Question

Initially the root mean square rms velocity of n2 molecules at certain temperature is you if the pressure is temperature is doubled and all the nitrogen molecules dissociate into nitrogen atoms the new rms velocity will be

Answer 1

Answer:

Initial root mean square velocity of N2, Vrms = u

Let the initial temperature be “T1”.

Let the molecular mass of N2 be “M1” = 28 g/mol and molecular mass of N atoms be “M2” = 14 g/mol.

After temperature is doubled, it the new temperature “T2” becomes equal to “2 * T1” and N2 molecules dissociate to become N atoms.

We know,

Vrms = √[3RT1/M1] = √[3RT1/28] = u ….. (i)

Where R = ideal gas constant, T= absolute temperature  

Now,  

New Vrms  = √[3RT2/M2]  

⇒ New Vrms = √[3R(2* T1)/14]

On multiplying and dividing by √2 on the R.H.S, we get

⇒ New Vrms = √[3 * 2 * R * (2* T1) / (2 * 14)]  

⇒New Vrms =2 *  √[3RT1/28]  = 2*u …… [from eq. (i) √[3RT1/28] = u]

Thus, the new rms velocity will be 2u.

Answer 2

above answer is right

thx