Gases and their Properties Goals: 1.Use the gas law equations. 2.Apply the gas laws to stoichiometric calculations. 3.Describe the states of matter based on kinetic- molecular theory. 4.Recognize why real gases do not behave like ideal gases.

THREE STATES OF MATTER

General Properties of Gases There is a lot of “free” space in a gas. Gases can be expanded __________________. Gases occupy containers uniformly and completely. Gases diffuse and mix ___________.

Properties of Gases Gas properties can be modeled using math. Model depends on— V = T = n = P =

Pressure Pressure – force per unit area (force divided by the area over which the force is exerted). SI unit is the pascal (Pa) Pressure = Force Area SI units – Pascal (Pa) Newton (N) Square meter (m 2 ) 1 Pa = 1 N/m 2 Newton- the force that will give a 1-kg mass an acceleration of 1 meter per second (m/s).

Barometer Barometer – device to measure atmospheric pressure. Invented in 1643 by Evangelista Torricelli. It is based on a column filled with mercury (Hg). On average, at sea level, the column supported by the air pressure is 760 mm Hg high – 1 atmosphere (atm). 1 atm = ______ mm Hg = _____ Torr Why Hg and not H 2 O?

Pressure Units 1 atm = 760 mm Hg = 760 Torr 1 atm = ________Pa (pascal) = ______ kPa (kilopascal) SI unit Units used in weather reports Units used by engineers 1 atm = inHg (inches of mercury) = mbar (millibars) 1 atm = lb/in 2 or psi (pounds per square inch)

Gas Laws Gas law EquationVariablesConstant(s) Boyle’s law Charles’s law Avogadro’s law Combined gas law Ideal gas law P- pressureT- temperatureR- universal gas constant V- volumen- number of moles

Boyle’s Law: P-V relationship Robert Boyle, 1662 : Boyle’s Law –For a given amount of gas at a constant temperature, the volume of the gas varies inversely with its pressure. V  1 / P V = a / P VP = a a- proportionality constant

Boyle’s Law: P-V relationship V = a / P VP = a If V increases P must decrease, and viceversa. P1P1 P2P2 V1V1 V2V2 A gas is enclosed in a 10.1 L tank at 1208 mmHg. Which of the following is a reasonable value for the pressure when the gas is transferred to a 30.0 L tank? 400 mmHg3600 mmHg

Boyle’s Law A bicycle pump is a good example of Boyle’s law. As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire.

Boyle’s Law: P-V relationship Any units can be used for P and V (same units used throughout the calculation). Same sample of a confined gas at constant T. V 1 P 1 = V 2 P 2 A sample of air occupies 73.3 mL at 98.7 atm and 0 o C. What volume will the air occupy at 4.02 atm and 0 o C?

Boyle’s Law P  1 / V V  1 / P P  1 / V P = a / V VP = a a- proportionality constant = slope of the line

Charle’s Law: T-V relationship Jacques Charles, 1787 : Charles’s Law –For a given amount of gas at a constant pressure, the volume of the gas varies directly with its temperature. –For each o C rise in T, V expands by 1/273 of its V at 0 o C; for each o C drop in T, V contracts by 1/273 of its V at 0 o C o CIs the absolute 0 in Kelvin scale - Volume of a fixed amount of a gas at a constant P is directly proportional to its Kelvin T. V  T V = bT V / T = b b - proportionality constant V2V2 V1V1 T1T1 T2T2

Charle’s Law: T-V relationship If T increases, V must also increase; when T decreases, V must also decrease. For the same sample of trapped gas at a constant pressure. For all gas law calculations involving T, absolute T in kelvins must be used. 0 o C 40 o C V1V1 V2V2 T1T1 T2T2 =

Charle’s Law: T-V relationship A sample of oxygen gas occupies a volume of 2.10 L at 25 o C. What volume will this sample occupy at 150 o C? (Assume no change in pressure.)

