For a given gas at temperature T,the root mea | vedclass

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

Question

(B) The expressions for the different velocities of gas molecules are given by:$V_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{3} \cdot \sqrt{\frac{RT}{M}} \a

Vedclass · Accepted Answer

$
u_{rms} > 
u_{av} > 
u_{mp}$

Vedclass · Answer

(B) The expressions for the different velocities of gas molecules are given by:$V_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{3} \cdot \sqrt{\frac{RT}{M}} \approx 1.732 \sqrt{\frac{RT}{M}}$$V_{av} = \sqrt{\frac{8RT}{\pi M}} = \sqrt{\frac{8}{\pi}} \cdot \sqrt{\frac{RT}{M}} \approx 1.596 \sqrt{\frac{RT}{M}}$$V_{mp} = \sqrt{\frac{2RT}{M}} = \sqrt{2} \cdot \sqrt{\frac{RT}{M}} \approx 1.414 \sqrt{\frac{RT}{M}}$Comparing the coefficients,we find that $1.732 > 1.596 > 1.414$.Therefore,the correct relation is $
u_{rms} > 
u_{av} > 
u_{mp}$.

For a given gas at temperature $T$,the root mean square velocity is $\nu_{rms}$,the average velocity is $\nu_{av}$,and the most probable velocity is $\nu_{mp}$. Which of the following relations is correct?

Similar Questions

If the mass of a $He$ atom is $4$ times that of a hydrogen atom,then the mean velocity of $He$ is:

$N$ molecules of gas $A$,each having mass $m$,and $2N$ molecules of gas $B$,each of mass $2m$,are contained in the same vessel at a constant temperature $T$. The mean square velocity of $B$ is $V^2$ and the mean square of the $x$-component of $A$ is $\omega^2$. The value of $\frac{\omega^2}{V^2}$ is

$V_{rms}$ denotes the root mean square speed,$\bar{V}$ the average speed,and $V_m$ the most probable speed of gas molecules. Which of the following is the correct relationship?

The absolute temperature of a gas is determined by

At what temperature is the root mean square velocity of gaseous hydrogen molecules equal to that of oxygen molecules at $47^{\circ}C$ (in $; K$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module