GAS LAWS Chapter 10

THE LAWS: Find and define Boyle’s law Charles’ law Lussac’s law Avogadro’s law Combined gas law Ideal gas law Dalton’s law Graham’s law

1 atmosphere = 760 mmHg = 760 Torr = 101325 Pascals = VARIABLES PRESSURE (P) Defined as force per unit area Caused by collisions of gas molecules with the container walls Measured using a barometer Invented by Torricelli Measured atmospheric pressure at sea level using a column of mercury which rose 760 mm Above sea level, atmospheric pressure decreases Units and equivalents 1 atmosphere = 760 mmHg = 760 Torr = 101325 Pascals = 101.325 kPa = 29.92 inHg = 14.7 psi

VARIABLES ?atm = 30.o8 in Hg × 1 atm = 1.005 atm 1 29.92 in Hg Example: Convert 30.08 inHg to atmospheres and pascals ?atm = 30.o8 in Hg × 1 atm = 1.005 atm 1 29.92 in Hg ?Pa = 30.08 in Hg × 101325 Pa = 101866 Pa 1 29.92 in Hg sig figs = 101900 Pa

VARIABLES VOLUME (V) TEMPERATURE (T) NUMBER OF MOLES (n) Units: liters Molar volume of a gas at STP = 22.4 L STP = 1 atm; 0˚C TEMPERATURE (T) Units: Kelvin Recall: Degrees C + 273.15 = Kelvin NUMBER OF MOLES (n) Units: moles What if you’re given grams? Use MM

CONSTANT THE UNIVERSAL GAS CONSTANT R Depends on pressure units. R = 0.0821 L atm/ mol K or R = 8.314 L kPa/ mol K

COMBINED GAS LAW P1V1 = P2V2 T1 T2 Combines Boyle’s, Charles, and Lussac’s laws Assumes a constant number of moles Example: 5.0 moles of gas are held in a flexible container with an initial volume of 10.0 ml at 25.0 degrees C and 1.0 atm. If the pressure changes to 755 mmHg and the temperature is 295 K, what is the new volume of the container?

COMBINED GAS LAW P1=1.00 atm P2=755 mm Hg V1=10.0 mL V2=? T1=25.0˚C T2=295 K 25.0˚C + 273.15 = 298.2 K 755 mm Hg× 1 atm = .993 atm 760 mm Hg (1.00 atm) (10.0 mL) = (.993 atm)(V2) 298.2 K 295 K V2=9.96 mL

IDEAL GAS LAW Assume the gas behaves ideally. No attraction or repulsion between particles. Also ignores the volume of the gas molecules. Real gases behave most like ideal gases when the temperature is high and the pressure is low. Less interaction between gas particles allows the gas molecules to be treated independently and the calculations to be simplified.

IDEAL GAS LAW PV = nRT Example: 3.00 moles of a gas are held in a 20.0 L container at 300.0 degrees C, what is the pressure inside the container? P= (3.00 moles) (.0821 L atm/mol K) (573.2 K) = 7.06 atm 20.0 L

IDEAL GAS LAW A gas at 105.0 kPa and 25.0 degrees C is held in a 250. ml container. How many moles are present? PV =n (105.0 K Pa) (.250L) = 1.06×10-2 moles RT (8.314 L K Pa/mol K) (298.2 K)

IDEAL GAS LAW A cylinder of gas holds 5.0 moles at 120. psi and 25.0 degrees C, what is the volume of the cylinder? 120. Psi × 1 atm =8.16 atm 14.7 psi V= (5.0 moles) (.0821 L atm/mol K) (298.2 K) =15L 8.16 atm What is the temperature of 1.00 mole of gas if 22.4 L has a pressure of 760.0 mmHg? 760.0 mm Hg × 1 atm =1atm 1 760.0 mm Hg T= PV = (1.000atm) (22.4 L) =273 K 273 K=0˚C nR (1.00 mol) (.0821 L atm/mol K)

IDEAL GAS LAW DERIVATIONS Including density and molar mass MEMORIZE! MM = g R T g=mass in grams PV MM = d R T d=density (g/L) P

DENSITY and MOLAR MASS Example: What is the molar mass of a gas if .250 grams of gas are held in a 250.0 ml container at 25.0 degrees C and 1.0 atmosphere of pressure? MM=gRT= (.250 g) (.0821 L atm/mol K) (298.2 K) =2.448 g/mol PV (1 atm) (.2500 L) sig figs=2.4 g/mol Example: What is the density of carbon dioxide gas at 30.0 degrees celsius and 29.80 inHg? MM= dRT MM P=d= (44.01 g/mol) (.9960 atm) =1.76 g/L P RT (1.0821 L atm/mol K) (303.2 K)