Charle’s Law

Practice What effect will the following changes have on the V of a fixed amount of gas? a. an increase in P at constant T b. a decrease in T at constant P c. a decrease in P coupled with an increase in T What effect will the following changes have on the P of a fixed amount of gas? a. an increase in T at constant V b. a decrease in V at constant T c. an increase in T coupled with a decrease in V

Avogadro’s Hypothesis Avogadro’s Hypothesis – equal number of molecules of different gases compared at the same T and P occupy equal V. Avogadro’s Law – at fixed T and P, the V of a gas is directly proportional to the number of molecules of a gas or to the number of moles of gas, n. –If the number of moles of gas doubles, the volume doubles. –If the mass of a gas doubles, the volume doubles. V  n V = cn

Avogadro’s Hypothesis The gases in this experiment are all measured at the same T and P. 2 H 2 (g) + O 2 (g) 2 H 2 O(g)

Molar Volume Molar Volume of a gas – is the volume occupied at STP by 1 mol of any gas and it is 22.4 L. Standard conditions of T and P (STP): T = 0 o C (273K) P = 1 atm (760 Torr) 1 mol gas = 22.4 L (at STP)

Molar Volume 1 mol of N 2 = 28 g N 2, will occupy 22.4 L at STP. 1 mol of CO 2 = 44 g CO 2, will occupy 22.4 L at STP. Calculate the volume occupied by 4.11 g of methane (CH 4 ) gas at STP. Strategy: Convert mass of gas to moles of gas and use the molar volume relationship as conversion factor to find out the volume at STP. 1 mol gas = 22.4 L (at STP)

Practice Solve the Charle’s Law for T 2 Students should be familiar rearranging gas laws equations and applying them in problems.

Gas Laws Gas law EquationVariablesConstant(s) Boyle’s law P 1 V 1 = P 2 V 2 P, Vn, T Charles’s law V 1 / T 1 = V 2 / T 2 V, Tn, P Avogadro’s law V 1 / n 1 = V 2 / n 2 V, nP, T Combined gas law Ideal gas law P- pressureT- temperatureR- universal gas constant V- volumen- number of moles

Combined Gas Law V proportional to 1/P V proportional to T V proportional to n Therefore, V prop. to nT/P A fixed amount of gas under a set of initial conditions is changed to a set of final conditions. V  nT P PV = constant nT V1V1 V2V2 T1T1 T2T2 = V 1 P 1 = V 2 P 2 V1P1V1P1 V2P2V2P2 T1T1 T2T2 = PV = nRT

What volume will occupy a 5 mL bubble of air exhaled by a deep-sea diver at 1.8 atm and at 5 o C when it reaches the surface of the sea, at 0.98 atm and at 25 o C?

Ideal Gas Law For varying amounts of gas (number of moles, n). Units: P in atmospheres (atm) V in liters (L) T in kelvins (K) Universal gas constant, R Used to calculate any of the four quantities – P, V, n or T if the other three are known. PV = nRT R = L atm mol K

How much N 2 (in Kg) is req’d to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 o C?

Gas Density Reported in g/L at STP. If you know the density, you can calculate the molar volume. From the molar volume of gas, you can calculate the density of the gas using the molar mass. 1 mol gas = 22.4 L (at STP)

Gas Density Density is proportional to molecular mass. Which balloon(s) is(are) filled with the higher density gas?

The density of air at 15 o C and 1.00 atm is 1.23 g/L. What is the molar mass of air?

Dalton’s Law Dalton’s experimental observation – adding H 2 O vapor to a certain pressure to dry air, the P exerted by the air would increase by an amount equal to the P of the water vapor. Dalton’s conclusion – each gas in a mixture of gases behaves independently, exerting its own pressure. Dalton’s Law – The total P of the mixture is equal to the sum of the partial pressures exerted by separate gases. P total = P 1 + P 2 + P 3 + ….

