Chapter 5: the Gaseous state Vanessa Prasad-Permaul Valencia Community College CHM 1045

GAS LAWS: PRESSURE AND MEASUREMENT Elements that exist as gases at 250C and 1 atmosphere

GAS LAWS: PRESSURE AND MEASUREMENT Physical Characteristics of Gases 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.

GAS LAWS: PRESSURE AND MEASUREMENT a) Gas is a large collection of particles moving at random throughout a volume b) Collisions of randomly moving particles with the walls of the container exert a force per unit area that we perceive as gas pressure

GAS LAWS: PRESSURE AND MEASUREMENT HOW IS PRESSURE DEFINED? The force the gas exerts on a given area of the container in which it is contained. The SI unit for pressure is the Pascal, Pa. Pressure = Force Area If you’ve ever inflated a tire, you’ve probably made a pressure measurement in pounds (force) per square inch (area).

GAS LAWS: PRESSURE AND MEASUREMENT Barometer 760mm mercury Height is proportional to the barometric pressure Units of Pressure 1 pascal (Pa) = 1 kg/m·s2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa 1 bar = 105 Pa 1 atm = 14.69 lb/in2 Hg is used instead of H2O : more dense better visibility accuracy

GAS LAWS: PRESSURE AND MEASUREMENT EXERCISE 5.1 A GAS CONTAINER HAD A MEASURED PRESSURE OF 57kPa. CALCULATE THE PRESSURE IN UNITS OF ATM AND mmHg. First, convert to atm (57 kPa = 57 x 103 Pa). 57 x 103 Pa x 1 atm = 0.562 = 0.56 atm 1.01325 x 105 Pa Next, convert to mmHg. 57 x 103 Pa x 760 mmHg = 427.5 = 4.3 x 102 mmHg

GAS LAWS: EMPIRICAL GAS LAWS You can predict the behavior of gases based on the following properties: Pressure Volume Amount (moles) Temperature * If two of these physical properties are held constant, it is possible to show a simple relationship between the other two…

GAS LAWS: EMPIRICAL GAS LAWS Boyle’s experiment: A manometer to study the relationship between pressure (P) & Volume (V) of a gas Hg Hg As P (h) increases V decreases

GAS LAWS: EMPIRICAL GAS LAWS BOYLE’S LAW the volume of a sample of gas at a given temperature varies inversely with the applied pressure So…. For a given amount of gas (n) & @ constant temperature (T) : If pressure (P) increases, the volume (V) of the gas decreases P1 * V1 = P2 * V2

Pressure–Volume Law (Boyle’s Law):

GAS LAWS: EMPIRICAL GAS LAWS Boyle’s Law P a 1/V P * V = constant P1 * V1 = P2 * V2 Constant temperature Any given amount of gas

Application of Boyle’s law gives GAS LAWS: EMPIRICAL GAS LAWS EXERCISE 5.2 A volume of carbon dioxide gas equivalent to 20.0 L was collected @ 23oC and 1.00atm pressure. What would be the volume of gas at constant temperature and 0.830atm? P1 * V1 = P2 * V2 Application of Boyle’s law gives V2 = V1 x P1 = 20.0 L x 1.00atm = 24.096 = 24.1 L P2 0.830atm

Charles’ Law: relating volume and temperature GAS LAWS: EMPIRICAL GAS LAWS Charles’ Law: relating volume and temperature As T increases V increases

Temperature–Volume Law (Charles’ Law):

GAS LAWS: EMPIRICAL GAS LAWS Charles’ Law Variation of gas volume with temperature V a T For a given amount of gas at constant pressure Temperature must be in Kelvin V / T = constant T (K) = t (0C) + 273.15

GAS LAWS: EMPIRICAL GAS LAWS CHARLES’ LAW the volume occupied by any sample of gas at a constant pressure is directly proportional to the absolute temperature So…. For a given amount of gas (n) & @ constant pressure (P) : If temperature (T) increases, the volume (V) of the gas increases V1 V2 T1 = T2

GAS LAWS: EMPIRICAL GAS LAWS EXERCISE 5.3 A chemical reaction is expected to produce 4.3dm3 of oxygen at 19oC and 101kPa. What will the volume be at constant pressure and 25oC? First, convert the temperatures to the Kelvin scale. Ti = (19 + 273) = 292 K Tf = (25 + 273) = 298 K Following is the data table. Vi = 4.38 dm3 Pi = 101 kPa Ti = 292 K Vf = ? Pf = 101 kPa Tf = 298 K Apply Charles’s law to obtain Vf = Vi x Tf = 4.38 dm3 x 298K = 4.470 = 4.47 dm3 Ti 292K

