What is the rms speed of a gas molecule of mass 10-22 g at room temperature 27- C according to the kinetic theory of gases-

Kinetic energy of gas per molecule =3RT2N_0 nbsp; [where N_0 is Avogadro's number] Therefore, 12mv^2 nbsp;=3RT2N_0 (where v is rms speed) Substituting the ...

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Standard XIII

Chemistry

Relating Translational KE and T

What is the r...

Question

What is the rms speed of a gas molecule of mass $10^{- 22} g$ at room temperature $27^{\circ} C$ according to the kinetic theory of gases?

42.7 m/s

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203.1 m/s

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0.70 m/s

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352.5 m/s

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Solution

The correct option is D $352.5 m/s$
Kinetic energy of gas per molecule $= \frac{3 R T}{2 N_{0}}$
[where $N_{0}$ is Avogadro's number]
Therefore, $\frac{1}{2} m v^{2} = \frac{3 R T}{2 N_{0}}$
(where $v$ is rms speed)
Substituting the given values, $m = 10^{- 25} k g, T = 300 K, R = 8.314 J / K . m o l$
So, $v_{r m s} = \sqrt{\frac{3 R T}{m N_{0}}} = 352.5 m / s$

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What is the rms speed of a gas molecule of mass 10−22 g at room temperature 27∘ C according to the kinetic theory of gases?

The correct option is D 352.5 m/sKinetic energy of gas per molecule =3RT2N0 [where N0 is Avogadro's number] Therefore, 12mv2=3RT2N0 (where v is rms speed) Substituting the given values, m=10−25 kg,T=300K,R=8.314 J/K.mol So, vrms=√3RTmN0=352.5 m/s

What is the rms speed of a gas molecule of mass $10^{- 22} g$ at room temperature $27^{\circ} C$ according to the kinetic theory of gases?