MEASUREMENT OF GAS VOLUMES

Standard 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

(1) v = V + t/273 V, 

or (2) v = V(1 + (t/273)). 

Since 1/273 = 0.00366, the formula may be written

(3) v = V(1 + 0.00366t). 

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:

(4) V = v/(1 + 0.00366t). 

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

V = 750/(1 + 0.00366 × 20) = 698.9 cc. 

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[Pg 25] 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:

(5) V_s = vp/(760(1 + 0.00366t)),

in which V_s 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.

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  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

(6) V_s = v(p - a)/(760(1 + 0.00366t)).

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

(6) V_s = 500(740 - 17.4)/(760(1 + 0.00366 × 20)) = 442.9 cc. 

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.