Question

at what temperature will be r.m.s velocity of oxygen be one and half time's of its value at NTP.?

Answer 1

solution

$\frac{1}{2} {mv}^{2} = \frac{3}{2}k T$

supposed the temperature required is T' then velocity will be 3 / 2 V.

$\frac{1}{2} {mv'}^{2} = \frac{3}{2}kT'$

3/ 2 V / v = √ T'/ T

or T'

9 / 4 × 273

= 614.25K

be brainly

Answer 2

The correct answer to the question is 660.375 K

CALCULATION:

Let us consider a gas whose molecular mass  is M.

The rms velocity of that gas at any temperature is calculated -

                          rms velocity =

Here, k stands for Boltzmann's constant and T is the temperature of the gas in kelvin.

At NTP, the temperature T = 293.5 K.

Hence, the rms velocity of the oxygen at NTP is

Let the temperature of oxygen gas is T' where the rms velocity is one and half of its value at NTP.

The rms velocity of oxygen at T' =

As per our question,

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