COMBINED GAS LAW Relating Volume, Temperature and Pressure GAS LAWS: EMPIRICAL GAS LAWS COMBINED GAS LAW Relating Volume, Temperature and Pressure Taking Boyle’s Law and Charles’ Law: The volume occupied by a given amount of gas is proportional to the absolute temperature divided by the pressure: V = constant x T or PV = constant (for a given amount of gas) P T P1V1 = P2V2 T1 T2

Use Kelvins for temp, any pressure, any volume Combined Gas Law We can combine Boyle’s and charles’ law to come up with the combined gas law Use Kelvins for temp, any pressure, any volume

EXERCISE 5.4 A balloon contains 5.41dm3 of helium at 24oC and 10.5kPa. GAS LAWS: EMPIRICAL GAS LAWS EXERCISE 5.4 A balloon contains 5.41dm3 of helium at 24oC and 10.5kPa. Suppose the gas in the balloon is heated to 35oC and the pressure is now 102.8kPa, what is the volume of the gas? First, convert the temperatures to kelvins. Ti = (24 + 273) = 297 K Tf = (35 + 273) = 308 K Following is the data table. Vi = 5.41 dm3 Pi = 101.5 kPa Ti = 297 K Vf = ? Pf = 102.8 kPa Tf = 308 K Apply both Boyle’s law and Charles’s law combined to get Vf = Vi x Pi x Tf =5.41 dm3 x 101.5kPa x 308K = 5.539 = 5.54 dm3 Pf Ti 102.8kPa 297K

AVOGADRO’S LAW relating volume and amount GAS LAWS: EMPIRICAL GAS LAWS AVOGADRO’S LAW relating volume and amount equal volumes of any two gases at the same temperature & pressure contain the same number of molecules V a number of moles (n) V = constant x n V1/n1 = V2/n2 Constant temperature Constant pressure

The Volume–Amount Law (Avogadro’s Law):

V1 = V2 n1 n2 The Volume–Amount Law (Avogadro’s Law): At constant pressure and temperature, the volume of a gas is directly proportional to the number of moles of the gas present. Use any volume and moles V1 = V2 n1 n2

GAS LAWS: EMPIRICAL GAS LAWS Avogadro’s Number = one mole of any gas contains the same number of molecules 6.023 x 1023 Must occupy the same volume at a given temperature and pressure The conditions 0 0C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.4 L.

IDEAL GAS LAW GAS LAWS: THE IDEAL GAS LAW Boyle’s law: V a (at constant n and T) 1 P Charles’ law: V a T (at constant n and P) Avogadro’s law: V a n (at constant P and T) V a nT P V = constant x = R nT P R is the molar gas constant PV = nRT

IDEAL GAS EQUATION GAS LAWS: THE IDEAL GAS LAW PV = nRT PV nT = (1 atm)(22.414L) (1 mol)(273.15 K) R = 0.082057 L • atm / (mol • K) R = 8.3145 J / (mol • K) R = 1.9872 cal / (mol • K) *The units of pressure times volume are the units of energy joules (J) or calories (cal)

1 mole of an ideal gas occupies 22.414 L at STP The Ideal Gas Law Ideal gases obey an equation incorporating the laws of Charles, Boyle, and Avogadro. 1 mole of an ideal gas occupies 22.414 L at STP STP conditions are 273.15 K and 1 atm pressure The gas constant R = 0.08206 L·atm·K–1·mol–1 P has to be in atm V has to be in L T has to be in K

Or, expressing this as a proportion, GAS LAWS: THE IDEAL GAS LAW EXERCISE 5.5 Show that the moles of gas are proportional to the pressure for constant volume and temperature Use the ideal gas law, PV = nRT, and solve for n: n = PV RT n = V x P Note that everything in parentheses is constant. Therefore, you can write n = constant x P Or, expressing this as a proportion, n  P

GAS LAWS: THE IDEAL GAS LAW EXERCISE 5.6 What is the pressure in a 50.0L gas cylinder that contains 3.03kg of oxygen at 23oC? T = 23oC + 273K = 296K V = 50.0L R = 0.08206 L . atm/K . mol n = 3.03kg x 1000g x 1mol = 94.688 mol O2 1 kg 31.998g P = ? PV= nRT P = nRT = 0.0347mol x 0.08206 L . atm x 296K V K . mol = 46.0 atm 50.0L

Density and Molar Mass Calculations: The Ideal Gas Law Density and Molar Mass Calculations: You can calculate the density or molar mass (MM) of a gas. The density of a gas is usually very low under atmospheric conditions.

