Unit 8: Gas Laws

Comparison of Solids, Liquids, & Gases The density of gases is much less than that of solids or liquids. Gases are easily compressed and they completely fill any container in which they occupy Tells us that gas molecules are far apart & interactions among them are weak Gases exert pressure on their surroundings; in turn, pressure must be exerted to confine a gas Gases diffuse into one another (i.e. they are miscible, unless they react with one another) Mix completely e.g. air is a mixture of gases Conversely, different gases in a mixture do not separate on standing

Pressure Pressure is force per unit area. lb/in2 commonly known as psi Atmospheric pressure is measured using a barometer. Definitions of standard pressure 1 atmosphere 760 mm Hg 76 cm Hg 760 torr 101.3 kPa Mercury barometer: the air pressure is measured in terms of the height of the mercury column i.e. the vertical distance between the surface of the mercury in the open dish and that inside the closed tube. The pressure exerted by the atmosphere is the pressure exerted by the column of mercury Hg density = 13.6 g/mL

Boyle’s Law: Inverse relationship of pressure & volume 25°C 10 L p = 1 atm 5 L 2.5 L p = 2 atm p = 4 atm Pressure doubled Volume reduced by half

Boyle’s Law: The Volume-Pressure Relationship Graphical Representation of Boyle’s Law At constant temp and with equal # mols, volume of a gas varies inversely with pressure:- V α 1  V = 1 k1  PV = k1 P P The product of volume (V) and pressure (P) is a constant This allows us to equate initial and final conditions to solve practical problems. Fig. 12-4, p. 405

Boyle’s Law: The Volume-Pressure Relationship Used to calculate:- Volume resulting from pressure change Pressure resulting from volume change P1V1 = k1 for one sample of a gas. P2V2 = k2 for a second sample of a gas. k1 = k2 for the same sample of a gas at the same T. Thus we can write Boyle’s Law mathematically as: P1V1 = P2V2

Boyle’s Law: The Volume-Pressure Relationship Example 1: At 25oC a sample of He has a volume of 4.00 x 102 mL under a pressure of 7.60 x 102 torr. What volume would it occupy under a pressure of 2.00 atm at the same T?

Boyle’s Law: The Volume-Pressure Relationship Notice that in Boyle’s law we can use any pressure or volume units as long as we consistently use the same units for both P1 and P2 or V1 and V2. Use your intuition to help you decide if the volume will go up or down as the pressure is changed and vice versa.

Charles’ Law: The Volume-Temperature Relationship p = 1 atm 1 L 2 L 4 L Temperature (K) doubled Volume doubles 273 K 546 K 1092 K

Charles’ Law: The Volume-Temperature Relationship Charles’s law states that the volume of a gas is directly proportional to the absolute temperature at constant pressure. Gas laws must use the Kelvin scale to be correct. Relationship between Kelvin and centigrade.

Charles’ Law: The Volume-Temperature Relationship Volume of a gas varies directly with temperature if the pressure and # of mols of gas remain constant V α T  V = Tk2  V = k2 T Ratio of volume (V) and temperature (T) is a constant This allows us to equate initial and final conditions to solve practical problems

Charles’ Law: The Volume-Temperature Relationship Example 2: A sample of hydrogen, H2, occupies 1.00 x 102 mL at 25.0oC and 1.00 atm. What volume would it occupy at 50.0oC under the same pressure? T1 = 25 + 273 = 298 T2 = 50 + 273 = 323

Standard Temperature and Pressure Have seen that both temperature & pressure affect the volume of a gas Often convenient to choose some “standard” temperature and pressure as a reference point for discussing gasses. Standard temperature and pressure is given the symbol STP. Standard P  1 atm or 101.3 kPa Standard T  273 K or 0 oC

The Combined Gas Law Equation Boyle’s and Charles’ Laws combined into one statement is called the combined gas law equation. Useful when the V, T, and P of a gas are changing.

The Combined Gas Law Equation Example 3: A sample of nitrogen gas, N2, occupies 7.50 x 102 mL at 75.00C under a pressure of 8.10 x 102 torr. What volume would it occupy at STP?

Avogadro’s Law Mathematically Avogadro’s law can be stated as: At constant temperature & pressure, the volume, V, occupies by a gas is directly proportional to the number of mols, n, of gas V α n or V = kn or V = k (constant temp & pressure) n For 2 samples of a gas at the same temperature & pressure, the relation between volumes and number of moles can be represented as: V1 = V2 (constant T , P) n1 n2

Using Avogadro’s Law Example 4: If 5.50 mol of CO occupy 20.6 L, how many liters will 16.5 mol of CO occupy at the same temperature and pressure? What do we know? n1 = 5.50 mol n2 = 16.5 mol V1 = 20.6 L V2 = ? L V2 = V1n2 = (20.6 L)(16.5 mol) n1 (5.50 mol) = 61.8 L CO

Avogadro’s Law and the Standard Molar Volume Avogadro’s Law states that at the same temperature and pressure, equal volumes of two gases contain the same number of molecules (or moles) of gas. If we set the temperature and pressure for any gas to be STP, then one mole of that gas has a volume called the standard molar volume. The standard molar volume is 22.4 L at STP. This is another way to measure moles. For gases, the volume is proportional to the number of moles. 11.2 L of a gas at STP = 0.500 mole 44.8 L = ? moles