DALTON’S LAW Partial pressure – the pressure that an individual gas exerts in a mixture of gases. P total = P1 + P2 + P3 + P4 … Pn Example: A container of gas at 1.0 atm contains O2 at 100.0 mmHg, N2 at 450. torr and CO2. What is the pressure of the CO2 in atmospheres? Ptotal= PCO2 + PO2 + PN2 1.0 atm = PCO2 + (100.0 mm Hg× 1 atm ) + (450 torr × 1 atm ) 1 760 mm Hg 1 760 torr 1.O atm = PCO2 + .1316 atm + .592 atm PCO2 = .28 atm

DALTON’S LAW Necessary for gases that have been collected by water displacement. H2O evaporates and causes water vapor to be mixed with the gas. In order to find the pressure of the gas alone (dry gas), the water vapor pressure must be subtracted from the total pressure. P atm = P total = P gas + P H20 Water vapor pressure increases with temperature, since more water evaporates at higher temperatures. See table of water vapor pressure in the appendix.

DALTON’S LAW Oxygen gas was collected by water displacement and the water level inside the collection flask was equal to the level in the water trough to ensure that the pressure inside the flask = atmospheric pressure. The water bath was at 23.0 degrees Celsius and the barometric pressure was 28.90 inHg. What is the pressure of dry oxygen gas? PO2+PH2O= P atm (table value) 21.1 mm Hg = (28.90 in Hg×760 mm Hg) 1 29.92 in Hg PO2= 734.2 mm Hg – 21.1 mm Hg = 713.1 mm Hg

PRACTICE PROBLEM A gas was collected by water displacement over a water bath at 21.0 degrees C. 350.o ml of gas were collected and the barometric pressure was 755 mmHg. If the gas weighed 1.00 grams, what is the molar mass of the gas? Is there any possibility that the gas is CO? MM= gRT (1.00 g) (.0821 L atm/mol K) (294.2 K) = 71.2 g/mol PV (.986 atm) (.350 L) Pdry gas = P atm – PH2O Pdry gas = 755 mm Hg – 18.6 mm Hg = 736.4 mm Hg× 1 atm = .986 atm 1 760 mm Hg NO there is no possibility that the gas is CO!

MOLE FRACTION Defined as: the ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture  (chi) = mole fraction  = n1 ntotal  = p1 ptotal Multiplying the mole fraction by __100_ results in a __percent__ composition of the gas in the mixture. Multiplying the mole fraction by _the total pressure_ results in the __partial pressure__ of the gas.

MOLE FRACTION Examples: A mixture of gases includes 3.0 moles of O2, 2.0 moles of N2, and 4.0 moles of CO2, what is the mole fraction of O2 in the mixture? 3.0 moles =.33 3.0+4.0+2.0 moles

MOLE FRACTION If the total pressure of the mixture above is 2.0 atm. What is the partial pressure of the O2 gas? PO2 = .33 PO2= .67 atm 2.0 atm

MOLE FRACTION A mixture of gases has a total pressure of 760 mmHg. If the mixture contains oxygen, hydrogen, and nitrogen, each with equal pressures, What is the partial pressure of O2? X+X+X=760 mmHg PO2= 253.33= 260 mmHg What is the mole fraction of O2? 260 mmHg= .34 760 mmHg What is the % of O2 in the mixture? 34%

GRAHAM’S LAW Effusion – movement of a gas through a small orifice Diffusion – spontaneous mixing of gas molecules The rate at which a gas effuses or diffuses is inversely proportional to the square root of the molar mass of the gas. Rate gas 1 = √MM gas 2 Rate gas 2 √MM gas 1 Example: How much faster does helium travel compared to carbon dioxide? Rate He = √MM CO2 rate = √(44.01 g/mol ÷ 4.00 g/mol) Rate CO2 √MM He ratio He travels 3.32 times faster than CO2

GRAHAM’S LAW A gas travels 3.0 times faster than N2. What is the molar mass of the gas? 3.0 = rate gas = √28.02 g/mol 3.0 √MM gas = √28.02 rate N2 √MM gas 3.0 √MM gas = √28.02 (√28.02)2 = 3.1 g/mol ( 3.0 )2 If nitrogen gas travels at 30.0 m/sec at a certain temperature, how fast does carbon dioxide travel at the same temperature? 30.0 m/s = √44.01 g/mol rate CO2=23.9 m/s rate CO2 √28.02 g/mol