1. The Gas State  Gases are everywhere – atmosphere, environmental processes, industrial processes, bodily functions  Gases have unique properties from liquids and solids  Gases are compressible (very important for storage)  Gas particles are widely separated and move at very fast speeds  Most gases have relatively low densities  Gas have relatively low viscosity (resistance to movement) allowing them move freely through pipes and even small orifices 111/9/2015

2. The Gas State  Chemical behavior of gases depends on composition  Physical behavior of all gases is similar  Gases are miscible mixing together in any proportion  Behavior of gases described by ideal gas law and kinetic-molecular theory, the cornerstone of this chapter  Gas volume changes greatly with pressure  Gas volume changes greatly with temperature  Gas volume is a function of the amount of gas 11/9/20152

3. Cylinders of Gas 11/9/20153

4. The Empirical Gas Laws  Gas behavior can be described by pressure, temperature, volume, and molar amount  Holding any two constant allows relations between the other two  Boyle’s Law : The volume of a sample of gas at a given temperature varies inversely with the applied pressure Vα 1/P PV= constant P 1 V 1 =P 2 V 2 11/9/20154

5. The Empirical Gas Laws Boyle’s Law Gas Pressure-Volume Relationship 5

6. Practice Problem Boyle’s Law A sample of chlorine gas has a volume of 1.8 L at 1.0 atm. If the pressure increases to 4.0 atm (at constant temperature), what would be the new volume? using P 1 V 1 =P 2 V 2 V 2 = P 1 V 1/ P 2 = (1.0 atm) × (1.8 L) / (4.0 atm) V 2 = 0.45 L 11/9/20156

7. The Empirical Gas Laws Charles’s Law The volume occupied by any sample of gas at constant pressure is directly proportional to its absolute temperature. Vα T abs Assumes constant moles and pressure Temperature on absolute scale ( o C + 273.15) V/T = constant V 1 /T 1 = V 2 /T 2 11/9/20157 (T abs (K) = o C + 273.15)

8. The Empirical Gas Laws Charles’s Law: Linear Relationship of Gas Volume and Temperature at Constant Pressure 11/9/20158 V = at + b 0 = a(-273.15) + b b = 273.15b V = at + 273.15a = a(t + 273.15)

9. Practice Problem Charles’s Law A sample of methane gas that has a volume of 3.8 L at 5.0°C is heated to 86.0°C at constant pressure. Calculate its new volume. using V 1 /T 1 = V 2 /T 2 Convert temperature (°C ) to absolute (kelvin) 50 °C = 5.0 +273.15 = 278.15 K 86.0 °C = 86.0 + 273.15 = 359.15 K V 2 = (T 2 /T 1 ) V 1 = (359.15 K /278.15 K ) × 3.8 L V2 = 4.9 L 11/9/20159

10. 10 The Combined Gas Law Combined Gas Law: In the event that all three parameters, P, V, and T, are changing, their combined relationship is defined as follows assuming the mass of the gas (number of moles) is constant. V 1 / V 2 = T 1 / T 2 (fixed P,n) P 1 V 1 = P 2 V 2 (fixed T,n) Boyle’s Law Charles’ Law V x P = const V / T = const 1662 1787 P 1 V 1 /T 1 = P 2 V 2 /T 2

11. Avagadro’s Law Avogadro’s Law  The volume of a sample of gas is directly proportional to the number of moles of gas, n Vα n V/n = constant  Equal volumes of any two gases at the same temperature and pressure contain the same number of molecules (mol) V 1 / n 1 = V 2 / n 2 11/9/201511

12. Avagadro’s Law Avogadro’s Law  The volume of one mole of gas is called the: molar gas volume, Vm  Volumes of gases are often compared at standard temperature and pressure (STP), chosen to be  0 oC (273.15 oK) and 1 atm pressure  At STP, the molar volume, Vm, that is, the volume occupied by one mole of any gas, is 22.4 L/mol V STP / N STP = V m = 22.4 L ( at STP ) 11/9/2015 12

13. The Ideal Gas Law From the empirical gas laws, we see that volume varies in proportion to pressure, absolute temperature, and the mass of gas (moles) present Boyle law Vα 1/P V= constant x 1/P Charles law Vα T abs V = constant x T Avogadros law Vα n V = constant x n 11/9/201513

