Question

If the rms velocity of nitrogen and oxygen molecule are same at two different temperatures and same pressure, then:

• A. the average speed of molecules is also same
• B. the density (gm/L) of nitrogen and oxygen is also equal
• C. the number of moles of each gas is also equal
• D. the most probable velocity of molecules is also equal

Answer 1

Answer: (A), (B), (D)

The correct options are
A the density (gm/L) of nitrogen and oxygen is also equal
B the average speed of molecules is also same
D the most probable velocity of molecules is also equal

(A) $(V_{rms}{)}_{N_2}=(V_{rms}{)}_{O_2}$
$\sqrt{\frac{3R{T}_{N_2}}{M_{N_2}}}=\sqrt{\frac{3R{T}_{O_2}}{M_{O_2}}}$
$\frac{T_{N_2}}{M_{N_2}}=\frac{T_{O_2}}{M_{O_2}}$
Then, $V_{av}$ and $V_{rms}$ is also same.
(B) $d_{N_2}=\frac{P_{N_2}{M}_{N_2}}{R{T}_{N_2}}$ $d_{O_2}=\frac{P_{O_2}{M}_{N_2}}{R{T}_{O_2}}$
If $P_{N_2}=P_{O_2}$, then
$d_{N_2}=d_{O_2}$
(C) weight of the nitrogen and oxygen is not given, hence no. of moles of gas can't be predicted. So, option C is incorrect.
(D) Most probable velocity is given by $V_m=\sqrt{\frac{2RT}{M}}$

since, $\frac{T_{N_2}}{M_{N_2}}=\frac{T_{O_2}}{M_{O_2}}$
$\therefore V_m\left(N_2\right)=V_m\left(O_2\right)$

Option A, B and D are correct.