A smoke particle has a mass of 1.89 10-17 kg and it is randomly moving about in thermal equilibrium with room temperature air at 25° C. What is the rms speed of the particle?
cm/s
Vrms = sqrt [ 3 k T /m ]
=sqrt (3*1.38*10^-23 *298 / 1.89*10^-17)
=sqrt(653 *10^-6)
25.5*10^-3
= .0255 m.s
To calculate the rms (root mean square) speed of a particle, we can use the equation:
v_rms = √(3kT / m)
Where:
- v_rms is the rms speed of the particle,
- k is the Boltzmann constant (1.38 × 10^-23 J/K),
- T is the temperature in Kelvin,
- m is the mass of the particle.
First, let's convert the temperature from Celsius to Kelvin:
T (in Kelvin) = T (in Celsius) + 273.15
T = 25 + 273.15 = 298.15 K
Now, we can substitute the values into the equation and calculate the rms speed:
v_rms = √(3 * (1.38 × 10^-23 J/K) * (298.15 K) / (1.89 × 10^-17 kg))
v_rms = √((4.14 × 10^-23 J) / (1.89 × 10^-17 kg))
To simplify the units, let's convert the joules to cm^2/s^2:
1 J = 1 kg * m^2 / s^2
1 J = (10^3 g) * (10^4 cm^2) / s^2
1 J = 10^7 g * cm^2 / s^2
v_rms = √((4.14 × 10^-23 J) / (1.89 × 10^-17 kg)) * sqrt((10^7 g * cm^2) / (10^7 g * cm^2))
v_rms = √( (4.14 × 10^-23 J) * (10^7 g * cm^2) / (1.89 × 10^-17 kg * 10^7 g * cm^2))
Now, let's simplify the units further:
1 g = 10^-3 kg
1 cm^2 = 10^-4 m^2
v_rms = √( (4.14 × 10^-23 J) * (10^7 * 10^-3 kg * 10^-4 m^2) / (1.89 × 10^-17 kg * 10^7 * 10^-3 kg * 10^-4 m^2))
v_rms = √( (4.14 × 10^-23 J) / (1.89 × 10^-17 kg))
v_rms = √(2.19 × 10^6 cm^2/s^2)
Finally, let's convert the units back to cm/s:
1 cm^2/s^2 = 1 cm/s * cm/s
v_rms = √(2.19 × 10^6) cm/s
v_rms ≈ 1480.4 cm/s
So, the rms speed of the smoke particle in thermal equilibrium with room temperature air is approximately 1480.4 cm/s.