Question

a smoke particle has a mass of 1.38*10^-17 kg and it is randomly moving about in thermal equilibirum with room temperature air at 27

Answer 1

Answer:

$v_{rsm}= 3.015\times10^{-2}m/s$

Explanation:

Let us suppose that we have to find the rsm speed of the smoke particles

According to Kinetic theory of gases RMS speed  

v_rms   is found from the expression

=$\sqrt{\frac{3k_{B}T}{m} }$

where  k_B  is Boltzmann constant =1.38×10^{−23} JK^−1

,T is temperature in  K   and  m= 1.38×10^{-17} kg  is mass of particle.We know that  NTP  is Normal Temperature and Pressure and is defined as air at  27°C(300.15K)and  1 atm

pressure. Inserting given values in above expression we get

=$\sqrt{\frac{3\times1.38\times10^{-23}\times303.15}{1.38\times10^{-17}} }$

=0.03015 m/s

$v_{rsm}= 3.015\times10^{-2}$

Answer 2

Answer:

The rms speed of the particle is 0.03 m/s.

Explanation:

Given that,

Mass of smoke particle $m=1.38\times10^{-17}\ kg$

Temperature =27°C

Suppose we need to find the rms speed of the particle

We need to calculate the rms speed of the particle

Using formula of rms speed

$v=\sqrt{\dfrac{3KT}{m}}$

Where, K = Boltzmann constant

T = temperature

m = mass

Put the value into the formula

$v=\sqrt{\dfrac{3\times1.38\times10^{-23}\times(27+273)}{1.38\times10^{-17}}}$

$v=0.03\ m/s$

Hence, The rms speed of the particle is 0.03 m/s.