclass Test Series

Exam Paper Generator

Online Exam Module