14. The Ideal Gas Law This implies that there must exist a proportionality constant governing these relationships Combining the three proportionalities, we can obtain the following relationship. V m = “R” x n x (T abs/P) where “R” is the proportionality constant referred to as the Ideal Gas Constant, which relates Molar Volume (V) to the ratio of Temperature to Pressure T/P 11/9/201514

15. The Ideal Gas Law The ideal gas equation is usually expressed in the following form: PV = nRT 11/9/201515 P = Pressure (in atm) V=Volume (in liters) n = Number of atoms (in moles) R=Universal gas constant 0.0821 L.atm/mol.K T=Temperature (in 0 Kelvin = °C + 273.15)

16. The Ideal Gas Law What is R, universal gas constant? the R is independent of the particular gas studied PV = nRT

17. Practice Problem A steel tank has a volume of 438 L and is filled with 0.885 kg of O 2. Calculate the pressure of oxygen in the tank at 21 o C use PV = nRT V = 438 L, R = 0.0821 L.atm/mol.K, T = 21 + 273.15 = 294.15 K, n = 885/32 = 27.7 mol. So the pressure = nRT/V = 27.7 mol x 0.0821 L.atm/mol.K x 294.15 K/483 L = 1.53 atm

18. Mixtures of Gases Dalton’s Law of Partial Pressures The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases.

19. The Ideal Gas Law is an empirical relationship based on experimental observations. Boyle, Charles and Avogadro. Kinetic Molecular Theory is a simple model that attempts to explain the behavior of gases. The Kinetic Molecular Theory of Gases

20. 1. A pure gas consists of a large number of identical molecules separated by distances that are large compared with their size. The volumes of the individual particles can be assumed to be negligible (zero). 2. The molecules of a gas are constantly moving in random directions with a distribution of speeds. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas. 3. The molecules of a gas exert no forces on one another except during collisions, so that between collisions they move in straight lines with constant velocities. The gases are assumed to neither attract or repel each other. The collisions of the molecules with each other and with the walls of the container are elastic; no energy is lost during a collision. 4. The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas.

21. Diffusion: is the transfer of a gas through space or another.gas over time Effusion: Is the process in which individual molecules flow through a hole without collisions between molecules.

22. According to Graham's law, the rate at which gases effuse (i.e., how many molecules pass through the hole per second) is dependent on their molecular weight; gases with a lower molecular weight effuse more quickly than gases with a higher molecular weight. The equation for effusion is given as Where M 1 and M 2 are molecular masses of gases 1 and 2.

23. Real Gases: Deviations from Ideality  Real gases behave ideally at ordinary temperatures and pressures.  At low temperatures and high pressures real gases do not behave ideally.  The reasons for the deviations from ideality are: 1.The molecules are very close to one another, thus their volume is important. 2.The molecular interactions also become important.

24. Real Gases: Deviations from Ideality  van der Waals’ equation accounts for the behavior of real gases at low temperatures and high pressures.  The van der Waals constants a and b take into account two things: a accounts for intermolecular attraction For nonpolar gases the attractive forces are London Forces For polar gases the attractive forces are dipole-dipole attractions or hydrogen bonds. b accounts for volume of gas molecules  At large volumes a and b are relatively small and van der Waal’s equation reduces to ideal gas law at high temperatures and low pressures.

25. Real Gases: Deviations from Ideality  Calculate the pressure exerted by 84.0 g of ammonia, NH 3, in a 5.00 L container at 200. o C using the ideal gas law. PV = nRT P = nRT/V n = 84.0g * 1mol/17 g T = 200 + 273 P = (4.94mol)(0.08206 L atm mol -1 K -1 )(473 K) (5 L) P = 38.3 atm

26. Real Gases: Deviations from Ideality  Calculate the pressure exerted by 84.0 g of ammonia, NH 3, in a 5.00 L container at 200. o C using the van der Waal’s equation. The van der Waal's constants for ammonia are: a = 4.17 atm L 2 mol -2 b = 3.71x10 -2 L mol -1 n = 84.0g * 1mol/17 g T = 200 + 273 P = (4.94mol)(0.08206 L atm mol -1 K -1 )(473K) (4.94 mol) 2 *4.17 atm L 2 mol -2 5 L – (4.94 mol*3.71 E -2 L mol -1 ) (5 L) 2 P = 39.81 atm – 4.07 atm = 35.74 P = 38.3 atm 7% error