Dalton’s Law

Vapor Pressure Vapor pressure – is the partial pressure exerted by the molecules of a substance in the gas phase above the liquid phase of the substance. Vapor pressure is T dependent (increasing T, increase the vapor pressure). For water:T (oC)Vapor pressure (mmHg)

Oxygen is collected over water at 20 o C. The total P inside the collection jar is 740 mmHg. What is the pressure due to the oxygen alone?

Partial Pressures and Respiration Exchange of gases (CO 2 and O 2 ) between: alveoli and blood capillaries; arteriole and extracellular fluid; extracellular fluid and cells occurs by diffusion. Gases go from higher partial pressure to lower partial pressure. Diffusion – the flow of a substance from a region of higher concentration to a region of lower concentration. CO 2 is the trigger; if too high we take a breath, if too low, there is no breathing.

Gases and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) 1.1 g of H 2 O 2 decompose in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O?

2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O?

Kinetic-Molecular Theory Theory used to explain gas laws. KMT assumptions are Gases consist of _____________________ __________________________. P arises from _______________________. No __________ or __________ forces between molecules. Collisions elastic. Volume of molecules is ____________.

Kinetic-Molecular Theory Because we assume molecules are in motion, they have a kinetic energy. KE = (1/2)(mass)(speed) 2 At the same T, all gases have the same average KE. As T goes up for a gas, KE also increases — and so does speed.

Kinetic-Molecular Theory At the same T, all gases have the same average KE. As T goes up, KE also increases — and so does speed.

Maxwell’s equation where u is the speed and M is the molar mass. –speed INCREASES with T –speed DECREASES with M

Speed’s Distribution Boltzmann plots Named for Ludwig Boltzmann doubted the existence of atoms. This played a role in his suicide in 1906.

Velocity of Gas Molecules Molecules of a given gas have a range of speeds. Average velocity decreases with increasing mass.

Gas Diffusion and Effusion DIFFUSION is the gradual mixing of molecules of different gases. EFFUSION is the movement of molecules through a small hole into an empty container.

Gas Effusion Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is proportional to T inversely proportional to M. Therefore, He effuses more rapidly than O 2 at same T.

Graham’s Law Rate of effusion is inversely proportional to its molar mass. –HCl and NH 3 diffuse from opposite ends of tube. –Gases meet to form NH 4 Cl –HCl heavier than NH 3 –Therefore, NH 4 Cl forms closer to HCl end of tube.

Kinetic-Molecular Theory Model used to explain de behavior of gases. Postulates: 1. All matter is composed of tiny, discrete particles called molecules (N 2, Ar). 2. The molecules are in rapid, constant motion and move in straight lines (strike the walls of the container, exert pressure). 3. The molecules of a gas are small compared with the distances between them (low density, compressibility). 4. There is little attraction between molecules of a gas. 5. Molecules collide with one another, and energy is conserved in these collisions (elastic- the sum of the kinetic energies before and after collision is equal). 6. Temperature is a measure of the average kinetic energy of the gas molecules (on average, the particles of a cold sample move more slowly than the particles of a hot sample).

Deviations from Ideal Gas Law Real molecules have volume. There are intermolecular forces. –Otherwise a gas could not become a liquid. Account for volume of molecules and intermolecular forces with VAN DER WAALS’s EQUATION.

Van Der Waals’s Equation Cl 2 gas has a = 6.49, b = For 8.0 mol Cl 2 in a 4.0 L tank at 27 o C. P (ideal) = nRT/V = 49.3 atm P (van der Waals) = 29.5 atm Measured V = V(ideal) Measured P intermol. forces vol. correction nRT V - nb V 2 n 2 a P )( P V = n R T

Deviations from Ideal Gas Law PV = nRT PV = constant PV P y = mx + b m = slope correction to Volume m = -1.5 Vc = V – m Vc = = 21.5 mL

Remember Go over all the contents of your textbook. Practice with examples and with problems at the end of the chapter. Practice with OWL tutor. Work on your OWL assignment for Chapter 12. Practice with the quiz on CD of Chemistry Now.