GAS LAWS: THE IDEAL GAS LAW EXERCISE 5.7 Calculate the density of helium (g/L) at 21oC and 752mmHg. The density of air under these conditions is 1.188g/L. What is the difference in mass between 1 liter of air and 1 liter of helium? Variable Value P 752 mmHg x 1 atm = 0.98947 atm 760mmHg V 1 L (exact number) T (21 + 273) = 294 K n ? 32

GAS LAWS: THE IDEAL GAS LAW Using the ideal gas law, solve for n, the moles of helium. n = PV = 0.98947 atm x 1L = RT 0.08206 L . atm/ K . mol x 294K 0.4101 mol Now convert mol He to grams. 0.04101 mol He x 4.00g He = 0.16404g He = 0.164g/L 1.00 mol He 1 L Therefore, the density of He at 21°C and 752 mmHg is 0.164 g/L. The difference in mass between one liter of air and one liter of He: mass air − mass He = 1.188 g − 0.16404 g = 1.02396 = 1.024 g difference

GAS LAWS: THE IDEAL GAS LAW EXERCISE 5.8 A sample of a gaseous substance at 25oC and 0.862 atm has a density of 2.26 g/L. What is the molecular mass of the substance? Variable Value P 0.862 atm V 1 L (exact number) T (25 + 273) = 298 K n ? From the ideal gas law, PV = nRT, you obtain n = PV = 0.862 atm x 1 L = 0.03525 mol RT 0.08206 L . atm/K . Mol x 298K 34 34

GAS LAWS: THE IDEAL GAS LAW Dividing the mass of the gas by moles gives you the mass per mole (the molar mass). Molar mass = grams of gas = 2.26g = 64.114 g/mol moles of gas 0.03525mol Therefore, the molecular mass is 64.1 amu.

Dalton’s Law of Partial Pressures GAS LAWS: GAS MIXTURES Dalton’s Law of Partial Pressures V and T are constant P1 P2 Ptotal = P1 + P2

Dalton’s Law of Partial Pressures For a two-component system, the moles of components A and B can be represented by the mole fractions (XA and XB). Mole fraction is related to the total pressure by:

Each gas obeys the ideal gas law. GAS LAWS: GAS MIXTURES EXERCISE 5.10 A 10.0L flask contains 1.031g O2 and 0.572g CO2 at 18oC. What are the partial pressures of each gas? What is the total pressure? What is the mole fraction of oxygen in this mixture? Each gas obeys the ideal gas law. 1.031 g O2 x 1 mol = 0.0322188 mol O2 32.00g P = nRT = 0.0322mol x 0.0821L.atm/K.mol x 291K = 0.076936 atm V 10.0L 0.572 g CO2 x 1 mol = 0.012997 mol CO2 44.01g P = nRT = 0.0130mol x 0.0821L.atm/K.mol x 291K = 0.031036 atm

The total pressure is equal to the sum of the partial pressures: GAS LAWS: GAS MIXTURES The total pressure is equal to the sum of the partial pressures: PT = PO2 + PCO2 = 0.076936atm + 0.031036atm = 0.10797 = 0.1080 atm The mole fraction of oxygen in the mixture is Mole fraction O2 = PO2 = 0.076936atm = 0.7122 = 0.712 PT 0.1080atm 0.712 x 100% = 71.2 mole % of O2 in this gas mixture

Kinetic Molecular Theory This theory presents physical properties of gases in terms of the motion of individual molecules. Average Kinetic Energy  Kelvin Temperature Gas molecules are points separated by a great distance Particle volume is negligible compared to gas volume Gas molecules are in constant random motion Gas collisions are perfectly elastic Gas molecules experience no attraction or repulsion

Kinetic Molecular Theory

Average Kinetic Energy (KE) is given by: U = average speed of a gas particle R = 8.314 J/K mol m = mass in kg MM = molar mass, in kg/mol NA = 6.022 x 1023

The Root–Mean–Square Speed: is a measure of the average molecular speed. Taking square root of both sides gives the equation

Example 17: Calculate the root–mean–square speeds of helium atoms and nitrogen molecules in m/s at 25°C.

Maxwell speed distribution curves. Kinetic Molecular Theory Maxwell speed distribution curves.

Graham’s Law Diffusion is the mixing of different gases by random molecular motion and collision.

Graham’s Law Effusion is when gas molecules escape without collision, through a tiny hole into a vacuum.

Graham’s Law: Rate of effusion is proportional to its rms speed, urms. For two gases at same temperature and pressure:

Behavior of Real Gases At higher pressures, particles are much closer together and attractive forces become more important than at lower pressures.

Behavior of Real Gases The volume taken up by gas particles is actually less important at lower pressures than at higher pressure. As a result, the volume at high pressure will be greater than the ideal value.