Summary of Gas Laws: The Ideal Gas Law Boyle’s Law - V  1/P (at constant T & n) Charles’ Law – V  T (at constant P & n) Avogadro’s Law – V  n (at constant T & P) Combine these three laws into one statement V  nT/P Convert the proportionality into an equality. V = nRT/P This provides the Ideal Gas Law. PV = nRT R = 0.0821 L . atm / mol . K It is a proportionality constant called the universal gas constant. An ideal gas is one that exactly obeys these gas laws. Many gases show slight deviations from ideality, but at normal temperatures, & pressures the deviations are usually small enough to ignore

Summary of Gas Laws: The Ideal Gas Law Example 5: What volume would 50.0 g of ethane, C2H6, occupy at 1.40 x 102 oC under a pressure of 1.82 x 103 torr? To use the ideal gas law correctly, it is very important that all of your values be in the correct units! T = 140 + 273 = 413 K P = 1820 torr (1 atm/760 torr) = 2.39 atm 50 g (1 mol/30 g) = 1.67 mol

Summary of Gas Laws: The Ideal Gas Law PV = nRT 

Dalton’s Law of Partial Pressures An illustration of Dalton’s law. When the 2 gases A and B are mixed in the same container at the same temperature, they exert a total pressure equal to the sum of their partial pressures.

Dalton’s Law of Partial Pressures Dalton’s law states that the pressure exerted by a mixture of gases is the sum of the partial pressures of the individual gases. Ptotal = PA + PB + PC + .... Where: PA = nA RT PB = nBRT PC = nCRT etc V V V The pressure that each gas exerts in a mixture is called its partial pressure. No way has been devised to measure the pressure of an individual gas in a mixture; it must be calculated from other quantities.

Dalton’s Law of Partial Pressures Example 6: If 100 mL of hydrogen, measured at 25.0 oC and 3.00 atm pressure, and 100 mL of oxygen, measured at 25.0 oC and 2.00 atm pressure, were forced into one of the containers at 25.0 oC, what would be the pressure of the mixture of gases?

Dalton’s Law of Partial Pressures Example 7: A 10.0 L flask contains 0.200 mol of methane, 0.300 mol of hydrogen and 0.400 mol of nitrogen at 250C (a) What is the pressure (in atm) inside the flask? (b) What is the partial pressure of each component of the mixture of gases? (a) Solution: Given # mols of each component. Can use ideal gas equation to find total pressure from # mols: ntotal = 0.200 + 0.300 + 0.400 = 0.900 mols V= 10.0L T = 25 + 273 = 298K Ptotal = nRT = (0.900mol)(0.0821L.atm / mol.K) (298K) = 2.20atm V 10.0L

Dalton’s Law of Partial Pressures (b) Solution: Now we find partial pressures for each component: PCH4 = (nCH4)RT = (0.200)(0.0821)(298) = 0.489 atm V 10.0 Similar calculations for N2 and H2 give: PH2 = 0.734 atm PN2 = 0.979 atm As a check, the sum of all the partial pressures should be equal to the total pressure (Dalton’s Law): 0.489 + 0.743 + 0.979 = 2.20 atm Problem solving tip: Sometimes the amount of gas is expressed in other units that can be converted to mols. E.g. molar mass. Can then convert mass  mols

Mole Fraction and Partial Pressure Can describe the composition of any gas mixture in terms of the mole fraction of each component . The mole fraction, XA, of component A in a mixture is defined as: XA = no. mol A total no. mol of all components Like any other fraction, mole fraction is dimensionless. Can relate the mole fraction of each component to its partial pressure

Mole Fraction and Partial Pressure XA = PA similarly, XB = PB and so on Ptotal Ptotal We can rearrange these equations to give another statement of Dalton’s Law of Partial Pressures: PA = XA Ptotal PB = XB Ptotal and so on The partial pressure of each gas is equal to the mole fraction in the gas mixture times the toal pressure of the mixture

Mole Fraction and Partial Pressure Example 14: Find the mole fractions of the gases in example 13. A 10.0 L flask contains 0.200 mol of methane, 0.300 mol of hydrogen and 0.400 mol of nitrogen at 250C Solution, using the mols given: X methane = nmethane / ntotal = 0.200 /0.900 = 0.222 X hydrogen = nhydrogen / ntotal = 0.300/ 0.900 = 0.333 X nitrogen = nnitrogen / ntotal = 0.400/ 0.900 = 0.444 Alternatively, could use partial & total pressure in Example 13 (b): X methane = Pmethane / P total = 0.489 atm / 2.20 atm = 0.222 X hydrogen = Phydrogen / P total = 0.734 atm / 2.20atm = 0.334 X nitrogen = Pnitrogen / P total = 0.979 atm / 2.20 atm = 0.445

The Kinetic-Molecular Theory The basic assumptions of the kinetic-molecular theory for an ideal gas (one that obeys the gas laws):- Gases consist of discrete molecules which are small and very far apart relative to their own size, and occupy no volume (they can be considered as points) The observation that gases can be easily compressed supports this.

The Kinetic-Molecular Theory The gas molecules are in continuous random, straight-line motion with varying speeds. The collisions between gas molecules and with the walls of the container are elastic, (that is, no energy is gained or lost during the collision). At any given instant only a small fraction of the gas molecules are involved in collisions.

The Kinetic-Molecular Theory There are no attractive or repulsive forces between molecules. NOTE: at HIGH pressures and LOW temperatures (conditions under which a gas liquefies attractions and repulsions between gas molecules become significant and the gas behaves non-ideally. A real gas is one that does not behave as an ideal gas due to interactions between gas molecules.