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 2 of 4: Average kinetic energy at 27 °C
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Ek=12mv2=12(5.2×10−16 kg)(0.50×10−2 m s−1)2=6.9×10−21 JE_k = \tfrac{1}{2} m v^2 = \tfrac{1}{2}(5.2\times10^{-16}\ \text{kg})(0.50\times10^{-2}\ \text{m s}^{-1})^2 = 6.9\times10^{-21}\ \text{J}
Analysis

Using the root-mean-square speed v=0.50 cm s−1=5.0×10−3v = 0.50\ \text{cm s}^{-1} = 5.0\times10^{-3} m s−1^{-1}, Ek=12mv2=6.9×10−21E_k = \frac{1}{2}mv^2 = 6.9\times10^{-21} J.