For 1 mole of gas, the average kinetic energy is given as $E$ . The $u_{r m s}$ of gas is:

{[\frac{2 E}{M}]}^{1 / 2}

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

{[\frac{3 E}{M}]}^{1 / 2}

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

{[\frac{2 E}{3 M}]}^{1 / 2}

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

{[\frac{3 E}{2 M}]}^{1 / 2}

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

Open in App

Solution

The correct option is A ${[\frac{2 E}{M}]}^{1 / 2}$
As we know,
Average kinetic energy, $E = \frac{3}{2} R T$
Also,
$u_{r m s} = \sqrt{\frac{3 R T}{M}}$
By putting value of $E$ , we get:
$u_{r m s} = \sqrt{\frac{2 E}{M}}$

Suggest Corrections

Similar questions

Q. For 1 mole of gas, the average kinetic energy is given as

E

. The

u_{r m s}

of gas is

{[\frac{2 E}{M}]}^{1 / n}

.
value of

n

is.....

The root mean square velocity of one mole of a monoatomic gas having molar mass M is $U_{r m s}$ . The relation between the average kinetic energy (E) of the gas and $U_{r m s}$ is :

For a monoatomic gas, kinetic energy = E. The relation with rms velocity is:

Q. For a monatomic gas, kinetic energy is

E

. The relation of E with rms velocity is

Q. For one mole of gas the average kinetic energy is given as E. The

U_{r m s}

of gas is:

Join BYJU'S Learning Program

For 1 mole of gas, the average kinetic energy is given as E. The urms of gas is:

The correct option is A [2EM]1/2As we know,Average kinetic energy, E=32RTAlso,urms=√3RTMBy putting value of E, we get:urms=√2EM

For 1 mole of gas, the average kinetic energy is given as $E$ . The $u_{r m s}$ of gas is:

The correct option is A ${[\frac{2 E}{M}]}^{1 / 2}$
As we know,
Average kinetic energy, $E = \frac{3}{2} R T$
Also,
$u_{r m s} = \sqrt{\frac{3 R T}{M}}$
By putting value of $E$ , we get:
$u_{r m s} = \sqrt{\frac{2 E}{M}}$