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Mullis1 Characteristics of Gases ► Vapor = term for gases of substances that are often liquids/solids under ordinary conditions ► Unique gas properties.

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Presentation on theme: "Mullis1 Characteristics of Gases ► Vapor = term for gases of substances that are often liquids/solids under ordinary conditions ► Unique gas properties."— Presentation transcript:

1 Mullis1 Characteristics of Gases ► Vapor = term for gases of substances that are often liquids/solids under ordinary conditions ► Unique gas properties 1.Highly compressible 2.Inverse pressure-volume relationship 3.Form homogeneous mixtures with other gases

2 Mullis2 Pressures of Enclosed Gases and Manometers ► Barometer: Used to measure atmospheric pressure* ► Manometer: Used to measure pressures of gases not open to the atmosphere ► Manometer is a bulb of gas attached to a U-tube containing Hg.  If U-tube is closed, pressure of gas is the difference in height of the liquid.  If U-tube is open, add correction term:  If P gas < P atm then P gas + P h = P atm  If P gas > P atm then P gas = P h + P atm * Alternative unit for atmospheric pressure is 1 bar = 10 5 Pa

3 Mullis3 Kinetic Molecular Theory ► Number of molecules ► Temp ► Volume ► Pressure ► Number of dancers ► Beat of music ► Size of room ► Number and force of collisions

4 Mullis4 Kinetic Molecular Theory ► Accounts for behavior of atoms and molecules ► Based on idea that particles are always moving ► Provides model for an ideal gas ► Ideal Gas = Imaginary: Fits all assumptions of the K.M theory ► Real gas = Does not fit all these assumptions

5 Mullis5 5 assumptions of Kinetic-molecular Theory 1. Gases = large numbers of tiny particles that are far apart. 2. Collisions between gas particles and container walls are elastic collisions (no net loss in kinetic energy). 3. Gas particles are always moving rapidly and randomly. 4. There are no forces of repulsion or attraction between gas particles. 5. The average kinetic energy of gas particles depends on temperature.

6 Mullis6 Physical Properties of Gasses ► Gases have no definite shape or volume – they take shape of container. ► Gas particles glide rapidly past each other (fluid). ► Gases have low density. ► Gases are easily compressed. ► Gas molecules spread out and mix easily

7 Mullis7 ► Diffusion = mixing of 2 substances due to random motion. ► Effusion = Gas particles pass through a tiny o p e n i n g……..…

8 Mullis8 Real Gases ► Real gases occupy space and exert attractive forces on each other. ► The K-M theory is more likely to hold true for particles which have little attraction for each other. ► Particles of N 2 and H 2 are nonpolar diatomic molecules and closely approximate ideal gas behavior. ► More polar molecules = less likely to behave like an ideal gas. Examples of polar gas molecules are HCl, ammonia and water.

9 Mullis9 Gas Behavior ► Particles in a gas are very far apart. ► Each gas particle is largely unaffected by its neighbors. ► Gases behave similarly at different pressures and temperatures according to gas laws. ► To identify a gas that is “most” ideal, choose one that is light, nonpolar and a noble gas if possible. ► Ex: Which gas is most likely to DEVIATE from the kinetic molecular theory, or is the “least” ideal: N 2, O 2, He, Kr, or SO 2 ? Answer: sulfur dioxide due to relative polarity and mass.

10 Mullis10 Boyle’s Law ► Pressure goes up if volume goes down. ► Volume goes down if pressure goes up. ► The more pressure increases, the smaller the change in volume.

11 Mullis11 Boyle’s law ► Pressure is the force created by particles striking the walls of a container. ► At constant temperature, molecules strike the sides of container more often if space is smaller. V 1 P 1 = V 2 P 2 ► Squeeze a balloon: If reduce volume enough, balloon will pop because pressure inside is higher than the walls of balloon can tolerate.

12 Mullis12 Charles’ Law ► As temperature goes up, volume goes up. ► Assumes constant pressure. V 1 = V 2 T 1 T 2 T, V

13 Mullis13 Charles’ law ► As temperature goes up volume goes up. ► Adding heat energy causes particles to move faster. ► Faster-moving molecules strike walls of container more often. The container expands if walls are flexible. ► If you cool gas in a container, it will shrink. ► Air-filled, sealed bag placed in freezer will shrink.

14 Mullis14 Gay-Lussac’s Law ► As temperature increases, pressure increases. ► Assumes volume is held constant. P 1 = P 2 T 1 T 2 ► A can of spray paint will explode near a heat source. ► Example is a pressure cooker.

