The rms velocity of air at NTP will be .................. if density of air is 1.29 kg/m $^{3}$ .

0.485 m s^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.485 \times 10^{2} m s^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4.85 \times 10^{2} m s^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

10^{4} m s^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $4.85 \times 10^{2} m s^{- 1}$
The rms velocity of air at NTP(298K and 1atm)
$u_{r m s} = \sqrt{(\frac{3 R T}{M}}) = \sqrt{\frac{3 P}{ρ}} = \sqrt{\frac{3 \times (1.013 \times 10^{5})}{1.29}} = 4.85 \times 10^{2} m / s$

Suggest Corrections

Similar questions

The rms velocity of air at NTP will be if density of air is 1.29 Kg/m $^{3}$

Q. Calculate the rms speed of

O_{2}

273 K

and

1 \times 10^{4} Pa

pressure. The density of

O_{2}

under these conditions is

1.50 {kg m}^{- 3}

The kinetic energy of an electron is $4.55 \times 10^{- 25} J$ . The mass ofelectron is $9.1 \times 10^{- 31} k g$ . Calculate velocity of the electron.

Q. A ball is thrown upwards with a velocity of

2 m s^{- 1}

from a height of

5 m

from the ground. What will be the speed of the ball when it touches the ground? Take the acceleration due to gravity to be

10 m s^{- 2}

Q. A proton enters a magnetic field of flux density 1.5 T velocity of

20 \times 10^{2} m s^{- 1}

at an angle of

30^{\circ}

with the field. The force on the position is

[e_{p} = 1.6 \times 10^{- 19} C]

Join BYJU'S Learning Program

The rms velocity of air at NTP will be .................. if density of air is 1.29 kg/m3.

The correct option is A 4.85×102ms−1The rms velocity of air at NTP(298K and 1atm)urms=√(3RTM)=√3Pρ=√3×(1.013×105)1.29=4.85×102m/s

The rms velocity of air at NTP will be .................. if density of air is 1.29 kg/m $^{3}$ .

The correct option is A $4.85 \times 10^{2} m s^{- 1}$
The rms velocity of air at NTP(298K and 1atm)
$u_{r m s} = \sqrt{(\frac{3 R T}{M}}) = \sqrt{\frac{3 P}{ρ}} = \sqrt{\frac{3 \times (1.013 \times 10^{5})}{1.29}} = 4.85 \times 10^{2} m / s$