IChO 2010 · Theory problem 1

Lord Rayleigh and the discovery of argon

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In 1894, Lord Rayleigh reported that the mass of chemically prepared nitrogen was different from that of nitrogen extracted from the atmosphere, as shown in Tables 1 and 2. Later, this difference was attributed to the presence of argon in atmospheric nitrogen. The masses of gases were measured by using a glass vessel with a known volume under atmospheric pressure (\(1.013\cdot10^5\ \mathrm{Pa}\)).

Table 1. Mass of Chemical Nitrogen in the Vessel

From nitric oxide 2.3001 g
From nitrous oxide 2.2990 g
From ammonium nitrite purified at a red heat 2.2987 g
From urea 2.2985 g
From ammonium nitrite purified in the cold 2.2987 g
Mean 2.2990 g

Table 2. Mass of Atmospheric Nitrogen in the Vessel

\(\ce{O2}\) was removed by hot copper (1892) 2.3103 g
\(\ce{O2}\) was removed by hot iron (1893) 2.3100 g
\(\ce{O2}\) was removed by ferrous hydrate (1894) 2.3102 g
Mean 2.3102 g

1.1 Calculate the volume \(V\) [\(\mathrm{m^3}\)] of the vessel used by Rayleigh from the mean mass of chemical nitrogen, which must have been pure nitrogen. Assume that the measurements were carried out at a temperature of \(15.0\,^\circ\mathrm{C}\).

1.2 Estimate the mole fraction \(x\) of argon in Rayleigh's atmospheric nitrogen, by assuming that argon and nitrogen were the only constituents. Use the mean masses of the atmospheric and chemical nitrogen for the calculation.

Ramsay and Clève discovered helium in cleveite (a mineral consisting of uranium oxide and oxides of lead, thorium, and rare earths; an impure variety of uraninite) independently and virtually simultaneously in 1895. The gas extracted from the rock showed a unique spectroscopic line at around 588 nm (indicated by \(\mathrm{D}_3\) in Figure 1), which was first observed in the spectrum of solar prominence during a total eclipse in 1868, near the well-known \(\mathrm{D}_1\) and \(\mathrm{D}_2\) lines of sodium.

Figure 1

Figure 1. Spectral lines around 588 nm

1.3 Calculate the energy \(E\) [J] of a photon with the wavelength of the \(\mathrm{D}_3\) line of helium shown in Figure 1.

Figure 2

Figure 2. Energy diagram of atomic orbitals of helium when an electron resides in the 1s orbital.

Figure 2 shows an energy diagram of the atomic orbitals of helium. The arrows indicate the "allowed" transitions according to the spectroscopic principle.

1.4 Identify the transition relevant to the \(\mathrm{D}_3\) line of helium among the transitions [A] to [E] indicated in Figure 2. Mark one of the following:

\(\square\) [A] \(\square\) [B] \(\square\) [C] \(\square\) [D] \(\square\) [E]

1.5 Which equation explains the occurrence of helium in cleveite among [A] to [D] below? Mark one.

Figure 3

Argon is also found in minerals such as malacon.

1.6 Which equation explains the occurrence of argon in rocks among [A] to [D] below? Mark one.

Figure 4

One of the strongest evidences for the monoatomicity of argon and helium is the ratio of the heat capacity under constant pressure to that at constant volume, \(\gamma=C_p/C_V\), which is exactly \(5/3\) (\(1.67\pm0.01\)) for a monoatomic gas. The ratio was derived from the measurement of speed of sound \(v_s\) by using the following equation, where \(f\) and \(\lambda\) are the frequency and wavelength of the sound, and \(R\), \(T\), and \(M\) denote the molar gas constant, absolute temperature, and molar mass, respectively.

\[ v_s=f\lambda=\sqrt{\frac{\gamma RT}{M}} \]

For an unknown gas sample, the wavelength of the sound was measured to be \(\lambda=0.116\ \mathrm{m}\) at a frequency of \(f=3520\ \mathrm{Hz}\) (\(\mathrm{Hz}=\mathrm{s^{-1}}\)) and temperature of \(15.0\,^\circ\mathrm{C}\) and under atmospheric pressure (\(1.013\cdot10^5\ \mathrm{Pa}\)). The density \(\rho\) of the gas for these conditions was measured to be \(0.850\pm0.005\ \mathrm{kg\,m^{-3}}\).

