Gas Laws What to do when conditions are ideal

Boyle’s Law What was the relationship between pressure and volume? When P Then V Algebraically this is written as P=k/V When you solved for k; PV=k Therefore P 1 V 1 =P 2 V 2

Gas Laws – Boyle’s Law Constant: Temperature Relationship: Pressure is inversely proportional to volume Pressure a 1/volume Written As: P 1 V 1 = P 2 V 2 Pressure is typically in atm or torr

Charles’ Law What was the relationship between Temperature and Volume? When T Then V Algebraically this is written as V=kT When you solved for k; V/T=k Therefore V1/T1=V2/T2

Gas Laws – Charles’ Law Constant: Pressure Relationship: Temperature is directly proportional to volume Temp a Volume Written As: V 1 /T 1 = V 2 /T 2

Charles’ Law What unit of measure is needed for these calculations? C or K? Temperature is in K (K = C)

Gas Laws – Gay-Lussac’s Law Constant: Volume Relationship: Pressure is directly proportional to temperature Pressure a Temperature Written As: P 1 /T 1 = P 2 /T 2

The IDEAL GAS Law –this is what we will use When we put all three laws together: PV a nT (n= number of moles) PV = nRT (R= ideal gas law constant) R=62.4 L torr/K mol or L atm/K mol

Lecture PLUS Timberlake Ideal Gases Behave as described by the ideal gas equation; no real gas is actually ideal Within a few %, ideal gas equation describes most real gases at room temperature and pressures of 1 atm or less In real gases, particles attract each other reducing the pressure Real gases behave more like ideal gases as pressure approaches zero.

Lecture PLUS Timberlake PV = nRT R is known as the universal gas constant Using STP conditions P V R = PV = (1.00 atm)(22.4 L) nT (1mol) (273K) n T = L-atm mol-K

LecturePLUS Timberlake 11 Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 Isolate V 2 P 1 V 1 T 2 = P 2 V 2 T 1 V 2 = P 1 V 1 T 2 P 2 T 1

Lecture PLUS Timberlake Learning Check G15 What is the value of R when the STP value for P is 760 mmHg?

Lecture PLUS Timberlake Solution G15 What is the value of R when the STP value for P is 760 mmHg? R = PV = (760 mm Hg) (22.4 L) nT (1mol) (273K) = 62.4 L-mm Hg mol-K

Lecture PLUS Timberlake Learning Check G16 Dinitrogen monoxide (N 2 O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mmHg) in the tank in the dentist office?

Lecture PLUS Timberlake Solution G16 Set up data for 3 of the 4 gas variables Adjust to match the units of R V = 20.0 L20.0 L T = 23°C K n = 2.86 mol2.86 mol P = ? ?

Lecture PLUS Timberlake Rearrange ideal gas law for unknown P P = nRT V Substitute values of n, R, T and V and solve for P P = (2.86 mol)(62.4L-mmHg)(296 K) (20.0 L) (K-mol) = 2.64 x 10 3 mm Hg

Lecture PLUS Timberlake Learning Check G17 A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?

Lecture PLUS Timberlake Solution G17 Solve ideal gas equation for n (moles) n = PV RT = (735 mmHg)(5.0 L)(mol K) (62.4 mmHg L)(293 K) = mol O 2 x 32.0 g O 2 = 6.4 g O 2 1 mol O 2

LecturePLUS Timberlake 19 Learning Check C1 Solve the combined gas laws for T 2.

LecturePLUS Timberlake 20 Solution C1 Solve the combined gas law for T 2. (Hint: cross-multiply first.) P 1 V 1 = P 2 V 2 T 1 T 2 P 1 V 1 T 2 = P 2 V 2 T 1 T 2 = P 2 V 2 T 1 P 1 V 1

LecturePLUS Timberlake 21 Combined Gas Law Problem A sample of helium gas has a volume of L, a pressure of atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

LecturePLUS Timberlake 22 Data Table Set up Data Table P 1 = atm V 1 = L T 1 = 302 K P 2 = 3.20 atm V 2 = 90.0 mL T 2 = ?? ??

LecturePLUS Timberlake 23 Solution Solve for T 2 Enter data T 2 = 302 K x atm x mL = K atm mL T 2 = K = °C

LecturePLUS Timberlake 24 Calculation Solve for T 2 T 2 = 302 K x 3.20 atm x 90.0 mL = 604 K atm mL T 2 = 604 K = 331 °C

LecturePLUS Timberlake 25 Learning Check C2 A gas has a volume of 675 mL at 35°C and atm pressure. What is the temperature in °C when the gas has a volume of L and a pressure of 802 mm Hg?

LecturePLUS Timberlake 26 Solution G9 T 1 = 308 KT 2 = ? V 1 = 675 mLV 2 = L = 315 mL P 1 = atm P 2 = 802 mm Hg = 646 mm Hg T 2 = 308 K x 802 mm Hg x 315 mL 646 mm Hg 675 mL P inc, T inc V dec, T dec = 178 K = - 95°C

LecturePLUS Timberlake 27 Volume and Moles How does adding more molecules of a gas change the volume of the air in a tire? If a tire has a leak, how does the loss of air (gas) molecules change the volume?

