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IPhO 1979, Теория и експеримент — Задача 2

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Problems of the XI International Olympiad, Moscow, 1979

Problems of the XI International Olympiad, Moscow, 1979

The publication has been prepared by Prof. S. Kozel and Prof. V.Orlov (Moscow Institute of Physics and Technology)

The XI International Olympiad in Physics for students took place in Moscow, USSR, in July 1979 on the basis of Moscow Institute of Physics and Technology (MIPT). Teams from 11 countries participated in the competition, namely Bulgaria, Finland, Germany, Hungary, Poland, Romania, Sweden, Czechoslovakia, the DDR, the SFR Yugoslavia, the USSR. The problems for the theoretical competition have been prepared by professors of MIPT (V.Belonuchkin, I.Slobodetsky, S.Kozel). The problem for the experimental competition has been worked out by O.Kabardin from the Academy of Pedagogical Sciences.

It is pity that marking schemes were not preserved.

Условие

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 $P_h = 2\cdot 10^{3}$ Pa. The total atmospheric pressure ( $P = 10^{5}$ Pa) and the temperature ( $t$ =20° 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 mg ?

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

Отговори

Решение

A sample and weights are affected by the Archimede’s buoyancy force of either dry or humid air in the first and second cases, respectively. The difference in the scale indication $\Delta F$ is determined by the change of difference of these forces.

The difference of Archimede’s buoyancy forces in dry air:

\Delta F_1 = \Delta V\rho_a' g

Whereas in humid air it is:

\Delta F_2 = \Delta V\rho_a'' g

where $\Delta V$ - the difference in volumes between the sample and the weights, and $\rho_a'$ and $\rho_a''$ - densities of dry and humid air, respectively.

Then the difference in the scale indications $\Delta F$ could be written as follows:

\Delta F = \Delta F_1 - \Delta F_2 = \Delta Vg\left(\rho_a' - \rho_a''\right) \qquad (1)

According to the problem conditions this difference should be distinguished, i.e. $\Delta F \ge m_0 g$ or $\Delta Vg\left(\rho_a' - \rho_a''\right) \ge m_0$ , wherefrom

\Delta V \ge \frac{m_0}{\rho_a' - \rho_a''}. \qquad (2)

The difference in volumes between the aluminum sample and brass weights can be found from the equation

\Delta V = \frac{m}{\rho_1} - \frac{m}{\rho_2} = m\left(\frac{\rho_2 - \rho_1}{\rho_1\rho_2}\right), \qquad (3)

where $m$ is the sought mass of the sample. From expressions (2) and (3) we obtain

m = \Delta V\left(\frac{\rho_1\rho_2}{\rho_2 - \rho_1}\right) \ge \frac{m_0}{\rho_a' - \rho_a''}\left(\frac{\rho_1\rho_2}{\rho_2 - \rho_1}\right). \qquad (4)

To find the mass $m$ of the sample one has to determine the difference $\left(\rho_a' - \rho_a''\right)$ .

With the general pressure being equal, in the second case, some part of dry air is replaced by vapor:

\rho_a' - \rho_a'' = \frac{\Delta m_a}{V} - \frac{\Delta m_v}{V}.

Changes of mass of air $\Delta m_a$ and vapor $\Delta m_v$ can be found from the ideal-gas equation of state

\Delta m_a = \frac{P_aVM_a}{RT}, \qquad \Delta m_v = \frac{P_vVM_v}{RT},

wherefrom we obtain

\rho_a' - \rho_a'' = \frac{P_a\left(M_a - M_v\right)}{RT}. \qquad (5)

From equations (4) and (5) we obtain

m \ge \frac{m_0RT}{P_a\left(M_a - M_v\right)}\left(\frac{\rho_1\rho_2}{\rho_2 - \rho_1}\right). \qquad (6)

The substitution of numerical values gives the answer: $m \ge 0.0432$ kg $\approx$ 43 g.

Note. When we wrote down expression (3), we considered the sample mass be equal to the weights’ mass, at the same time allowing for a small error.

One may choose another way of solving this problem. Let us calculate the change of Archimede’s force by the change of the air average molar mass.

In dry air the condition of the balance between the sample and weights could be written down in the form of

\left(\rho_1 - \frac{M_aP}{RT}\right)V_1 = \left(\rho_2 - \frac{M_aP}{RT}\right)V_2. \qquad (7)

In humid air its molar mass is equal to

M = M_a\frac{P_a}{P} + M_v\frac{P - P_a}{P}, \qquad (8)

whereas the condition of finding the scale error could be written in the form

\left(\rho_1 - \frac{M_aP}{RT}\right)V_1 - \left(\rho_2 - \frac{M_aP}{RT}\right)V_2 \ge m_0. \qquad (9)

From expressions (7) –(9) one can get a more precise answer

m \ge \frac{m_0RT\rho_1\rho_2 - M_aP_a}{\left(M_a - M_v\right)\left(\rho_2 - \rho_1\right)P_a}. \qquad (10)

Since $M_aP_a \ll m_0\rho_1\rho_2RT$ , then both expressions (6) and (10) lead practically to the same quantitative result, i.e. m $\ge$ 43 g.

Acknowledgement

The authors would like to express their thanks and gratitude to Professor Waldemar Gorzkowski and Professor Ivo Volf for supplying the materials for the XI IphO in the Polish, Hungarian and Czech languages that have been of great help to the authors in their work at the problems of the Olympiad.

References:

O.Kabardin, V.Orlov, International Physics Olympiads for Pupuls, Nauka, Moskva 1985.
W.Gorzkowski, A.Kotlicki, Olimpiady Fizyczne XXVII i XXVIII, WsiP, Warszawa 1983
R.Kunfalvi, Collection of Competition Tasks from the 1 trough XV International Physics Olympiads, 1867-1984, Roland Eotvos Physical Society an UNESCO, Budapest 1985
V.Urumov, Megjunadodni Olimpijadi po Fisika, Prosvento Delo, Skopje 1999
D.Kluvanec, I.Volf, Mezinarodni Fysikalni Olympiady, MaFy, Hradec Kralowe 1993

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