Presentation on theme: "Gases Chapter 10. Elements that exist as gases 25 0 C and 1 atmosphere 10.1."— Presentation transcript:
Gases Chapter 10
Elements that exist as gases 25 0 C and 1 atmosphere 10.1
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 Physical Characteristics of Gases
Units of Pressure 1 pascal (Pa) = 1 N/m 2 1 bar = 10 5 Pascals 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa 1 atm = kPa 10.2 Barometer Pressure = Force Area (force = mass x acceleration)
10.3 As P (h) increases V decreases Apparatus for Studying the Relationship Between Pressure and Volume of a Gas
10.3 Boyles Law The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.
P 1/V P x V = constant P 1 x V 1 = P 2 x V Boyles Law Constant temperature Constant amount of gas
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 10.3 P x V = constant
As T increasesV increases 10.3
Variation of gas volume with temperature at constant pressure V T V = constant x T V 1 /T 1 = V 2 /T 2 T (K) = t ( 0 C) Charles & Gay-Lussacs Law Temperature must be in Kelvin
10.4 Charless Law The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature. A plot of V versus T will be a straight line. i.e., VTVT = k
A sample of carbon monoxide gas occupies 3.20 L at C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V 1 = 3.20 L T 1 = K V 2 = 1.54 L T 2 = ? T 2 = V 2 x T 1 V1V L x K 3.20 L = = 192 K 10.3 V 1 /T 1 = V 2 /T 2 T 1 = 125 ( 0 C) (K) = K
10.3 Avogadros Law The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas. Mathematically, this means V = kn
Avogadros Law V number of moles (n) V = constant x n V 1 / n 1 = V 2 / n Constant temperature Constant pressure
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 10.3
Final and Initial State Problems Use Combined Gas Law List all variables (P,V,n,T) in two columns… one for initial state, one for final state. Substitute and solve ( make sure units cancel)
Final and Initial State Problems A sample of gas occupies 355 ml and 15 o C and 755 torr of pressure. What temperature will the gas have at the same pressure if its volume increases to 453 ml?
Final and Initial State Problems A sample of gas at 30.0 o has a pressure of 1.23 bar and a volume of m 3. If the volume of the gas is compressed to m 3 at the same temperature, what is its pressure at this volume?
Final and Initial State Problems A Sample of gas at 27.0 o C has a volume of 2.08 L and a pressure of mm Hg. If the gas is in a sealed container, what is its pressure in bar when the temperature (in o C) doubles?
10.3 Ideal-Gas Equation V 1/P (Boyles law) V T (Charless law) V n (Avogadros law) So far weve seen that Combining these, we get V nT P
10.4 Ideal-Gas Equation The relationship then becomes nT P V nT P V = R or PV = nRT
10.4 Ideal-Gas Equation The constant of proportionality is known as R, the ideal gas constant.
Ideal Gas Equation 10.4 Charles law: V T (at constant n and P) Avogadros law: V n (at constant P and T) Boyles law: V (at constant n and T) 1 P V nT P V = constant x = R nT P P R is the gas constant PV = nRT
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)( K) R = L atm / (mol K) 10.4 Experiments show that at STP, 1 mole of an ideal gas occupies L.
What is the volume (in liters) occupied by 49.8 g of HCl at STP? PV = nRT V = nRT P T = 0 0 C = K P = 1 atm n = 49.8 g x 1 mol HCl g HCl = 1.37 mol V = 1 atm 1.37 mol x x K Latm molK V = 30.6 L 10.4
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 10.4
Single State Problems Use Ideal gas Law – youre given three for the four variables and R - solve for the unknown. PV = nRT
Single State Problems How many grams of oxygen gas in a 10.0 L container will exert a pressure of 712 torr at a temperature of 25.0 o C?
Single State Problems At what temperature will grams of sulfur dioxide in a ml container exert a pressure of atm?
Other calculations involving the Ideal Gas Law We can often determine the molar mass of a gas by measuring its mass at a given volume, temperature and pressure. Ideal Gas Law can also be used to determine the density of a gas at given conditions of temperature and pressure.
10.4 Densities of Gases If we divide both sides of the ideal-gas equation by V and by RT, we get nVnV P RT =
10.4 We know that –moles molecular mass = mass Densities of Gases So multiplying both sides by the molecular mass ( ) gives n = m P RT mVmV =
10.4 Densities of Gases Mass volume = density So, Note: One only needs to know the molecular mass, the pressure, and the temperature to calculate the density of a gas. P RT mVmV = d =
10.4 Molecular Mass We can manipulate the density equation to enable us to find the molecular mass of a gas: Becomes P RT d = dRT P =
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 10.4
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? 10.4 dRT P M = d = m V 4.65 g 2.10 L = = 2.21 g L M = 2.21 g L 1 atm x x K Latm molK M = 54.6 g/mol
When 4.93 grams of carbon tetrachloride gas are in a 1.oo liter container at K, the gas exerts a pressure of torr. What is the molar mass of carbon tetrachloride?
A gaseous compound with a mass of grams has a volume of 2236 mL at 27.0oC and 735 mm Hg. Calculate the molar mass of the compound.
Calculate the density of ammonia at 0oC and 1 atm of pressure.
Calculate the molar mass of a compound with a density of g/L at 25.0 o C and torr pressure.
Gas Stoichiometry What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO g C 6 H 12 O 6 1 mol C 6 H 12 O g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x = mol CO 2 V = nRT P mol x x K Latm molK 1.00 atm = = 4.76 L 10.5
Daltons Law of Partial Pressures V and T are constant P1P1 P2P2 P total = P 1 + P 2 5.6
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
A sample of natural gas contains 8.24 moles of CH 4, moles of C 2 H 6, and 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 = P T = 1.37 atm = P propane = x 1.37 atm= atm 5.6
2KClO 3 (s) 2KCl (s) + 3O 2 (g) Bottle full of oxygen gas and water vapor P T = P O + P H O
Chemistry in Action: Scuba Diving and the Gas Laws PV Depth (ft)Pressure (atm)
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
Kinetic theory of gases and … Compressibility of Gases Boyles Law P collision rate with wall Collision rate number density 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 5.7
Kinetic theory of gases and … Avogadros Law P collision rate with wall Collision rate number density Number density n P n Daltons 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
Apparatus for studying molecular speed distribution 5.7
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 u rms = 3RT M
Chemistry in Action: Super Cold Atoms Gaseous Rb Atoms 1.7 x K Bose-Einstein Condensate
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 =
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 = x 28 = x = 4.1 ~ 4
Deviations from Ideal Behavior 1 mole of ideal gas PV = nRT n = PV RT = Repulsive Forces Attractive Forces
Effect of intermolecular forces on the pressure exerted by a gas. 5.8
Van der Waals equation nonideal gas P + (V – nb) = nRT an 2 V2V2 () } corrected pressure } corrected volume