The Law of Combining Volumes

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 13:24

In effect, the preceding example used the factor P/RT to convert from volume to amount of gasA state of matter in which a substance occupies the full volume of its container and changes shape to match the shape of the container. In a gas the distance between particles is much greater than the diameters of the particles themselves; hence the distances between particles can change as necessary so that the matter uniformly occupies its container.. The reciprocal of this factor can be used to convert from amount of gas to volume. This is emphasized if we rewrite Eq. (2) from The Ideal Gas Equation as

V=\frac{RT}{P}\text{ }n\text{                 (1)}

This indicates that when we write a chemical equationA representation of a chemical reaction in which chemical symbols represent reactants on the left side and products on the right side. involving gases, the coefficients not only tell us what amount of each substanceA material that is either an element or that has a fixed ratio of elements in its chemical formula. is consumed or produced, they also indicate the relative volume of each gas consumed or produced. For example,

means that for every
2 mol H2(g)


will be
1 mol O2(g)


2 mol H2O(g) produced
It also implies that for every
\left( \text{2 mol }\times \frac{RT}{P} \right)\text{ dm}^{\text{3}}
will be
\left( \text{1 mol }\times \frac{RT}{P} \right)\text{ dm}^{\text{3}}
\left( \text{2 mol }\times \frac{RT}{P} \right)\text{ dm}^{\text{3}}
2H2(g) consumed O2(g) consumed 2H2O(g) produced

This is an example of the law of combining volumes, which states that when gases combine at constant temperatureA physical property that indicates whether one object can transfer thermal energy to another object. and pressureForce per unit area; in gases arising from the force exerted by collisions of gas molecules with the wall of the container., the volumes involved are always in the ratio of simple whole numbers. Since the factor RT/P would be the same for all three gases, the volume of O2(g) consumed must be half the volume of H2(g) consumed. The volume of H2O(g) produced would be only two-thirds the total volume [of H2(g) and O2(g)] con- sumed, and so at the end of the reaction the total volume must be less than at the beginning.

The law of combining volumes was proposed by Gay-Lussac at about the same time that Dalton published his atomic theory. Shortly thereafter, Avogadro suggested the hypothesis that equal volumes of gases contained equal numbers of molecules. Dalton strongly opposed Avogadro’s hypothesis because it required that some molecules contain more than the minimum number of atomsThe smallest particle of an element that can be involved in chemical combination with another element; an atom consists of protons and neutrons in a tiny, very dense nucleus, surrounded by electrons, which occupy most of its volume..

For example, according to Dalton, the formula for hydrogen gas should be the simplest possible, e.g., H. Similarly, Dalton proposed the formula O for oxygen gas. His equation for formation of water vaporThe gaseous state of a substance that typically exists as a liquid or solid; a gas at a temperature near or below the boiling point of the corresponding liquid. was

 \text{1 volume} \ 
\end{smallmatrix}}{\mathop{\text{H}}}\,\text{ + }\underset{\begin{smallmatrix} 
 \text{1 volume} \ 
\end{smallmatrix}}{\mathop{\text{O}}}\,\text{ }\to \text{     }\underset{\begin{smallmatrix} 
 \text{1 volume} \ 
 \text{water vapor} 

But experiments showed that twice as great a volume of hydrogen as of oxygen was required for complete reaction. Furthermore, the volume of water vapor produced was twice the volume of oxygen consumed. Avogadro proposed (correctly, as it turned out) that the formulas for hydrogen, oxygen, and water were H2, O2 and H2O, and he explained the volume data in much the same way as we have done for Eq. (2).

Dalton, who had originally conceived the idea of atoms and molecules, was unwilling to concede that substances such as hydrogen or water might have formulas more complicated than was absolutely necessary. Partly as a result of Dalton’s opposition, it took almost half a century before Avogadro’s Italian countryman Stanislao Cannizzaro (1826 to 1910) was able to convince chemists that Avogadro’s hypothesis was correct. The blindness of chemists to Avogadro’s ideas for so long makes one wonder whether today’s Nobel prize winners might not be equally wrong about some other aspect of chemistry. Who knows but that some forgotten Argentinian Avogadro is still waiting for a Cannizzaro to explain his or her ideas to the scientific world.

Because the amount of gas is related to volume by the ideal gas law, it is possible to calculate the volume of a gaseous substance consumed or produced in a reaction. Molar massThe mass of a mole of substance; the same as molecular weight for molecular substances. and stoichiometric ratio are employed in the same way as in Sec. 3.1, and the factor RT/P is used to convert from amount of gas to volume.

EXAMPLE 1 Oxygen was first prepared by Joseph Priestly-by heating red “calx of mercury“ [mercury(II)oxide, HgO] according to the equation

2HgO(s) → 2Hg(l) + O2(g)

What volume (in cubic centimeters) of O2 can be prepared from 1.00 g HgO?

The volume is measured at 20°C and 0.987 atmAbbreviation for atmosphere, a unit of pressure equal to 101.325 kPa or 760 mmHg..

SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. The mass of HgO can be converted to amount of HgO and this can be converted to amount of O2 by means of a stoichiometric ratio. Finally, the ideal gas law is used to obtain the volume of O2. Schematically,

m_{\text{HgO}}\xrightarrow{M_{\text{HgO}}}n_{\text{HgO}}\xrightarrow{S\left( \text{O}_{\text{2}}\text{/HgO} \right)}n_{\text{O}_{\text{2}}}\xrightarrow{RT/P}V_{\text{O}_{\text{2}}}

V_{\text{O}_{\text{2}}}=\text{1 g HgO }\times \text{ }\frac{\text{1 mol HgO}}{\text{216}\text{.59 g HgO}}\text{ }\times \text{ }\frac{\text{1 mol O}_{\text{2}}}{\text{2 mol HgO}}\text{ }

\times \text{ }\frac{\text{0}\text{.0820 liter atm}}{\text{1 K mol O}_{\text{2}}}\text{ }\times \text{ }\frac{\text{293}\text{.15 K}}{\text{0}\text{.987 atm}}=\text{0}\text{.0562 liter}=\text{56}\text{.2 cm}^{\text{3}}