Chemistry Labs

Problem 1

Avogadro's number. Spherical water droplets are dispersed in argon gas. At 27 ∘C27\ ^\circ\mathrm{C}, each droplet is 1.0 μ\mum in diameter and undergoes collisions with argon; assume inter-droplet collisions do not occur. The root-mean-square speed of these droplets was determined to be 0.50 cm s−10.50\ \mathrm{cm\ s^{-1}} at 27 ∘C27\ ^\circ\mathrm{C}. The density of a water droplet is 1.0 g cm−3^{-3}. (1.1) Calculate the average kinetic energy (mv2/2m v^2/2) of this droplet at 27 ∘C27\ ^\circ\mathrm{C}. The volume of a sphere is 43πr3\frac{4}{3}\pi r^3. If the temperature changes, droplet size and speed also change. The average kinetic energy between 0 ∘C0\ ^\circ\mathrm{C} and 100 ∘C100\ ^\circ\mathrm{C} is linear in temperature; assume it remains linear below 0 ∘C0\ ^\circ\mathrm{C} and vanishes at absolute zero (0 K=−273 ∘C0\ \mathrm{K} = -273\ ^\circ\mathrm{C}). At thermal equilibrium, the average kinetic energy is the same irrespective of particle masses (equipartition theorem). The specific heat capacity at constant volume of argon gas (atomic weight 40) is 0.31 J g−1 K−10.31\ \mathrm{J\ g^{-1}\ K^{-1}}. (1.2) Calculate Avogadro's number without using the ideal gas law, the gas constant, or Boltzmann's constant.
Step 1 of 4: Mass of a water droplet
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m=43πr3ρ=43π(0.5×10−6 m)3(1.0×103 kg m−3)=5.2×10−16 kgm = \tfrac{4}{3}\pi r^3 \rho = \tfrac{4}{3}\pi (0.5\times10^{-6}\ \text{m})^3(1.0\times10^3\ \text{kg m}^{-3}) = 5.2\times10^{-16}\ \text{kg}
Analysis

The droplet radius is 0.50 μm=0.50×10−60.50\ \mu\text{m} = 0.50\times10^{-6} m; the volume is 43πr3=5.24×10−19\frac{4}{3}\pi r^3 = 5.24\times10^{-19} m3^3 and the density is 1000 kg m−3^{-3}, giving m=5.2×10−16m = 5.2\times10^{-16} kg.