Chemistry
Graham's Law
Graham's Law states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of its molar mass. In simpler terms, lighter gases will diffuse or effuse faster than heavier gases at the same temperature. This law is important in understanding the behavior of gases and is used in various applications, such as in the study of gas mixtures.
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7 Key excerpts on "Graham's Law"
eBook - PDF- Arthur Adamson(Author)
- 2012(Publication Date)
- Academic Press(Publisher)
Graham studied the rate of effusion of gases, that is, the rate of escape of gas through a small hole or orifice. He found that for a given temperature and pressure difference, the rate of effusion of a gas is inversely proportional to the square root of its density. For two different gases, then, we have ^ = ( -^ ) ' (2-47) n 2 P l J where η denotes the number of moles escaping per unit time. Since the comparison is at a given pressure and temperature, an alternative form of Graham's Law is Mo 1/2 9 χ (2-48) Mj ' where M is molecular weight. Graham's Law follows directly from Eq. (2-45) if it is assumed that in an effusion experiment the rate of passage of molecules through the hole is proportional to the rate at which they would be hitting the area of surface corresponding to the area of the hole. Effusion rates are thus often assumed to be given directly by Z. There is a problem in that, unless the hole is very small, there may be both a pressure and a temperature drop in its vicinity, which introduces a correction term dependent on the effusion rate and hence on the molecular weight of the gas. It is also important in effusion that the flow be molecular, that is, the molecules should escape directly through the hole without collisions with the sides of the hole or with each other. Should such collisions occur, then some molecules will be reflected back into the vessel whence they came. The limiting case in which many molecular collisions occur as the gas flows through a channel is one of diffusion, a much slower process than effusion; and the limiting case in which molecules make many collisions with the sides of the hole, but not with each other, is called Knudsen flow, again a slower process than effusion. Example. A sample calculation is appropriate to illustrate the use of units and to give an order-of-magnitude appreciation of Z. Consider the case of a water surface at 25°C and in equilibrium with its vapor.
eBook - ePubChemistry
Concepts and Problems, A Self-Teaching Guide
- Richard Post, Chad Snyder, Clifford C. Houk(Authors)
- 2020(Publication Date)
- Jossey-Bass(Publisher)
In the next section you will learn how to determine the formula weight of an unknown gas. Graham's Law Earlier in this chapter, we mentioned that gases readily diffuse (mix spontaneously throughout each other). Even with no breeze blowing the smell of ammonia (a gas) or the scent of a surprised skunk (gaseous vapor) rapidly diffuses through the air. While diffusion denotes the mixing of gases, effusion is the process of a gas going through a small opening into a vacuum. (A vacuum is the absence of a solid, liquid, or gas.) Diffusion indicates one gas mixing into another gas. Effusion indicates a gas moving into a vacuum. A scientist named Graham noted a relationship between the formula weight of a gas and the velocity or speed with which it effuses. Such a relationship is very useful for determining the formula weights of unknown gases. It is also useful in the separation of two gases of different formula weights. Although Graham's Law predicts accurately the velocity of a gas during effusion, the same law can be used as a reasonable approximation of the velocity of a gas during diffusion. According to Graham's Law, at a constant temperature, the rate (velocity) of effusion of a gas is inversely proportional to the square root of its formula weight. In the equation, a gas of known molecular weight is compared to an unknown gas. In all calculations with Graham's Law, it is assumed that the temperature and pressure are the same for both gases: Here is a practical example using Graham's Law. Let oxygen gas (O 2) be the known gas. Oxygen has an effusion rate of 10 milliliters (mL) per second and a formula weight of 32 amu. An unknown gas has an effusion rate (v X) of 5 mL per second. Determine the formula weight (M X) of the unknown gas. (Note that the equation can be modified as follows.) Answer: In another situation involving the use of Graham's Law, helium gas (He) has a fairly high velocity of effusion, 50 mL per second, and has a formula weight of 4 amu
eBook - PDF- Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson(Authors)
- 2015(Publication Date)
- Openstax(Publisher)
The Kinetic-Molecular Theory Explains the Behavior of Gases, Part II According to Graham’s law, the molecules of a gas are in rapid motion and the molecules themselves are small. The average distance between the molecules of a gas is large compared to the size of the molecules. As a consequence, gas molecules can move past each other easily and diffuse at relatively fast rates. The rate of effusion of a gas depends directly on the (average) speed of its molecules: effusion ate ∝ u rms Using this relation, and the equation relating molecular speed to mass, Graham’s law may be easily derived as shown here: u rms = 3RT m m = 3RT u rms 2 = 3RT u ¯ 2 effusion ate A effusion ate B = u rms A u rms B = 3RT m A 3RT m B = m B m A The ratio of the rates of effusion is thus derived to be inversely proportional to the ratio of the square roots of their masses. This is the same relation observed experimentally and expressed as Graham’s law. Link to Learning 498 Chapter 9 | Gases This OpenStax book is available for free at http://cnx.org/content/col11760/1.9 9.6 Non-Ideal Gas Behavior By the end of this section, you will be able to: • Describe the physical factors that lead to deviations from ideal gas behavior • Explain how these factors are represented in the van der Waals equation • Define compressibility (Z) and describe how its variation with pressure reflects non-ideal behavior • Quantify non-ideal behavior by comparing computations of gas properties using the ideal gas law and the van der Waals equation Thus far, the ideal gas law, PV = nRT, has been applied to a variety of different types of problems, ranging from reaction stoichiometry and empirical and molecular formula problems to determining the density and molar mass of a gas. As mentioned in the previous modules of this chapter, however, the behavior of a gas is often non-ideal, meaning that the observed relationships between its pressure, volume, and temperature are not accurately described by the gas laws.
