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 4 of 4: Avogadro's number
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NA=MAr×cva=40 g mol−1×0.31 J g−1K−12.3×10−23 J K−1=5.6×1023 mol−1N_A = M_{\ce{Ar}}\times\frac{c_v}{a} = 40\ \text{g mol}^{-1}\times\frac{0.31\ \text{J g}^{-1}\text{K}^{-1}}{2.3\times10^{-23}\ \text{J K}^{-1}} = 5.6\times10^{23}\ \text{mol}^{-1}
Analysis

The bulk heat capacity of 1 g of argon is cv=0.31c_v = 0.31 J g−1^{-1} K−1=a N^{-1} = a\,N, where NN is the number of atoms in 1 g: N=0.31/(2.3×10−23)=1.4×1022N = 0.31/(2.3\times10^{-23}) = 1.4\times10^{22} g−1^{-1}. Multiplying by the molar mass 40 g mol−1^{-1} gives NA=5.6×1023N_A = 5.6\times10^{23} mol−1^{-1}, obtained purely from Brownian motion and calorimetry.