The kinetic molecular theory attributes an average translational kinetic energy of =(3RT/2N) to each particle. What rms speed would a mist particle of mass 10-12 g have at room temperature (27Â°C) according to kinetic theory of gases?

Question

The kinetic molecular theory attributes an average translational kinetic energy of =(3RT/2N) to each particle. What rms speed would a mist particle of mass 10-12 g have at room temperature (27Â°C) according to kinetic theory of gases? (A) 0.41 cm s-1 (B) 0.35 cm s-1 (C) 0.21 cm s-1 (D) 0.11 cm s-1

Tardigrade Admin · Accepted Answer

K.E. per molecule =(3RT/2N) ⇒ (1/2)mu2=(3RT/2N) or u=√(3RT/m× N) =√(3×8.314×107×300/6.023×1023×10-12)=0.35 cm-1

Q. The kinetic molecular theory attributes an average translational kinetic energy of $= \frac{3 RT}{2 N}$ to each particle. What rms speed would a mist particle of mass $1 0^{- 12} g$ have at room temperature $(2 7^{°} C)$ according to kinetic theory of gases?

$0.41 c m s^{- 1}$

$0.35 c m s^{- 1}$

$0.21 c m s^{- 1}$

$0.11 c m s^{- 1}$

K.E. per molecule $= \frac{3 RT}{2 N}$
$\Rightarrow \frac{1}{2} m u^{2} = \frac{3 RT}{2 N}$ or $u = \frac{3 RT}{m \times N}$
$= \frac{3 \times 8.314 \times 1 0 ^{7} \times 300}{6.023 \times 1 0 ^{23} \times 1 0 ^{- 12}} = 0.35 c m^{- 1}$

Q. The kinetic molecular theory attributes an average translational kinetic energy of =2N3RT​ to each particle. What rms speed would a mist particle of mass 10−12g have at room temperature (27°C) according to kinetic theory of gases?

0.41cms−1

0.35cms−1

0.21cms−1

0.11cms−1