LecturePLUS Timberlake 28 Learning Check C3 True (1) or False(2) 1.___The P exerted by a gas at constant V is not affected by the T of the gas. 2.___ At constant P, the V of a gas is directly proportional to the absolute T 3.___ At constant T, doubling the P will cause the V of the gas sample to decrease to one-half its original V.

LecturePLUS Timberlake 29 Solution C3 True (1) or False(2) 1. (2)The P exerted by a gas at constant V is not affected by the T of the gas. 2. (1) At constant P, the V of a gas is directly proportional to the absolute T 3. (1) At constant T, doubling the P will cause the V of the gas sample to decrease to one-half its original V.

LecturePLUS Timberlake 30 Avogadro’s Law When a gas is at constant T and P, the V is directly proportional to the number of moles (n) of gas V 1 = V 2 n 1 n 2 initial final

LecturePLUS Timberlake 31 STP The volumes of gases can be compared when they have the same temperature and pressure (STP). Standard temperature 0°C or 273 K Standard pressure 1 atm (760 mm Hg)

LecturePLUS Timberlake 32 Learning Check C4 A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C? P 1 = V 1 = T 1 = K P 2 = V 2 = ?? T 2 = K V 2 = 15 L x atm x K = 6.8 L atm K

LecturePLUS Timberlake 33 Solution C4 P 1 = 1.0 atm V 1 = 15 L T 1 = 273 K P 2 = 2.0 atm V 2 = ?? T 2 = 248 K V 2 = 15 L x 1.0 atm x 248 K = 6.8 L 2.0 atm 273 K

LecturePLUS Timberlake 34 Molar Volume At STP 4.0 g He 16.0 g CH g CO 2 1 mole 1 mole1mole (STP) (STP)(STP) V = 22.4 L V = 22.4 L V = 22.4 L

LecturePLUS Timberlake 35 Molar Volume Factor 1 mole of a gas at STP = 22.4 L 22.4 L and 1 mole 1 mole 22.4 L

LecturePLUS Timberlake 36 Learning Check C5 A.What is the volume at STP of 4.00 g of CH 4 ? 1) 5.60 L2) 11.2 L3) 44.8 L B. How many grams of He are present in 8.0 L of gas at STP? 1) 25.6 g2) g3) 1.43 g

LecturePLUS Timberlake 37 Solution C5 A.What is the volume at STP of 4.00 g of CH 4 ? 4.00 g CH 4 x 1 mole CH 4 x 22.4 L (STP) = 5.60 L 16.0 g CH 4 1 mole CH 4 B. How many grams of He are present in 8.0 L of gas at STP? 8.00 L x 1 mole He x 4.00 g He = 1.43 g He 22.4 He 1 mole He

LecturePLUS Timberlake 38 Daltons’ Law of Partial Pressures Partial Pressure Pressure each gas in a mixture would exert if it were the only gas in the container Dalton's Law of Partial Pressures The total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture. P T = P 1 + P 2 + P

LecturePLUS Timberlake 39 Gases in the Air The % of gases in air Partial pressure (STP) 78.08% N mmHg 20.95% O mmHg 0.94% Ar 7.1 mmHg 0.03% CO mmHg P AIR = P N + P O + P Ar + P CO = 760 mmHg Total Pressure760 mm Hg

LecturePLUS Timberlake 40 Learning Check C6 A. If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O 2 in the air? 1) ) 156 3) 760 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N 2 in the air? 1) 557 2) 9.143) 0.109

LecturePLUS Timberlake 41 Solution C6 A. If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O 2 in the air? 2) 156 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N 2 in the air? 1) 557

LecturePLUS Timberlake 42 Partial Pressures The total pressure of a gas mixture depends on the total number of gas particles, not on the types of particles. P = 1.00 atm 0.5 mole O mole He mole Ar 1 mole H 2

LecturePLUS Timberlake 43 Health Note When a scuba diver is several hundred feet under water, the high pressures cause N 2 from the tank air to dissolve in the blood. If the diver rises too fast, the dissolved N 2 will form bubbles in the blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O 2 in scuba tanks used for deep descents.

LecturePLUS Timberlake 44 Learning Check C7 A 5.00 L scuba tank contains 1.05 mole of O 2 and mole He at 25°C. What is the partial pressure of each gas, and what is the total pressure in the tank?

LecturePLUS Timberlake 45 Solution C7 P = nRT P T = P O + P He V 2 P T = 1.47 mol x L-atm x 298 K 5.00 L(K mol) =7.19 atm

Lecture PLUS Timberlake Molar Mass of a gas What is the molar mass of a gas if g of the gas occupy 215 mL at atm and 30.0°C? n = PV = (0.813 atm) (0.215 L) = mol RT ( L-atm/molK) (303K) Molar mass = g = g = 35.6 g/mol mol mol

Lecture PLUS Timberlake Density of a Gas Calculate the density in g/L of O 2 gas at STP. From STP, we know the P and T. P = 1.00 atm T = 273 K Rearrange the ideal gas equation for moles/L PV = nRTPV = nRT P = n RTV RTV RT V

Lecture PLUS Timberlake Substitute (1.00 atm ) mol-K = mol O 2 /L ( L-atm) (273 K) Change moles/L to g/L mol O 2 x 32.0 g O 2 = 1.43 g/L 1 L 1 mol O 2 Therefore the density of O 2 gas at STP is 1.43 grams per liter