Dust particles are pulverized rock, which has density 2500 kg/m³. They are approximately spheres 20 μm in diameter. Treating dust as an ideal gas, what is the root-mean-square speed (thermal speed) of a dust particle at 400°C? (The Boltzmann constant is 1.38 × 10^-23 J/K.) A) 5.2 × 10^-5 m/s B) 1.7 × 10^-5m/s C) 3.0 × 10^-5m/s D) 7.3 × 10^-5m/s

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Quizplus · Accepted Answer

The root-mean-square speed of a gas particle is given by:$$v_{rms}=\sqrt{\frac{3kT}{m}}$$where k is the Boltzmann constant, T is the temperature in Kelvin, and m is the mass of the particle.The mass of a dust particle can be calculated from its density and volume:$$m= ho V= ho \frac{4}{3}\pi r^3$$where $ ho$ is the density, V is the volume, and r is the radius of the particle.Substituting the given values into the equations above, we get:$$m=2500 ext{ kg/m}^3 imes\frac{4}{3}\pi (10^{-5} ext{ m})^3=4.19 imes10^{-15} ext{ kg}$$$$v_{rms}=\sqrt{\frac{3(1.38 imes10^{-23} ext{ J/K})(400+273.15 ext{ K})}{4.19 imes10^{-15} ext{ kg}}}=5.2 imes10^{-5} ext{ m/s}$$Therefore, the root-mean-square speed of a dust particle at 400°C is 5.2 × 10^-5 m/s.

Dust Particles Are Pulverized Rock, Which Has Density 2500 Kg/m3

Dust Particles Are Pulverized Rock, Which Has Density 2500 Kg/m³