1.7 Calculate the molar mass \(M\) [\(\mathrm{kg\,mol^{-1}}\)] of this gas.

1.8 Calculate the heat capacity ratio \(\gamma\) for this gas sample.

1.9 Which is this gas? Choose and mark among [A] to [D] .

\(\square\) [A] \(\ce{HCl}\)

\(\square\) [B] \(\ce{HF}\)

\(\square\) [C] \(\ce{Ne}\)

\(\square\) [D] \(\ce{Ar}\)

Official English text of the problem.

Answers

  • 1.1: \(V=1.94\times10^{-3}\ \mathrm{m^3}\).
  • 1.2: \(x=0.0114\), corresponding to \(1.14\ \mathrm{mol}\%\) argon.
  • 1.3: \(E=3.38\times10^{-19}\ \mathrm{J}\) per photon.
  • 1.4: [E], the \(2p\leftrightarrow3d\) transition; emission is \(3d\to2p\).
  • 1.5: [A], alpha decay of uranium-238.
  • 1.6: [C], conversion of potassium-40 into argon-40.
  • 1.7: \(M=2.01\times10^{-2}\ \mathrm{kg\,mol^{-1}}\).
  • 1.8: \(\gamma=1.40\).
  • 1.9: [B], hydrogen fluoride, \(\ce{HF}\), SMILES [H]F.

Solution

The calculations use the ideal-gas model and \[ T=273.15+15.0=288.15\ \mathrm{K}. \] The problem does not print the physical constants; standard values used here are \[ R=8.3145\ \mathrm{J\,mol^{-1}\,K^{-1}},\qquad h=6.6261\times10^{-34}\ \mathrm{J\,s},\qquad c=2.9979\times10^8\ \mathrm{m\,s^{-1}}. \]

1.1

Pure nitrogen is molecular nitrogen, \(\ce{N2}\), SMILES N#N, with molar mass \(28.02\ \mathrm{g\,mol^{-1}}\). Its amount in the vessel is \[ n=\frac{m}{M} =\frac{2.2990\ \mathrm{g}}{28.02\ \mathrm{g\,mol^{-1}}} =0.08205\ \mathrm{mol}. \]

From \(pV=nRT\), \[ V=\frac{nRT}{p} =\frac{(0.08205)(8.3145)(288.15)} {1.013\times10^5}\ \mathrm{m^3} =\boxed{1.94\times10^{-3}\ \mathrm{m^3}}. \]

1.2

Argon, \(\ce{Ar}\), SMILES [Ar], has molar mass \(39.95\ \mathrm{g\,mol^{-1}}\).

At the same pressure, temperature, and vessel volume, the pure nitrogen and the mixture contain the same total amount of gas. If \(x\) is the argon mole fraction, the mixture’s mean molar mass is \[ \overline M=(1-x)M(\ce{N2})+xM(\ce{Ar}). \] Consequently, \[ \frac{m_{\mathrm{atmospheric}}}{m_{\mathrm{chemical}}} =\frac{\overline M}{M(\ce{N2})} =1+x\frac{M(\ce{Ar})-M(\ce{N2})}{M(\ce{N2})}. \]

Solving, \[ x= \left(\frac{2.3102}{2.2990}-1\right) \frac{28.02}{39.95-28.02} =0.011442\ldots \] and therefore \[ \boxed{x=0.0114\quad\text{or}\quad1.14\ \mathrm{mol}\%}. \]

This is the argon fraction in Rayleigh’s oxygen-removed atmospheric sample.

1.3

The helium line—helium, \(\ce{He}\), SMILES [He]—has wavelength approximately \(588\ \mathrm{nm}\). Its photon energy is \[ E=h\nu=\frac{hc}{\lambda}. \] Thus, \[ E= \frac{(6.6261\times10^{-34}\ \mathrm{J\,s}) (2.9979\times10^8\ \mathrm{m\,s^{-1}})} {588\times10^{-9}\ \mathrm{m}} =\boxed{3.38\times10^{-19}\ \mathrm{J}}. \]

Reading the line more closely as approximately \(587.7\ \mathrm{nm}\) gives the same result to three significant figures.

1.4

The photon energy must equal the separation between the two levels: \[ \Delta E=3.38\times10^{-19}\ \mathrm{J} =0.338\times10^{-18}\ \mathrm{J}. \]

From Figure 2, the \(2p\) and \(3d\) levels lie at approximately \(3.36\times10^{-18}\ \mathrm{J}\) and \(3.70\times10^{-18}\ \mathrm{J}\), respectively. Their separation is \[ \Delta E\approx(3.70-3.36)\times10^{-18} =3.4\times10^{-19}\ \mathrm{J}, \] which matches the photon energy.

