2 Relating temperature, pressure, volume and amount of gases Heat an aerosol can Blow up a balloon and place it in a freezer Adding 5 mols of nitrogen to a tank that already contains 10 mols of nitrogen Exploding tires on a hot summer day
3 Elements that exist as gases at 250C and 1 atmosphere
5 Physical Characteristics of Gases solidliquidgasGases 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.
7 (force = mass x acceleration) AreaPressure =(force = mass x acceleration)Units of Pressure1 pascal (Pa) = 1 N/m21 atm = 760 mmHg = 760 torr1 atm = 101,325 PaBarometer
8 1 atm = 760 mm Hg 1 atm 760 mmHg P = 749 mmHg x = 0.986 atm The pressure outside of a jet plane flying at high altitude falls considerably below standard atmospheric pressure. Therefore the air inside of the cabin must be pressurized to protect the passengers. What is the pressure in the cabin if the barometer reading is 749 mmHg?1 atm = 760 mm Hg1 atm760 mmHgP =749 mmHg x= atm
10 1 atm = 760 mm Hg = 101.325 kPa 101.325 kPa 744 mmHg x = 99.2 kPa P = The atmospheric pressure in New York City on a given day is 744 mmHg. What was the pressure in kPa?1 atm = 760 mm Hg = kPakPa760 mmHg744 mmHg xP == 99.2 kPa
11 Problem 5.14The atmospheric pressure at the summit of Mt. McKinley is 606 mmHg on a certain day. What is the pressure in atm? In kPa?
12 Manometers Used to Measure Gas Pressures closed-tubeopen-tube
14 The Gas Laws Boyle’s Law Robert Boyle (1627-1691) Relates pressure and volume at constant temperatureFor a given mass of a gas at constant temperature, the volume of the gas varied inversely with pressure.
15 Apparatus for Studying the Relationship Between Pressure and Volume of a GasAs P (h) increasesV decreases
16 Boyle’s Law P a 1/V Constant temperature P x V = k Constant amount of gasP x V = kP1 x V1 = P2 x V2
17 P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL 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 x V = constantP1 x V1 = P2 x V2P1 = 726 mmHgP2 = ?V1 = 946 mLV2 = 154 mLP1 x V1V2726 mmHg x 946 mL154 mL=P2 == 4460 mmHg
18 The Gas Laws Charles's Law Jacques Charles (1746-1823) Relates temperature and volume at constant pressureFor a given mass of a gas at constant pressure, the volume of the gas varied directly with temperature.V = b x TV1 = V2T1 T2
19 Variation in Gas Volume with Temperature at Constant Pressure As T increasesV increases
20 Variation of Gas Volume with Temperature at Constant PressureCharles’ LawV a TTemperature must bein KelvinV = constant x TV1/T1 = V2 /T2T (K) = t (0C)
21 A sample of carbon monoxide gas occupies 3. 20 L at 125 0C A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?V1 /T1 = V2 /T2V1 = 3.20 LV2 = 1.54 LT1 = KT2 = ?T1 = 125 (0C) (K) = KV2 x T1V11.54 L x K3.20 L=T2 == 192 K
22 Problem 5.24Under constant pressure conditions a sample of hydrogen gas initially at 88ºC and 9.6 L is cooled until its final volume is 3.4 L. What is the final temperature (in K)?V = b x TV1 = V2T1 T2
23 The Gas Laws Gay-Lussac's Law Joseph Gay-Lussac (1778-1850) Gay-Lussac studied the effect of temperature on the pressure of a gas at constant volumeFor a given mass of a gas at constant volume, the pressure of the gas varied directly with temperature.P1 = P2T1 T2
24 P1 /T1 = P2 /T2 P1 = 2.50 kPa P2 = 1.54 kPa T1 = 298.15 K T2 = ? A sample of nitrogen gas has a pressure of 2.50 kPa at 25 0C. At what temperature will the gas have a pressure of 1.54 kPa if the volume remains constant?P1 /T1 = P2 /T2P1 = 2.50 kPaP2 = 1.54 kPaT1 = KT2 = ?T1 = 25 (0C) (K) = KP2 x T1P11.54 kPa x K2.50 kPa=T2 == 184 K
25 Problem 5.20At 46ºC a sample of ammonia gas exerts a pressure of 5.3 atm. What is the temperature when the pressure of the gas is reduced to one-tenth of the original pressure at the same volume?
26 Problem 5.22A sample of air occupies 3.8 L when the pressure is 1.2 atm. (a) What volume (in L) does it occupy at 5010mmHg if the temperature remains constant? (b) At the original pressure and volume the temperature was 101ºC. What temperature is necessary to increase the pressure to 5010mmHg, but keep the volume constant?
27 Explain the following. State the gas law. Helium balloon pops at a party on a summer dayPredict how the size of exhaled bubbles change as a scuba diver approaches the surface of the waterAerosol cans have a warning not to put near a flame.
