IPhO 1971 · Theory problem 2

Hydrogen in a tube closed by a mercury column

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A vertical glass tube of cross section \(S = 1.0\ \mathrm{cm}^2\) contains unknown amount of hydrogen. The upper end of the tube is closed. The other end is opened and is immersed in a pan filled with mercury. The tube and the pan are placed in a sealed chamber containing air at temperature \(T_0 = 273\ \mathrm{K}\) and pressure \(P_0 = 1.334\times10^5\ \mathrm{Pa}\). Under these conditions the height of mercury column in the tube above the mercury level in the pan is \(h_0 = 0.70\ \mathrm{m}\).

One of the walls of the chamber is a piston, which expands the air isothermally to a pressure of \(P_1 = 8.00\times10^4\ \mathrm{Pa}\). As a result the height of the mercury column in the tube decreases to \(h_1 = 0.40\ \mathrm{m}\). Then the chamber is heated up at a constant volume to some temperature \(T_2\) until the mercury column rises to \(h_2 = 0.50\ \mathrm{m}\). Finally, the air in the chamber is expanded at constant pressure and the mercury level in the tube settles at \(h_3 = 0.45\ \mathrm{m}\) above the mercury level in the pan.

Provided that the system is in mechanical and thermal equilibrium during all the processes calculate the mass \(m\) of the hydrogen, the intermediate temperature \(T_2\), and the pressure \(P\) in the final state.

The density of mercury at temperature \(T_0\) is \(\rho_0 = 1.36\times10^4\ \mathrm{kg/m}^3\), the coefficient of expansion for mercury \(\beta = 1.84\times10^{-4}\ \mathrm{K}^{-1}\), and the gas constant \(R = 8.314\ \mathrm{J/(mol\times K)}\). The thermal expansion of the glass tube and the variations of the mercury level in the pan are not considered.

Hint. If \(\Delta T\) is the interval of temperature variations of the system then \(\beta\Delta T = x \ll 1\). In that case you can use the approximation: \[ \frac{1}{1+x}\approx 1-x. \]

Official English text of the problem.

Answers

Question 2 has no numbered subparts. In the requested order:

  • Hydrogen mass: \(\boxed{m \approx 2.1\times10^{-6}\ \mathrm{kg}=2.1\ \mathrm{mg}}\).
  • Intermediate temperature: \(\boxed{T_2\approx358\ \mathrm{K}}\).
  • Final hydrogen pressure: \(\boxed{P_{\mathrm{H}_2}\approx4.7\times10^4\ \mathrm{Pa}}\). The final chamber-air pressure is \(\boxed{P_{\mathrm{air}}\approx1.05\times10^5\ \mathrm{Pa}}\); the statement does not say whether “the pressure \(P\) in the final state” means the hydrogen pressure or the chamber-air pressure, so both are given.

Solution

Treat hydrogen and air as ideal gases, with negligible hydrostatic pressure variations within the gases. Use the standard values \[ g=9.81\ \mathrm{m\,s^{-2}},\qquad M_{\mathrm{H}_2}=2.0\times10^{-3}\ \mathrm{kg\,mol^{-1}}. \]

Let \(H\) be the height of the tube’s closed end above the mercury surface in the pan. Since glass expansion and changes of the pan level are neglected, \(H\) is constant. At state \(i\), the hydrogen volume is \[ V_i=S(H-h_i). \]

Write \(P_i\) for the chamber-air pressure and \(p_i\) for the hydrogen pressure. Hydrostatic equilibrium in the connected mercury gives \[ p_i=P_i-\rho_i gh_i. \] Thermal equilibrium means that both gases and the mercury have the same temperature at each state.

Hydrogen mass \(m\).