IntermolecularAttractions Behavior of Real Gases Corrections for non-ideality require van der Waals equation. Excluded Volume IntermolecularAttractions

Example 1: Boyle’s Law A sample of argon gas has a volume of 14.5 L at 1.56 atm of pressure. What would the pressure be if the gas was compressed to 10.5 L? (at constant temperature and moles of gas)

Example 2: Charles’ Law A sample of CO2(g) at 35C has a volume of 8.56 x10-4 L. What would the resulting volume be if we increased the temperature to 85C? (at constant moles and pressure)

Example 3: Avogadro’s Law 6.53 moles of O2(g) has a volume of 146 L. If we decreased the number of moles of oxygen to 3.94 moles what would be the resulting volume? (constant pressure and temperature)

Example 4:Combined Gas Law Oxygen gas is normally sold in 49.0 L steel containers at a pressure of 150.0 atm. What volume would the gas occupy if the pressure was reduced to 1.02 atm and the temperature raised from 20oC to 35oC?

Example 5: Gas Laws An inflated balloon with a volume of 0.55 L at sea level, where the pressure is 1.0 atm, is allowed to rise to a height of 6.5 km, where the pressure is about 0.40 atm. Assuming that the temperature remains constant, what is the final volume of the balloon?

Example 6: Ideal Gas Law Sulfur hexafluoride (SF6) is a colorless, odorless, very unreactive gas. Calculate the pressure (in atm) exerted by 1.82 moles of the gas in a steel vessel of volume 5.43 L at 69.5°C.

What is the volume (in liters) occupied by 7.40 g of CO2 at STP? Example 7: Ideal Gas Law What is the volume (in liters) occupied by 7.40 g of CO2 at STP?

What is the molar mass of a gas with a density of 1.342 g/L at STP? Example 8: Density & MM What is the molar mass of a gas with a density of 1.342 g/L at STP?

Example 9: Density & MM What is the density of uranium hexafluoride, UF6, (MM = 352 g/mol) under conditions of STP?

Example 10: Density & MM The density of a gaseous compound is 3.38 g/L at 40°C and 1.97 atm. What is its molar mass?

Example 11: Dalton Exactly 2.0 moles of Ne and 3.0 moles of Ar were placed in a 40.0 L container at 25oC. What are the partial pressures of each gas and the total pressure?

Example 12: Dalton A sample of natural gas contains 6.25 moles of methane (CH4), 0.500 moles of ethane (C2H6), and 0.100 moles of propane (C3H8). If the total pressure of the gas is 1.50 atm, what are the partial pressures of the gases?

Example 13: Mole Fraction What is the mole fraction of each component in a mixture of 12.45 g of H2, 60.67 g of N2, and 2.38 g of NH3?

Example 14: Partial Pressure On a humid day in summer, the mole fraction of gaseous H2O (water vapor) in the air at 25°C can be as high as 0.0287. Assuming a total pressure of 0.977 atm, what is the partial pressure (in atm) of H2O in the air?

Gas Stoichiometry & Example In gas stoichiometry, for a constant temperature and pressure, volume is proportional to moles. Example: Assuming no change in temperature and pressure, calculate the volume of O2 (in liters) required for the complete combustion of 14.9 L of butane (C4H10): 2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(l)

Example 15: All of the mole fractions of elements in a given compound must add up to? 100 1 50 2

Zn(s) + 2 HCl(aq)  H2(g) + ZnCl2(aq) Example 16: Hydrogen gas, H2, can be prepared by letting zinc metal react with aqueous HCl. How many liters of H2 can be prepared at 742 mm Hg and 15oC if 25.5 g of zinc (MM = 65.4 g/mol) was allowed to react? Zn(s) + 2 HCl(aq)  H2(g) + ZnCl2(aq)

Example 18: Under the same conditions, an unknown gas diffuses 0.644 times as fast as sulfur hexafluoride, SF6 (MM = 146 g/mol). What is the identity of the unknown gas if it is also a hexafluoride?

Example 19: Diffusion What are the relative rates of diffusion of the three naturally occurring isotopes of neon: 20Ne, 21Ne, and 22Ne?

Deviations result from assumptions about ideal gases. Behavior of Real Gases Deviations result from assumptions about ideal gases. 1. Molecules in gaseous state do not exert any force, either attractive or repulsive, on one another. 2. Volume of the molecules is negligibly small compared with that of the container.

Example 20: Ideal Vs. Van Der Waals Given that 3.50 moles of NH3 occupy 5.20 L at 47°C, calculate the pressure of the gas (in atm) using the ideal gas equation (b) the van der Waals equation. (a = 4.17, b = 0.0371)

Example 21: Ideal Vs. Van Der Waals Assume that you have 0.500 mol of N2 in a volume of 0.600 L at 300 K. Calculate the pressure in atmospheres using both the ideal gas law and the van der Waals equation. For N2, a = 1.35 L2·atm mol–2, and b = 0.0387 L/mol.