An Elementary Study of Chemistry by William McPherson (best beach reads .txt) 📖
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Allotropic states or forms of matter. Other elements besides oxygen may exist in more than one form. These different forms of the same element are called allotropic states or forms of the element. These forms differ not only in physical properties but also in their energy contents. Elements often exist in a variety of forms which look quite different. These differences may be due to accidental causes, such as the size or shape of the particles or the way in which the element was prepared. Only such forms, however, as have different energy contents are properly called allotropic forms.
MEASUREMENT OF GAS VOLUMESStandard conditions. It is a well-known fact that the volume occupied by a definite weight of any gas can be altered by changing the temperature of the gas or the pressure to which it is subjected. In measuring the volume of gases it is therefore necessary, for the sake of accuracy, to adopt some standard conditions of temperature and pressure. The conditions agreed upon are (1) a temperature of 0°, and (2) a pressure equal to the average pressure exerted by the atmosphere at the sea level, that is, 1033.3 g. per square centimeter. These conditions of temperature and pressure are known as the standard conditions, and when the volume of a gas is given it is understood that the measurement was made under these conditions, unless it is expressly stated otherwise. For example, the weight of a liter of oxygen has been given as 1.4285 g. This means that one liter of oxygen, measured at a temperature of 0° and under a pressure of 1033.3 g. per square centimeter, weighs 1.4285 g.
The conditions which prevail in the laboratory are never the standard conditions. It becomes necessary, therefore, to find a way to calculate the volume which a gas will occupy under standard conditions from the volume which it occupies under any other conditions. This may be done in accordance with the following laws.
Law of Charles. This law expresses the effect which a change in the temperature of a gas has upon its volume. It may be stated as follows: For every degree the temperature of a gas rises above zero the volume of the gas is increased by 1/273 of the volume which it occupies at zero; likewise for every degree the temperature of the gas falls below zero the volume of the gas is decreased by 1/273 of the volume which it occupies at zero, provided in both cases that the pressure to which the gas is subjected remains constant.
If V represents the volume of gas at 0°, then the volume at 1° will be V + 1/273 V; at 2° it will be V + 2/273 V; or, in general, the volume v, at the temperature t, will be expressed by the formula
Since 1/273 = 0.00366, the formula may be written
Since the value of V (volume under standard conditions) is the one usually sought, it is convenient to transpose the equation to the following form:
The following problem will serve as an illustration of the application of this equation.
The volume of a gas at 20° is 750 cc.; find the volume it will occupy at 0°, the pressure remaining constant.
In this case, v = 750 cc. and t = 20. By substituting these values, equation (4) becomes
Law of Boyle. This law expresses the relation between the volume occupied by a gas and the pressure to which it is subjected. It may be stated as follows: The volume of a gas is inversely proportional to the pressure under which it is measured, provided the temperature of the gas remains constant.
If V represents the volume when subjected to a pressure P and v represents its volume when the pressure is changed to p, then, in accordance with the above law, V : v :: p : P, or VP = vp. In other words, for a given weight of a gas the product of the numbers representing its volume and the pressure to which it is subjected is a constant.
Since the pressure of the atmosphere at any point is indicated by the barometric reading, it is convenient in the solution of the problems to substitute the latter for the pressure measured in grams per square centimeter. The average reading of the barometer at the sea level is 760 mm., which corresponds to a pressure of 1033.3 g. per square centimeter. The following problem will serve as an illustration of the application of Boyle's law.
A gas occupies a volume of 500 cc. in a laboratory where the barometric reading is 740 mm. What volume would it occupy if the atmospheric pressure changed so that the reading became 750 mm.?
Substituting the values in the equation VP = vp, we have 500 × 740 = v × 750, or v = 493.3 cc.
Variations in the volume of a gas due to changes both in temperature and pressure. Inasmuch as corrections must be made as a rule for both temperature and pressure, it is convenient to combine the equations given above for the corrections for each, so that the two corrections may be made in one operation. The following equation is thus obtained:
in which Vs represents the volume of a gas under standard conditions and v, p, and t the volume, pressure, and temperature respectively at which the gas was actually measured.
The following problem will serve to illustrate the application of this equation.
A gas having a temperature of 20° occupies a volume of 500 cc. when subjected to a pressure indicated by a barometric reading of 740 mm. What volume would this gas occupy under standard conditions?