15 Mullis15 Combined Gas Law ► In real life, more than one variable may change. If have more than one condition changing, use the combined formula. ► In solving problems, use the combined gas law if you know more than 3 variables. V 1 P 1 = V 2 P 2 V 1 P 1 = V 2 P 2 T 1 T 2 T 1 T 2

16 Mullis16 Using Gas Laws ► Convert temperatures to Kelvin! ► Ensure volumes and/or pressures are in the same units on both sides of equation. ► STP = 0° C and 1 atm. ► Use proper equation to solve for desired value using given information. V 1 P 1 = V 2 P 2 V 1 = V 2 P 1 = P 2 V 1 P 1 = V 2 P 2 V 1 P 1 = V 2 P 2 V 1 = V 2 P 1 = P 2 V 1 P 1 = V 2 P 2 T 1 T 2 T 1 T 2 T 1 T 2 T 1 T 2 T 1 T 2 T 1 T 2

17 Mullis17 Gay Lussac’s law of combining volumes of gases ► When gases combine, they combine in simple whole number ratios. ► These simple numbers are the coefficients of the balanced equation. N 2 + 3H 2 2NH 3 ► 3 volumes of hydrogen will produce 2 volumes of ammonia

18 Mullis18 Avogadro’s Law and Molar Volume of Gases ► Equal volumes of gases (at the same temp and pressure) contain an equal number of molecules. In the equation for ammonia formation, 1 volume N 2 = 1 molecule N 2 = 1 mole N 2 ► One mole of any gas will occupy the same volume as one mole of any other gas ► Standard molar volume of a gas is the volume occupied by one mole of a gas at STP. Standard molar volume of a gas is 22.4 L.

19 Mullis19 Sample molar volume problem ► A chemical reaction produces 98.0 mL of sulfur dioxide gas at STP. What was the mass, in grams, of the gas produced? ***Turn mL to L first! (This way, you can can use 22.4 L) 98 mL 1 L 1 mol SO 2 64.07g SO 2 = 0.280g SO 2 1000 mL 22.4 L 1 mol SO 2 1000 mL 22.4 L 1 mol SO 2

20 Mullis20 Sample molar volume problem 2 What is the volume of 77.0 g of nitrogen dioxide gas at STP? 77.0 g NO 2 1 mol NO 2 22.4 L = 37.5 L NO 2 46.01g NO 2 1 mol NO 2 46.01g NO 2 1 mol NO 2

21 Mullis21 Ideal Gas Law ► Mathematical relationship for PVT and number of moles of gas PV = nRT n = number of moles R = ideal gas constant P = pressure V = volume in L T = Temperature in K R = 0.0821 if pressure is in atm R = 8.314 if pressure is in kPa R = 62.4 if pressure is in mm Hg

22 Mullis22 Sample Ideal Gas Law Problem ► What pressure in atm will 1.36 kg of N 2 O gas exert when it is compressed in a 25.0 L cylinder and is stored in an outdoor shed where the temperature can reach 59°C in summer? V = 25.0 LT = 59+273 = 332 K P = ? R = 0.0821 L-atm n = 1.36 kg converted to moles mol-K ► 1.36 kg N 2 O 1000 g 1 mol N 2 O = 30.90 mol N 2 O 1 kg 44.02 g N 2 O 1 kg 44.02 g N 2 O ► PV = nRT ► P = 30.90 mol x 0.0821 L-atm x 332 K= 33.7 atm 25.0 L mol-K

23 Mullis23 Volume-Volume Calculations ► Volume ratios for gases are expressed the same way as mole ratios we used in other stoichiometry problems. N 2 + 3H 2 2NH 3 Volume ratios are: 2 volumes NH 3 3 volumes H 2 2 volumes NH 3 3 volumes H 2 1 volume N 2 1 volume N 2

24 Mullis24 Sample Volume-Volume Problem ► How many liters of oxygen are needed to burn 100 L of carbon monoxide? 2CO + O 2 2CO 2 100 L CO 1 volume O 2 = 50 L O 2 2 volume CO

25 Mullis25 Sample Volume-Volume Problem 2 Ethanol burns according to the equation below. At 2.26 atm and 40° C, 55.8 mL of oxygen are used. What volume of CO 2 is produced when measured at STP? C 2 H 5 OH + 3O 2 2CO 2 + 3H 2 O Number moles oxygen under these conditions is? PV = nRT: 2.26 atm (.0558 L ) = n = 0.0049 mol O 2 ( 0.0821 L-atm )(313K ) ( 0.0821 L-atm )(313K ) mol-K mol-K 0.0049 mol O 2 2 mol CO 2 22.4 L =0.073 L CO 2 3 mol O 2 1 mol CO 2 3 mol O 2 1 mol CO 2

26 Mullis26 Gas Densities and Molar Mass ► Need units of mass over volume for density (d) ► Let M = molar mass (g/mol, or mass/mol) PV = nRT M PV = M nRT M P/RT = n M /V M P/RT = mol(mass/mol)/V M P/RT = density M = dRT P