The first expansion is isothermal at \(T_0\), so the mercury density remains \(\rho_0\). Thus \[ \begin{aligned} p_0&=P_0-\rho_0gh_0 =4.00088\times10^4\ \mathrm{Pa},\\ p_1&=P_1-\rho_0gh_1 =2.66336\times10^4\ \mathrm{Pa}. \end{aligned} \]

The fixed amount of hydrogen obeys Boyle’s law: \[ p_0S(H-h_0)=p_1S(H-h_1). \] Consequently, \[ H=\frac{p_0h_0-p_1h_1}{p_0-p_1} \approx1.29738\ \mathrm{m}. \]

Using \(S=1.0\times10^{-4}\ \mathrm{m^2}\), the amount of hydrogen is \[ n=\frac{p_0S(H-h_0)}{RT_0} \approx1.0530\times10^{-3}\ \mathrm{mol}. \] Therefore, \[ \boxed{m=nM_{\mathrm{H}_2}\approx2.1\times10^{-6}\ \mathrm{kg} =2.1\ \mathrm{mg}.} \]

For the remaining calculations, define the constant \[ C=\frac{nR}{S} \approx87.5475\ \mathrm{Pa\,m\,K^{-1}}. \] The hydrogen ideal-gas equation then becomes \[ p_i=\frac{CT_i}{H-h_i}. \]

Intermediate temperature \(T_2\).

Mercury’s volume expansion gives its temperature-dependent density: \[ \rho(T)=\frac{\rho_0}{1+\beta(T-T_0)} \approx\rho_0\left[1-\beta(T-T_0)\right], \] where the final expression uses the supplied first-order approximation.

During the constant-volume heating, the chamber air satisfies \[ P_2=P_1\frac{T_2}{T_0}. \] Combining this with hydrostatic equilibrium and the hydrogen gas law at \(h_2\), \[ \frac{CT_2}{H-h_2} = P_1\frac{T_2}{T_0} -\rho_0gh_2\left[1-\beta(T_2-T_0)\right]. \]

Collecting the terms containing \(T_2\) gives \[ \left[ \frac{P_1}{T_0} -\frac{C}{H-h_2} +\beta\rho_0gh_2 \right]T_2 = \rho_0gh_2(1+\beta T_0). \] Hence \[ T_2= \frac{\rho_0gh_2(1+\beta T_0)} {\displaystyle P_1/T_0-C/(H-h_2)+\beta\rho_0gh_2} \approx358.32\ \mathrm{K}, \] so \[ \boxed{T_2\approx358\ \mathrm{K}.} \]

The chamber-air pressure at this stage is \[ P_2=P_1\frac{T_2}{T_0} \approx1.05002\times10^5\ \mathrm{Pa}. \]

Pressure in the final state.

The final expansion occurs at constant chamber-air pressure. Therefore, \[ \boxed{P_3=P_2\approx1.05\times10^5\ \mathrm{Pa}.} \]

To determine the hydrogen pressure, its final temperature \(T_3\) must also be found. At \(h_3=0.45\ \mathrm{m}\), \[ p_3=\frac{CT_3}{H-h_3} = P_2-\rho_0gh_3\left[1-\beta(T_3-T_0)\right]. \] Rearranging, \[ \left[ \frac{C}{H-h_3}-\beta\rho_0gh_3 \right]T_3 = P_2-\rho_0gh_3(1+\beta T_0). \] Thus \[ T_3= \frac{P_2-\rho_0gh_3(1+\beta T_0)} {\displaystyle C/(H-h_3)-\beta\rho_0gh_3} \approx454.64\ \mathrm{K}. \]

The final hydrogen pressure is consequently \[ p_3=\frac{CT_3}{H-h_3} \approx4.6971\times10^4\ \mathrm{Pa}, \] or \[ \boxed{P_{\mathrm{H}_2}\approx4.7\times10^4\ \mathrm{Pa}.} \]

Finally, the expansion parameters are \[ \beta(T_2-T_0)\approx0.0157,\qquad \beta(T_3-T_0)\approx0.0334. \] Both are small, consistent with the approximation requested in the hint.

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.