IChO 2006 · Theory problem 1

Avogadro's number

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Spherical water droplets are dispersed in argon gas. At 27 °C, each droplet is 1.0 micrometer in diameter and undergoes collisions with argon. Assume that inter-droplet collisions do not occur. The root-mean-square speed of these droplets was determined to be \(0.50\ \mathrm{cm\,s^{-1}}\) at 27 °C. The density of a water droplet is \(1.0\ \mathrm{g\,cm^{-3}}\).

1.1 Calculate the average kinetic energy (\(mv^2/2\)) of this droplet at 27 °C. The volume of a sphere is given by \((4/3)\,\pi\,r^3\) where r is the radius.

If the temperature is changed, then droplet size and speed of the droplet will also change. The average kinetic energy of a droplet between 0 °C and 100 °C as a function of temperature is found to be linear. Assume that it remains linear below 0 °C.

Figure 1

At thermal equlibrium, the average kinetic energy is the same irrespective of particle masses (equipartition theorem).

The specific heat capacity, at constant volume, of argon (atomic weight, 40) gas is \(0.31\ \mathrm{J\,g^{-1}\,K^{-1}}\).

1.2. Calculate Avogadro's number without using the ideal gas law, the gas constant, Boltzmann’s constant).

Official English text of the problem.

Answers

  • 1.1: \(\boxed{\overline{E}_{\mathrm{k}}=6.5\times10^{-21}\ \mathrm{J}}\) per droplet.
  • 1.2: \(\boxed{N_{\mathrm A}=5.7\times10^{23}\ \mathrm{mol^{-1}}}\).

Solution

1.1

The droplet radius and density in SI units are

\[ r=\frac{1.0\times10^{-6}}{2} =5.0\times10^{-7}\ \mathrm m, \qquad \rho=1.0\times10^3\ \mathrm{kg\,m^{-3}}. \]

Its mass is therefore

\[ m=\rho\frac{4\pi r^3}{3} =(1.0\times10^3)\frac{4\pi}{3}(5.0\times10^{-7})^3 =5.236\times10^{-16}\ \mathrm{kg}. \]

The root-mean-square speed is

\[ v_{\mathrm{rms}} =0.50\ \mathrm{cm\,s^{-1}} =5.0\times10^{-3}\ \mathrm{m\,s^{-1}}. \]

Since \(v_{\mathrm{rms}}^2=\langle v^2\rangle\), the average translational kinetic energy is

\[ \begin{aligned} \overline{E}_{\mathrm{k}} &=\frac12m\langle v^2\rangle =\frac12m v_{\mathrm{rms}}^2\\ &=\frac12(5.236\times10^{-16})(5.0\times10^{-3})^2\\ &=6.545\times10^{-21}\ \mathrm J. \end{aligned} \]

To two significant figures,

\[ \boxed{\overline{E}_{\mathrm{k}}=6.5\times10^{-21}\ \mathrm J}. \]

1.2

In the classical treatment used in this problem, the mean translational kinetic energy extrapolates to zero at absolute zero. Consequently, the stated linear temperature dependence has the form

\[ \overline{E}_{\mathrm{k}}=aT, \]

where \(T\) is the absolute temperature. At 27 °C,

\[ T_0=300.15\ \mathrm K, \]

so the energy increase per particle per kelvin is

\[ a=\frac{d\overline{E}_{\mathrm{k}}}{dT} =\frac{\overline{E}_{\mathrm{k}}(T_0)}{T_0} =\frac{6.545\times10^{-21}}{300.15} =2.181\times10^{-23}\ \mathrm{J\,K^{-1}}. \]

By the stated equipartition principle, an argon atom has the same average translational kinetic energy as a droplet at the same temperature. Thus the same value of \(a\) applies to each argon atom.

Argon is monatomic, so its thermal energy in this classical treatment is translational kinetic energy. At constant volume, the molar heat capacity is therefore

\[ C_{V,m} =\frac{dU_m}{dT} =N_{\mathrm A}\frac{d\overline{E}_{\mathrm{k}}}{dT} =N_{\mathrm A}a. \]

The supplied specific heat capacity gives

\[ C_{V,m} =(0.31\ \mathrm{J\,g^{-1}\,K^{-1}}) (40\ \mathrm{g\,mol^{-1}}) =12.4\ \mathrm{J\,mol^{-1}\,K^{-1}}. \]

Hence

\[ \begin{aligned} N_{\mathrm A} &=\frac{C_{V,m}}{a} =\frac{C_{V,m}T_0}{\overline{E}_{\mathrm{k}}(T_0)}\\ &=\frac{(12.4\ \mathrm{J\,mol^{-1}\,K^{-1}}) (300.15\ \mathrm K)} {6.545\times10^{-21}\ \mathrm J}\\ &=5.687\times10^{23}\ \mathrm{mol^{-1}}. \end{aligned} \]

To two significant figures,

\[ \boxed{N_{\mathrm A}=5.7\times10^{23}\ \mathrm{mol^{-1}}}. \]

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.