28 Describe the relationships of… 1. Boyle’s Law 2. Charles’s Law 3. Gay-Lussac’s Law
29 Avogadro’s Law V a number of moles (n) V = a x n V1 / n1 = V2 / n2 Constant temperatureConstant pressureV a number of moles (n)V = a x nV1 / n1 = V2 / n2
30 4NH3 + 5O2 4NO + 6H2O 1 mole NH3 1 mole NO At constant T and P 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?4NH3 + 5O NO + 6H2O1 mole NH mole NOAt constant T and P1 volume NH volume NO
34 Ideal Gas Equation 1 Boyle’s law: P a (at constant n and T) V Charles’ law: V a T (at constant n and P)Avogadro’s law: V a n (at constant P and T)V anTPV = constant x = RnTPR is the gas constantPV = nRTR = L • atm / (mol • K)
35 When the moles are constant, The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).Calculate the volume of 1 mole of an ideal gas at STPPV = nRTR = L • atm / (mol • K)When the moles are constant,PV = nRTP1V1 = P2V2T T2Therefore
36 PV = nRT nRT V = P 1.37 mol x 0.0821 x 273.15 K V = 1 atm V = 30.7 L What is the volume (in liters) occupied by 49.8 g of HCl at STP?T = 0 0C = KP = 1 atmPV = nRTn = 49.8 g x1 mol HCl36.45 g HCl= 1.37 molV =nRTPV =1 atm1.37 mol x x KL•atmmol•KV = 30.7 L
37 Problem 5.40Calculate the volume (in L) of 88.4 g of CO2 at STP.
38 PV = nRT n, V and R are constant nR V = P T = constant P1 T1 P2 T2 = 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 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?PV = nRTn, V and R are constantnRV=PT= constantP1 = 1.20 atmT1 = 291 KP2 = ?T2 = 358 KP1T1P2T2=P2 = P1 xT2T1= 1.20 atm x358 K291 K= 1.48 atm
39 Extra Practice: Problem 5.36 The temperature of 2.5 L of a gas initially at STP is raised to 250ºC at constant volume. Calculate the final pressure of the gas in atm.
40 Extra Practice: Problem 5.32 Given that 6.9 moles of carbon monoxide are present in a container of volume 30.4 L, what is the pressure of the gas (in atm) if the temperature is 62ºC?
41 d is the density of the gas in g/L Density (d) Calculationsm is the mass of the gas in gd =mV=PMRTM is the molar mass of the gasn =mMMolar Mass (M ) of a Gaseous SubstancedRTPM =d is the density of the gas in g/L
42 dRT P M = d = m V = = 2.21 1 atm x 0.0821 x 300.15 K M = M = A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and C. What is the molar mass of the gas?dRTPM =d =mV4.65 g2.10 L== 2.21gL2.21gL1 atmx x KL•atmmol•KM =M =54.5 g/mol
43 Problem 5.44At 741 torr and 44ºC 7.10 g of a gas occupy a volume of 5.40 L. What is the molar mass of the gas (in g/mol)?
44 C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) Gas StoichiometryWhat is the volume of CO2 produced at 37 0C and 1.00 atm when 5.60 g of glucose are used up in the reaction:C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l)g C6H12O mol C6H12O mol CO V CO21 mol C6H12O6g C6H12O6x6 mol CO21 mol C6H12O6x5.60 g C6H12O6= mol CO20.187 mol x x KL•atmmol•K1.00 atm=nRTPV == 4.76 L
45 Problem 5.60Ethanol (C2H5OH) burns in air: C2H5OH(l) + O2(g) CO2(g) + H2O(l) Determine the volume of air in liters at 35.0ºC and 790 mmHg required to burn 227 g of ethanol. Assume that air is 21.0 percent O2 by volume.
46 Dalton’s Law of Partial Pressures V and T are constantP1P2Ptotal = P1 + P2Studying gaseous mixtures shows that each gasbehaves independantly of the others
47 PA = nART V PB = nBRT V XA = nA nA + nB XB = nB nA + nB PT = PA + PB Consider a case in which two gases, A and B, are in a container of volume V.PA =nARTVnA is the number of moles of APB =nBRTVnB is the number of moles of BXA =nAnA + nBXB =nBnA + nBPT = PA + PBPA = XA PTPB = XB PTPi = Xi PTmole fraction (Xi ) =ninT
48 Pi = Xi PT PT = 1.37 atm 0.116 8.24 + 0.421 + 0.116 Xpropane = A sample of natural gas contains 8.24 moles of CH4, moles of C2H6, and moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?Pi = Xi PTPT = 1.37 atm0.116Xpropane ==Ppropane = x 1.37 atm= atm
49 Problem 5.63A mixture of gases contains 0.31 mol CH4, 0.25 mol C2H6, and 0.29 mol C3H8. The total pressure is 1.50 atm. Calculate the partial pressure of each gas. = _____ = ______ = ______
50 Collecting a Gas over Water 2KClO3 (s) KCl (s) + 3O2 (g)PT = PO + PH O2
52 Problem 5.66A mixture of helium and neon gasses is collected over water at 28.0ºC and 745 mmHg. If the partial pressure of helium is 368 mmHg, what is the partial pressure of neon (in mmHg)?(vpH20 = 28.3 mmHg at 28.0ºC)
53 Problem 5.68A sample of zinc metal reacts completely with an excess of hydrochloric acid:Zn(s) + 2HCl(aq) The hydrogen gas produced is collected over water at 25.0ºC. The volume of the as is 7.80L, and the pressure is atm. Calculate the amount of zinc metal (in grams) consumed in the reaction.(vpH20 = 28.3 mmHg at 28.0ºC)