The Earth’s atmosphere consists primarily of oxygen (21 %) and nitrogen (78 %) . The rms speed of oxygen molecules (O2) in the atmosphere at a certain location is 535 m/s . The rms speed of the nitrogen molecules at this location will be [Given 1 amu=1.66× 10- 27 kg , molecular mass of O2=32 , molecular mass of N2=28 ]

Question

The Earth’s atmosphere consists primarily of oxygen (21 %) and nitrogen (78 %) . The rms speed of oxygen molecules (O2) in the atmosphere at a certain location is 535 m/s . The rms speed of the nitrogen molecules at this location will be [Given 1 amu=1.66× 10- 27 kg , molecular mass of O2=32 , molecular mass of N2=28 ] (A) 572 m/s (B) 437 m/s (C) 835 m/s (D) 715 m/s

Tardigrade Admin · Accepted Answer

The rms speed of N2,vr m s , N2=√(3 R T/MN2) For O2,vrms , O2=√(3 R T/MO2) Thus, (vr m s , N2/vr m s , O2)=√(MO2/MN2)=√(32/28) vr m s , N2=572m/s

$572 m / s$

$437 m / s$

$835 m / s$

$715 m / s$

The rms speed of $N_{2}, v_{r m s,_{N_{2}}} = \frac{3 RT}{M _{N_{2}}}$
For $O_{2}, v_{r m s,_{O_{2}}} = \frac{3 RT}{M _{O_{2}}}$
Thus,
$\frac{v _{r m s, N_{2}}}{v _{r m s, O_{2}}} = \frac{M _{O_{2}}}{M _{N_{2}}} = \frac{32}{28}$
$v_{r m s, N_{2}} = 572 m / s$

572m/s

437m/s

835m/s

715m/s

The rms speed of N2​,vrms,N2​​​=MN2​​3RT​​ For O2​,vrms,O2​​​=MO2​​3RT​​ Thus, vrms,O2​​vrms,N2​​​=MN2​​MO2​​​​=2832​​ vrms,N2​​=572m/s

$572 m / s$

$437 m / s$

$835 m / s$

$715 m / s$

The rms speed of $N_{2}, v_{r m s,_{N_{2}}} = \frac{3 RT}{M _{N_{2}}}$
For $O_{2}, v_{r m s,_{O_{2}}} = \frac{3 RT}{M _{O_{2}}}$
Thus,
$\frac{v _{r m s, N_{2}}}{v _{r m s, O_{2}}} = \frac{M _{O_{2}}}{M _{N_{2}}} = \frac{32}{28}$
$v_{r m s, N_{2}} = 572 m / s$