27 Mullis27 Sample Problem: Density 1.00 mole of gas occupies 27.0 L with a density of 1.41 g/L at a particular temperature and pressure. What is its molecular weight and what is its density at STP? M.W. = 1.41 g|27.0 L = 38.1 g___ L|1.0 molmol L|1.0 molmol M = dRT d= M P = 38.1 g (1 atm)______________ = 1.70 g/L PRT mol (0.0821 L-atm )(273K) PRT mol (0.0821 L-atm )(273K) ( mol-K ) ( mol-K ) OR…AT STP: 38.1 g | 1 mol = 1.70 g/L mol | 22.4 L mol | 22.4 L

28 Mullis28 Example: Molecular Weight A 0.371 g sample of a pure gaseous compound occupies 310. mL at 100. º C and 750. torr. What is this compound’s molecular weight? n=PV = (750 torr)(.360L) = 0.0116 mole RT62.4 L-torr(373 K) RT62.4 L-torr(373 K) mole-K mole-K MW = x g_= 0.371 g = 32.0 g/mol mol0.0116 mol mol0.0116 mol

29 Mullis29 Partial Pressures ► Gas molecules are far apart, so assume they behave independently. ► Dalton: Total pressure of a mixture of gases is sum of the pressures that each exerts if it is present alone. P t = P 1 + P 2 + P 3 + …. + P n P t = (n 1 + n 2 + n 3 +…)RT/V = n i RT/V ► Let n i = number of moles of gas 1 exerting partial pressure P 1 : P 1 = X 1 P 1 where X 1 is the mole fraction (n 1 /n t ) P 1 = X 1 P 1 where X 1 is the mole fraction (n 1 /n t )

30 Mullis30 Collecting Gases Over Water ► It is common to synthesize gases and collect them by displacing a volume of water. ► To calculate the amount of a gas produced, correct for the partial pressure of water: ► P total = P gas + P water ► The vapor pressure of water varies with temperature. Use a reference table to find.

31 Mullis31 Kinetic energy ► The absolute temperature of a gas is a measure of the average* kinetic energy. ► As temperature increases, the average kinetic energy of the gas molecules increases. ► As kinetic energy increases, the velocity of the gas molecules increases. ► Root-mean square (rms) speed of a gas molecule is u. ► Average kinetic energy, ε,is related to rms speed: ε = ½ mu 2 where m = mass of molecule *Average is of the energies of individual gas molecules.

32 Mullis32 Maxwell-Boltzmann Distribution ► Shows molecular speed vs. fraction of molecules at a given speed ► No molecules at zero energy ► Few molecules at high energy ► No maximum energy value (graph is slightly misleading: curves approach zero as velocity increases) ► At higher temperatures, many more molecules are moving at higher speeds than at lower temperatures (but you already guessed that) Just for fun: Link to mathematical details: http://user.mc.net/~buckeroo/MXDF.html http://user.mc.net/~buckeroo/MXDF.html Source: http://www.tannerm.com/maxwell_boltzmann.htm

33 Mullis33 Molecular Effusion and Diffusion ► Kinetic energy ε = ½ mu 2 ► u = 3RT Lower molar mass M, higher rms speed u M Lighter gases have higher speeds than heavier ones, so diffusion and effusion are faster for lighter gases.

34 Mullis34 Graham’s Law of Effusion ► To quantify effusion rate for two gases with molar masses M 1 and M 2 : r 1 = M 2 r 2 M 1 ► Only those molecules that hit the small hole will escape thru it. ► Higher speed, more likely to hit hole, so r 1 /r 2 = u 1 /u 2

35 Mullis35 Sample Problem: Molecular Speed Find the root-mean square speed of hydrogen molecules in m/s at 20º C. 1 J = 1 kg-m 2 /s 2 R = 8.314 J/mol-K R = 8.314 kg-m 2 /mol-K-s 2 u 2 = 3RT = 3(8.314 kg-m 2 /mol-K-s 2 )293K M 2.016 g |1 kg___ M 2.016 g |1 kg___ mol |1000g u 2 = 3.62 x 10 6 m 2 /s 2 u = 1.90 x 10 3 m/s

36 Mullis36 Example: Using Graham’s Law An unknown gas composed of homonuclear diatomic molecules effuses at a rate that is only 0.355 times that of O 2 at the same temperature. What is the unknown gas? r x = M O2 0.355 = 32.0 g/mol r O2 M x 1 M x Square both sides: 0.355 2 = 32.0 g/mol M x M x = 32.0 g/mol = 254 g/mol  Each atom is 127 g, 0.355 2 so gas is I 2 0.355 2 so gas is I 2

37 Mullis37 The van der Waals equation ► Add 2 terms to the ideal-gas equation to correct for 1.The volume of molecules (V-nb) 2.Molecular attractions (n 2 a/V 2 ) Where a and b are empirical constants. P + n 2 a (V – nb) = nRT V 2 V 2 ► The effect of these forces—If a striking gas molecule is attracted to its neighbors, its impact on the wall of its container is lessened.


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