IPhO 1979 · Theory problem 2

Weighing a sample in dry and in humid air

Download PDFIPhO 1979

Brass weights are used to weigh an aluminum-made sample on an analytical balance. The weighing is ones in dry air and another time in humid air with the water vapor pressure \(P_h=2\cdot10^3\ \mathrm{Pa}\). The total atmospheric pressure \((P=10^5\ \mathrm{Pa})\) and the temperature \((t=20^\circ\ \mathrm{C})\) are the same in both cases.

What should the mass of the sample be to be able to tell the difference in the balance readings provided their sensitivity is \(m_0=0.1\ \mathrm{mg}\) ?

Aluminum density \(\rho_1=2700\ \mathrm{kg/m^3}\), brass density \(\rho_2=.8500\ \mathrm{kg/m^3}\).

Official English text of the problem.

Answers

  • Problem 2. The sample must have a mass of at least approximately \[ \boxed{m_{\min}\simeq4.4\times10^{-2}\ \mathrm{kg}=44\ \mathrm{g}.} \] The balance reading is higher in humid air; at the threshold, the increase is \(0.1\ \mathrm{mg}\).

Solution

Problem 2.

Treat dry air and water vapor as ideal gases, and assume that the masses and volumes of the sample and weights remain unchanged. We use the brass density \(8500\ \mathrm{kg/m^3}\), interpreting the preceding dot in the statement as a printing error.

First find the change in air density. For an ideal gas of molar mass \(M\), \[ \rho=\frac{pM}{RT}. \] In humid air, Dalton’s law gives the dry-air partial pressure as \(P-P_h\). Thus, writing \(M_a\) and \(M_v\) for the molar masses of dry air and water vapor, \[ \rho_d=\frac{PM_a}{RT}, \qquad \rho_h=\frac{(P-P_h)M_a+P_hM_v}{RT}. \] Consequently, \[ \Delta\rho\equiv\rho_d-\rho_h =\frac{P_h(M_a-M_v)}{RT}. \] Water vapor has a smaller molar mass than dry air, so humid air is less dense at the same total pressure and temperature.

Using the standard approximate constants \[ M_a=29\times10^{-3}\ \mathrm{kg/mol},\qquad M_v=18\times10^{-3}\ \mathrm{kg/mol},\qquad R=8.314\ \mathrm{J/(mol\,K)}, \] and \(T=293.15\ \mathrm{K}\), we obtain \[ \Delta\rho =\frac{(2\times10^3)(29-18)\times10^{-3}} {8.314\times293.15} \simeq9.03\times10^{-3}\ \mathrm{kg/m^3}. \]

Next account for buoyancy on both the sample and the brass weights. Let \(m\) be the true sample mass and \(m_b\) the total brass mass required for balance in air of density \(\rho\). Their volumes are \(m/\rho_1\) and \(m_b/\rho_2\). Archimedes’ principle and equality of the apparent weights give \[ mg-\rho g\frac{m}{\rho_1} =m_bg-\rho g\frac{m_b}{\rho_2}. \] Therefore, the balance reading is \[ m_b(\rho)=m\,\frac{1-\rho/\rho_1}{1-\rho/\rho_2}. \]

The difference between the humid-air and dry-air readings is \[ \Delta m_b = \frac{m(\rho_d-\rho_h) \left(\dfrac1{\rho_1}-\dfrac1{\rho_2}\right)} {\left(1-\dfrac{\rho_d}{\rho_2}\right) \left(1-\dfrac{\rho_h}{\rho_2}\right)}. \] Here \(\rho_d\simeq1.190\ \mathrm{kg/m^3}\) and \(\rho_h\simeq1.181\ \mathrm{kg/m^3}\), so the denominator differs from unity by only about \(0.028\%\). To the required accuracy, \[ \Delta m_b \simeq m\,\Delta\rho \left(\frac1{\rho_1}-\frac1{\rho_2}\right) \simeq m(2.28\times10^{-6}). \] This difference is positive: the aluminum sample has a greater volume per unit mass, so reducing air density increases its apparent weight more than that of the brass weights.

For a detectable difference, require \[ \Delta m_b\ge m_0=0.1\ \mathrm{mg}=1.0\times10^{-7}\ \mathrm{kg}. \] Hence \[ m_{\min}\simeq \frac{1.0\times10^{-7}} {(9.03\times10^{-3}) \left(\dfrac1{2700}-\dfrac1{8500}\right)} \simeq4.38\times10^{-2}\ \mathrm{kg} \simeq\boxed{44\ \mathrm{g}}. \]

The solution was checked against the official solution.