In this problem v = 500, p = 740, and t = 20. Substituting these values in the above equation, we get
Variations in the volume of a gas due to the pressure of aqueous vapor. In many cases gases are collected over water, as explained under the preparation of oxygen. In such cases there is present in the gas a certain amount of water vapor. This vapor exerts a definite pressure, which acts in opposition to the atmospheric pressure and which therefore must be subtracted from the latter in determining the effective pressure upon the gas. Thus, suppose we wish to determine the pressure to which the gas in tube A (Fig. 8) is subjected. The tube is raised or lowered until the level of the water inside and outside the tube is the same. The atmosphere presses down upon the surface of the water (as indicated by the arrows), thus forcing the water upward within the tube with a pressure equal to the atmospheric pressure. The full force of this upward pressure, however, is not spent in compressing the gas within the tube, for since it is collected over water it contains a certain amount of water vapor. This water vapor exerts a pressure (as indicated by the arrow within the tube) in opposition to the upward pressure. It is plain, therefore, that the effective pressure upon the gas is equal to the atmospheric pressure less the pressure exerted by the aqueous vapor. The pressure exerted by the aqueous vapor increases with the temperature. The figures representing the extent of this pressure (often called the tension of aqueous vapor) are given in the Appendix. They express the pressure or tension in millimeters of mercury, just as the atmospheric pressure is expressed in millimeters of mercury. Representing the pressure of the aqueous vapor by a, formula (5) becomes
The following problem will serve to illustrate the method of applying the correction for the pressure of the aqueous vapor.
The volume of a gas measured over water in a laboratory where the temperature is 20° and the barometric reading is 740 mm. is 500 cc. What volume would this occupy under standard conditions?
The pressure exerted by the aqueous vapor at 20° (see table in Appendix) is equal to the pressure exerted by a column of mercury 17.4 mm. in height. Substituting the values of v, t, p, and a in formula (6), we have
Adjustment of tubes before reading gas volumes. In measuring the volumes of gases collected in graduated tubes or other receivers, over a liquid as illustrated in Fig. 8, the reading should be taken after raising or lowering the tube containing the gas until the level of the liquid inside and outside the tube is the same; for it is only under these conditions that the upward pressure within the tube is the same as the atmospheric pressure.
EXERCISES1. What is the meaning of the following words? phlogiston, ozone, phosphorus. (Consult dictionary.)
2. Can combustion take place without the emission of light?
3. Is the evolution of light always produced by combustion?
4. (a) What weight of oxygen can be obtained from 100 g. of water? (b) What volume would this occupy under standard conditions?
5. (a) What weight of oxygen can be obtained from 500g. of mercuric oxide? (b) What volume would this occupy under standard conditions?
6. What weight of each of the following compounds is necessary to prepare 50 l. of oxygen? (a) water; (b) mercuric oxide; (c) potassium chlorate.
7. Reduce the following volumes to 0°, the pressure remaining constant: (a) 150 cc. at 10°; (b) 840 cc. at 273°.
8. A certain volume of gas is measured when the temperature is 20°. At what temperature will its volume be doubled?
9. Reduce the following volumes to standard conditions of pressure, the temperature remaining constant: (a) 200 cc. at 740 mm.; (b) 500 l. at 380 mm.
10. What is the weight of 1 l. of oxygen when the pressure is 750 mm. and the temperature 0°?
11. Reduce the following volumes to standard conditions of temperature and pressure: (a) 340 cc. at 12° and 753 mm; (b) 500 cc. at 15° and 740 mm.
12. What weight of potassium chlorate is necessary to prepare 250 l. of oxygen at 20° and 750 mm.?
13. Assuming the cost of potassium chlorate and mercuric oxide to be respectively $0.50 and $1.50 per kilogram, calculate the cost of materials necessary for the preparation of 50 l. of oxygen from each of the above compounds.
14. 100 g. of potassium chlorate and 25 g. of manganese dioxide were heated in the preparation of oxygen. What products were left in the flask, and how much of each was present?
CHAPTER III HYDROGENHistorical. The element hydrogen was first clearly recognized as a distinct substance by the English investigator Cavendish, who in 1766 obtained it in a pure state, and showed it to be
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