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

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

Step-by-step explanation:

Given that,

Mass of smoke particle
`m=1.38*10^(-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{(3KT)/(m)}`

Where, K = Boltzmann constant

T = temperature

m = mass

Put the value into the formula

`v=\sqrt{(3*1.38*10^(-23)*(27+273))/(1.38*10^(-17))}`

`v=0.03\ m/s`

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

Answer 2

`v_(rsm)= 3.015*10^(-2)m/s`

Step-by-step 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{(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{(3*1.38*10^(-23)*303.15)/(1.38*10^(-17)) }`

=0.03015 m/s

`v_(rsm)= 3.015*10^(-2)`