eBook - PDF- William R. Robinson, Edward J. Neth, Paul Flowers, Klaus Theopold, Richard Langley(Authors)
- 2016(Publication Date)
- Openstax(Publisher)
The Kinetic-Molecular Theory Explains the Behavior of Gases, Part II According to Graham’s law, the molecules of a gas are in rapid motion and the molecules themselves are small. The average distance between the molecules of a gas is large compared to the size of the molecules. As a consequence, gas molecules can move past each other easily and diffuse at relatively fast rates. The rate of effusion of a gas depends directly on the (average) speed of its molecules: effusion ate ∝ u rms Using this relation, and the equation relating molecular speed to mass, Graham’s law may be easily derived as shown here: u rms = 3RT m m = 3RT u rms 2 = 3RT u ¯ 2 effusion ate A effusion ate B = u rms A u rms B = 3RT m A 3RT m B = m B m A The ratio of the rates of effusion is thus derived to be inversely proportional to the ratio of the square roots of their masses. This is the same relation observed experimentally and expressed as Graham’s law. Link to Learning 440 Chapter 8 | Gases This OpenStax book is available for free at http://cnx.org/content/col12012/1.7 8.6 Non-Ideal Gas Behavior By the end of this section, you will be able to: • Describe the physical factors that lead to deviations from ideal gas behavior • Explain how these factors are represented in the van der Waals equation • Define compressibility (Z) and describe how its variation with pressure reflects non-ideal behavior • Quantify non-ideal behavior by comparing computations of gas properties using the ideal gas law and the van der Waals equation Thus far, the ideal gas law, PV = nRT, has been applied to a variety of different types of problems, ranging from reaction stoichiometry and empirical and molecular formula problems to determining the density and molar mass of a gas. As mentioned in the previous modules of this chapter, however, the behavior of a gas is often non-ideal, meaning that the observed relationships between its pressure, volume, and temperature are not accurately described by the gas laws.
