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?

  • A
    $\frac{x}{4}$
  • B
    $2x$
  • C
    $3x$
  • D
    $4x$

Explore More

Similar Questions

Find the $v_{rms}$ of nitrogen gas at $300 \ K$.

Difficult
View Solution

If the temperature of a gas is increased from $27^{\circ} C$ to $159^{\circ} C$, the increase in the rms speed of the gas molecules is (in $\%$)

The root mean square $(r.m.s.)$ velocity of a gas particle is $v$ at pressure $P$. If the pressure is increased to $2P$ while keeping the temperature constant,the $r.m.s.$ velocity becomes:

At what temperature $(K)$ will the $r.m.s.$ velocity of hydrogen gas be equal to the escape velocity from the Earth?

Difficult
View Solution

Let $\bar{v}$,${v_{rms}}$ and ${v_p}$ respectively denote the mean speed,root mean square speed,and most probable speed of the molecules in an ideal monoatomic gas at absolute temperature $T$. The mass of a molecule is $m$. Then

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo