1. Gases Chapter 5 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. Topics: Gases & Pressure Gas Laws Ideal Gas Equation Molecular Weight/Density Gas Stoichiometry Gas mixtures Kinetic molecular theory Deviations from Ideal

3. Substances that exist as gases Elements that exist as gases at 25 0 C and 1 atmosphere 5.1

5. Gases assume the volume and shape of their containers. Gases are the most compressible state of matter. Gases will mix evenly and completely when confined to the same container. Gases have much lower densities than liquids and solids. 5.1 Physical Characteristics of Gases

6. Our Atmosphere: exerts pressure on earth more at sea level less on mountain top The air we breathe: 80% N2 20% O2 Sea level1 atm 4 miles0.5 atm 10 miles0.2 atm

7. SI Units of Pressure 1 pascal (Pa) = 1 N/m 2 [N = newton = 1 kg (m/s2)] Atmospheric pressure 1 atm = 760 mmHg = 760 torr (1 torr = 1mm Hg) 1 atm = 101,325 Pa 1 atm = 101 kPa 5.2 Barometer Pressure = Force Area (force = mass x acceleration) Pressure of Gas

8. 5.2 Manometers Used to Measure Gas Pressures Closed end P gas < 1atm Open end P gas> 1atm

9. The Gas Laws Pressure and Temperature Pressure = force/area (we will use torr, mm Hg, Pa & atm) Always use Kelvin temperature (K) K = ° C + 273 4 variables are involved: P = pressure V = volume n = # of moles T = temperature (in Kelvin)

10. The Pressure-Volume Relationship: Boyle’s Law P  1/V P x V = constant P 1 x V 1 = P 2 x V 2 If temperature T and amount of gas n does not change, pressure P is inversely proportional to volume V.

11. A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P 1 x V 1 = P 2 x V 2 P 1 = 726 mmHg V 1 = 946 mL P 2 = ? V 2 = 154 mL P 2 = P 1 x V 1 V2V2 726 mmHg x 946 mL 154 mL = = 4460 mmHg 5.3 P x V = constant

12. As T increasesV increases 5.3 Temperature-Volume Relationship: Charle’s Law

13. Variation of gas volume with temperature at constant pressure. 5.3 T (K) = t ( 0 C) + 273.15 Temperature must be in Kelvin Charles’ & Gay-Lussac’s Law V  TV  T V = constant x T V 1 /T 1 = V 2 /T 2 If n and P do NOT change, then V is proportional to T

14. What is the final temperature (in K), under constant- pressure condition, when a sample of hydrogen gas initially at 88 o C and 9.6 L is cooled until its finial volume is 3.4 L? V 1 = 9.6 L T 1 =(88+273.15) K=361.15K V 2 = 3.4 L T 2 = ? T 2 = V 2 x T 1 V1V1 3.4 L x 361.15 K 9.6 L = = 128K 5.3 V 1 /T 1 = V 2 /T 2 Group practice problem chap4-5:Q13

15. The Volume-amount Relationship: Avogadro’s Law V  number of moles (n) V = constant x n V 1 / n 1 = V 2 / n 2 5.3 If P and T do NOT change, then n is proportional to V

16. Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure? 4NH 3 + 5O 2 4NO + 6H 2 O 1 mole NH 3 1 mole NO At constant T and P 1 volume NH 3 1 volume NO 5.3

17. Ideal Gas Equation 5.4 Charles’ law: V  T  (at constant n and P) Avogadro’s law: V  n  (at constant P and T) Boyle’s law: V  (at constant n and T) 1 P V V  nT P V = constant x = R nT P P R is the gas constant PV = nRT Ideal gas is a hypothetical gas whose pressure-volume-temperature behavior can be completely accounted for by the ideal gas equation.

18. The conditions 0 0 C and 1 atm are called standard temperature and pressure (STP). PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)(273.15 K) R = 0.082057 L atm / (mol K)=8.314J/(K·mol) = 8.314 L·kPa/(K·mol) 5.4 Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. Molar volume of gas 1 mole of gas at STP = 22.4 Liters Example: 2 moles of gas at STP = 44.8 L In calculation, the units of R must match those for P,V,T and n.

19. What is the pressure of the gas (in atm) when 5.0 moles of CO gas are present in a container of 20.0 L at 27 o C? n= 5.0mole, V=20.0L, T= 27 o C=(27+273.15)K=300.15K PV=nRT P=nRT/V = 5.0mole*0.082 L atm / (mol K)*300.15K/20.0L =6.15L Group practice problem chap4-5:Q15

20. What is the volume (in liters) occupied by 49.8 g of HCl at STP? PV = nRT V = nRT P T = 0 0 C = 273.15 K P = 1 atm n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol V = 1 atm 1.37 mol x 0.0821 x 273.15 K Latm molK V = 30.6 L 5.4 Practice problem of chap4-5: Q14

21. Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR V = P T = constant P1P1 T1T1 P2P2 T2T2 = P 1 = 1.20 atm T 1 = 291 K P 2 = ? T 2 = 358 K P 2 = P 1 x T2T2 T1T1 = 1.20 atm x 358 K 291 K = 1.48 atm 5.4

22. P 1 V 1 n 1 T 1 R= The combined gas law P 2 V 2 n 2 T 2 P 1 V 1 n 1 T 1 = P 2 V 2 n 2 T 2 R= If n1=n2 P 2 V 2 T 2 P 1 V 1 T 1 =