eBook - ePub- Atef Korchef(Author)
- 2022(Publication Date)
- CRC Press(Publisher)
Graham’s law of effusion can be written by the following equation: R a t e A R a t e B = M B M A where Rate A and Rate B are the rates of effusion of gases A and B, respectively, and M A and M B are the molar masses of gases A and B, respectively. It is interesting to note that performing two identical effusion experiments measuring the rates of two different gases, one known and the other one unknown, allows the molar mass of the unknown gas to be determined. Practice 10.9 Calculate the ratio of the effusion rates of helium (He) and methane (CH 4) gases. Atomic masses: H = 1, He = 4, C = 12 Answer: We use Graham’s law of effusion to determine the ratio of effusion rates of He and CH 4. The molar masses of He and CH 4 are respectively, 4 g mol −1 and 16 g mol −1. We have: Rate He Rate CH 4 = M CH 4 M He Resolving gives Rate He Rate CH 4 = 16 4 = 2 This means that He diffuses two times faster than CH 4. Practice 10.10 An unknown gas (X) effuses 1.4 times faster than oxygen (O 2). What is the molar mass of this unknown gas? Atomic mass: O = 16 Answer: We use Graham’s law of effusion to determine the molar mass (M X) of the unknown gas X. We have: Rate X Rate O 2 = M O 2 M X Therefore, we have Rate X Rate O 2 2 = M O 2 M X Rearranging gives M X = M O 2 Rate X Rate O 2 2 Resolving gives M X = 32 1.4 2 = 16 g mol − 1. 10.11 Dalton’s Law of Partial Pressures In a mixture of non-reacting gases, each gas behaves as if it is the only gas present, and its pressure is called the partial pressure. The partial pressure of a gas (i) in a mixture of non-reacting gases is calculated from the ideal gas law P i V = n i RT, where P i and n i are the partial pressure and the number of moles of the gas (i), respectively. Dalton’s law of partial pressures states that the total gas pressure in a container is the sum of the partial pressures of all the gases present. Indeed, the collisions of the gaseous molecules with the container define the pressure of an ideal gas
eBook - ePub- Walter Kauzmann(Author)
- 2013(Publication Date)
- Dover Publications(Publisher)
rms .Equation (2-39) explains a phenomenon discovered by Graham in 1846. Let a gas be introduced into a container which has a small hole in its wall and let the space outside the container be evacuated (see Fig. 2-3 ). The hole must be sufficiently small that the molecule undergoes no collisions with other molecules in passing through it. (This will produce the result that the molecules will leave by effusion rather than by the flow of a stream or jet of gas.)FIG. 2-3 Effusion of gases. The gas is introduced into the effusion vessel through the stopcock , which is then closed . The rate at which the gas leaks or effuses through the orifice is measured by observing the rate at which the manometer pressure changes . After the rate has been measured the effusion vessel is evacuated and refilled through the stopcock with another gas .Then the rate at which gas leaves the container through the hole should be proportional to the average molecular velocity, which in turn would be expected to be inversely proportional to the square root of the molecular weight, because of Eq. (2-39). Therefore, if the effusion rates of two different gases through a given orifice at a given temperature and pressure are r 1 and r 2 (as measured by the rate of change of the manometer in Fig. 2-3 ) and if the molecular weights of the two gases are M 1 and M 2 , we should expect to findThis relationship was observed by Graham at about the same time that the kinetic theory was beginning to be rediscovered and it provided strong evidence for its validity.The effusion process may be used to separate the components of a gaseous mixture. If a mixture is placed in a container whose walls are porous, the constituent of lowest molecular weight should be able to leak through the walls most readily. Thus the gas remaining in the container after a period of time will be enriched in the heavier species, and the gas that has leaked through the wall will be enriched in the lighter species. This phenomenon was the basis of one of the processes developed in World War II at Oak Ridge for the manufacture of the uranium isotope 235 that was used in the atomic bombing of Hiroshima.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
A form of the ideal gas law that applies when the number of moles and the pressure are constant is called Charles’ law and is given by Equation 14.4. P i V i = P f V f (14.3) V i _ T i = V f _ T f (14.4) 14.3 Kinetic Theory of Gases The distribution of particle speeds in an ideal gas at constant temperature is the Maxwell speed distribu- tion (see Figure 14.9). The kinetic theory of gases indicates that the Kelvin temperature T of an ideal gas is related to the average trans- lational kinetic energy ¯ KE of a particle, according to Equation 14.6, where υ rms is the root-mean-square speed of the particles. ¯ KE = 1 _ 2 mυ rms 2 = 3 _ 2 kT (14.6) The internal energy U of n moles of a monatomic ideal gas is given by Equation 14.7. The internal energy of any type of ideal gas (e.g., monatomic, diatomic) is proportional to its Kelvin temperature. U = 3 _ 2 nRT (14.7) 14.4 Diffusion Diffusion is the process whereby solute molecules move through a solvent from a region of higher solute concentration to a region of lower solute concentration. Fick’s law of diffusion states that the mass m of solute that diffuses in a time t through the solvent in a channel of length L and cross-sectional area A is given by Equa- tion 14.8, where ΔC is the solute concentration difference between the ends of the channel and D is the diffusion constant. m = (DA ΔC)t _ L (14.8) 434 CHAPTER 14 The Ideal Gas Law and Kinetic Theory Focus on Concepts Additional questions are available for assignment in WileyPLUS. Section 14.1 Molecular Mass, the Mole, and Avogadro’s Number 1. All but one of the following statements are true. Which one is not true? (a) A mass (in grams) equal to the molecular mass (in atomic mass units) of a pure substance contains the same number of mol- ecules, no matter what the substance is. (b) One mole of any pure substance contains the same number of molecules. (c) Ten grams of a pure substance contains twice as many molecules as five grams of the substance.
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