Therefore, mark [E]. The upward arrow represents absorption, \(2p\to3d\); emission of the same spectral line corresponds to \(3d\to2p\). The other electron remains in the \(1s\) orbital.

1.5

Uranium-238 undergoes alpha decay: \[ \boxed{\ce{^{238}_{92}U -> ^{234}_{90}Th + ^{4}_{2}He}}. \]

Both nucleon number and nuclear charge are conserved: \[ 238=234+4,\qquad 92=90+2. \]

The emitted alpha particle is a helium-4 nucleus, \(\ce{^{4}He^{2+}}\), SMILES [4He+2]. After slowing down, it acquires two electrons: \[ \ce{^{4}He^{2+} + 2 e- -> ^{4}He}. \] This produces neutral helium-4, SMILES [4He], which can accumulate in the mineral.

Therefore, mark [A]. SMILES identifies the chemical species; the nuclear transformation is specified by the isotope equation.

1.6

Potassium-40 can form argon-40 by electron capture: \[ \boxed{\ce{^{40}_{19}K + ^{0}_{-1}e -> ^{40}_{18}Ar + \nu_e}}. \] Here \(\nu_e\) is an electron neutrino. The mass number remains 40, while the nuclear charge decreases from 19 to 18.

The positron-emission alternative included in option [C] is \[ \ce{^{40}_{19}K -> ^{40}_{18}Ar + ^{0}_{+1}e + \nu_e}. \]

Both processes produce argon-40, \(\ce{^{40}Ar}\), SMILES [40Ar], which can remain trapped in rocks. Therefore, mark [C].

1.7

Combining the ideal-gas equation with \(n=m/M\) and \(\rho=m/V\), \[ pV=\frac{m}{M}RT \quad\Longrightarrow\quad M=\frac{\rho RT}{p}. \] Therefore, \[ M= \frac{(0.850\ \mathrm{kg\,m^{-3}}) (8.3145\ \mathrm{Pa\,m^3\,mol^{-1}\,K^{-1}}) (288.15\ \mathrm{K})} {1.013\times10^5\ \mathrm{Pa}} =\boxed{2.01\times10^{-2}\ \mathrm{kg\,mol^{-1}}}. \]

Propagating only the stated density uncertainty, \[ \delta M=M\frac{\delta\rho}{\rho} \approx1.2\times10^{-4}\ \mathrm{kg\,mol^{-1}}. \] Thus a result displaying that uncertainty is \[ M=(0.02010\pm0.00012)\ \mathrm{kg\,mol^{-1}}. \]

1.8

The measured speed of sound is \[ v_s=f\lambda =(3520\ \mathrm{s^{-1}})(0.116\ \mathrm{m}) =408.32\ \mathrm{m\,s^{-1}}. \]

Rearranging the supplied relation and substituting \(M=\rho RT/p\), \[ \gamma=\frac{Mv_s^2}{RT} =\frac{\rho v_s^2}{p}. \] Hence, \[ \gamma= \frac{(0.850)(408.32)^2}{1.013\times10^5} =1.39898\ldots =\boxed{1.40}. \]

The density uncertainty alone gives \[ \delta\gamma=\gamma\frac{0.005}{0.850}\approx0.0082, \] or approximately \(\gamma=1.399\pm0.008\).

1.9

The measured molar mass is approximately \(20.1\ \mathrm{g\,mol^{-1}}\). This excludes hydrogen chloride, \(\ce{HCl}\), SMILES [H]Cl, and argon, \(\ce{Ar}\), SMILES [Ar], whose molar masses are approximately \(36.46\) and \(39.95\ \mathrm{g\,mol^{-1}}\), respectively.

Both hydrogen fluoride, \(\ce{HF}\), SMILES [H]F, with \(M\approx20.01\ \mathrm{g\,mol^{-1}}\), and neon, \(\ce{Ne}\), SMILES [Ne], with \(M\approx20.18\ \mathrm{g\,mol^{-1}}\), are compatible with the molar-mass measurement.

The heat-capacity ratio distinguishes them. A monatomic ideal gas such as neon has \[ C_V=\frac32R,\qquad C_p=\frac52R,\qquad \gamma=\frac53\approx1.67. \] A diatomic ideal gas with active translational and rotational motion, but negligible vibrational excitation, has \[ C_V=\frac52R,\qquad C_p=\frac72R,\qquad \gamma=\frac75=1.40. \]

The measured ratio agrees with the diatomic gas. Therefore, mark [B]: hydrogen fluoride, \(\ce{HF}\), SMILES [H]F.

The solution was checked against the official solution.