23. Density (d) Calculations d = m V = PMPM RT m is the mass of the gas in g M is the molar mass of the gas Molar Mass ( M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L 5.4

24. What is the density of HCl gas in grams per liter at 700 mmHg and 25 o C? d = m V = PMPM RT P=700mmHg=700/760atm=0.92atm T= 25 o C=25+273.15K=298.15K d=d= 0.92 atm x 36.45g/mol x 298.15 K 0.0821 Latm molK =1.37g/L Group practice problem chap4-5:Q18

25. What is the molar mass (g/mol) of 7.10 grams of gas whose volume is 5.40 L at 741 torr and 40 o C? 5.4 dRT P M = d = m V 7.10 g 5.40 L = = 1.31 g L M = 1.31 g L 0.975 atm x 0.0821 x 313.15 K Latm molK M = 34.6 g/mol T=313.15KP= 741torr=741/760atm=0.975atm Group practice problem chap4-5:Q17

26. Gas Stoichiometry The combustion process for methane is CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(l) If 15.0 moles of methane are reacted, what is the volume of carbon dioxide (in L) produced at 23.0 o C and 0.985 atm? V = nRT P 15mol x 0.0821 x 296.15 K Latm molK 0.985 atm = = 369.8 L 5.5 x 1CO2/1CH 4 15 mole CH 4 ----------------  15 mole CO2 Group practice problem chap4-5:Q20

27. Dalton’s Law of Partial Pressures Partial pressure is the pressure of the individual gas in the mixture. V and T are constant P1P1 P2P2 P total = P 1 + P 2 5.6

28. Consider a case in which two gases, A and B, are in a container of volume V. P A = n A RT V P B = n B RT V n A is the number of moles of A n B is the number of moles of B P T = P A + P B X A = nAnA n A + n B X B = nBnB n A + n B P A = X A P T P B = X B P T P i = X i P T 5.6 mole fraction (X i ) = nini nTnT

29. A sample of natural gas contains 8.24 moles of CH 4, 0.421 moles of C 2 H 6, and 0.116 moles of C 3 H 8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C 3 H 8 )? P i = X i P T X propane = 0.116 8.24 + 0.421 + 0.116 P T = 1.37 atm = 0.0132 P propane = 0.0132 x 1.37 atm= 0.0181 atm 5.6 Group practice problem chap4-5:Q21

30. Kinetic Molecular Theory of Gases 1.A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume. 2.Gas molecules are in constant motion in random directions, and they frequently collide with one another. Collisions among molecules are perfectly elastic. 3.Gas molecules exert neither attractive nor repulsive forces on one another. 4.The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy 5.7 KE = ½ mu 2 u 2 = (u 1 2 + u 2 2 + …+ u N 2 )/N KE  T Mean square speed

31. Application to the Gas Laws Compressibility of Gases (assumption 1) Boyle’s Law P  collision rate with wall Collision rate  number density (number of molecules per unit volume) Number density  1/V P  1/V Charles’ Law P  collision rate with wall Collision rate  average kinetic energy of gas molecules Average kinetic energy  T P  T  expand volume until gas pressure is balanced by the external constant pressure 5.7

32. Avogadro’s Law P  collision rate with wall Collision rate  number density Number density  n P  n Dalton’s Law of Partial Pressures Molecules do not attract or repel one another P exerted by one type of molecule is unaffected by the presence of another gas P total =  P i 5.7 Application to the Gas Laws

33. Distribution of molecular speeds 5.7 At constant temperature, the average kinetic energy and the mean square speed will remain unchanged. The motion of molecules are random. How many molecules are moving at a particular speed?

34. The distribution of speeds for nitrogen gas molecules at three different temperatures The distribution of speeds of three different gases at the same temperature 5.7 Maxwell speed distribution curve

35. Root-mean-square speed Total KE of one mole gas = (½ mu 2 )N A = 3 / 2 RT u rms = 3RT M  N A m= M

37. Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. 5.7 NH 3 17 g/mol HCl 36 g/mol NH 4 Cl r1r1 r2r2 M2M2 M1M1  = Under the same conditions of T and P, rates of diffusion for gases are inversely proportional to the square roots of their molar masses. Gramham’s Law

38. Gas effusion is the is the process by which gas under pressure escapes from one compartment of a container to another by passing through a small opening. 5.7 r1r1 r2r2 t2t2 t1t1 M2M2 M1M1  == Nickel forms a gaseous compound of the formula Ni(CO) x.What is the value of x given that under the same conditions methane (CH 4 ) effuses 3.3 times faster than the compound? r 1 = 3.3 x r 2 M 1 = 16 g/mol M 2 = r1r1 r2r2 ( ) 2 x M 1 = (3.3) 2 x 16 = 174.2 58.7 + x 28 = 174.2 x = 4.1 ~ 4 Group practice problems chap4-5:Q22

40. Deviations from Ideal Behavior 1 mole of ideal gas PV = nRT n = PV RT = 1.0 5.8 Repulsive Forces Attractive Forces Assumption of gas laws and kinetic theory: Gas molecules do not exert any force, either attractive or repulsive; The volume of the molecules are negligible;

41. Effect of intermolecular forces on the pressure exerted by a gas. 5.8 The speed of the gas molecule that is moving toward the wall is reduced by the attractive force of neighbor molecules. Consequently the measured pressure is lower than the pressure the gas would exert if it behave ideally.

42. 5.8 Van der Waals equation nonideal gas P + (V – nb) = nRT an 2 V2V2 () } corrected pressure } corrected volume