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## Chapter 5

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**Chapter 5**Gases**Properties of a gas**5.1 Pressure • Uniformly fills any container. • Mixes completely with any other gas • Exerts pressure on its surroundings. • Compressible • Gas pressure varies with altitudes and storms**Force (N)**P (Pa)= Area (m2) Measuring atmospheric pressureTorricellian barometer • Torricelli (1608-1647) studied the problem using mercury rather than H2O. • Mercury is denser than water, so the column wasn’t quite so high. Pascal (SI units) • Gas Pressure • Liquid Pressure = g ·h ·d**Pascal and Torricelli**Blaise Pascal (1623-1662) Evangelista Torricelli (1608-1647)**Barometer**760 mmHg atmospheric pressure P = d·g·h h d - density g - acc. of gravity atmospheric pressure**Units of Pressure**One atmosphere (1 atm) • Is the average pressure of the atmosphere at sea level • Is the standard atmospheric pressure Standard Atmospheric Pressure: 1 atm = 76 cm Hg = 760 mm Hg = 760 Torr = 101,325 Pa Very small unit, thus it is not commonly used**Example**A. What is 475 mm Hg expressed in atm? 485 mm Hg x 1 atm = 0.625 atm 760 mm Hg B. The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg? 29.4 psi x 1.00 atm x 760 mmHg = 1.52 x 103 mmHg 14.7 psi 1.00 atm**ManometerDevice for Measuring the Pressure of a Gas in a**Container**5.2 The Gas Laws of Boyle, Charles and Avogadro**• Boyle’s Law: PV = const • at constant n, T • Charles’ Law: V/T = const • at constant n, P • Avogadro’s Law: V/n = const • at constant P, T**Boyle’s Law**V Slope= 1/k PV =k (at constant T and n) P1V1 = P2V2**A Plot of PV Versus P for Several Gases at Pressure Below 1**atm Boyle’s holds Only at very Low pressures A gas strictly obeys Boyle’s law is called Ideal gas**V/T = b**• V = bT • (constant P & n) • V1/T1 = V2/T2 Charles’s Law**Plots of V Versus T(Celsius) for Several Gases**Volume of a gas Changes by When the temp. Changes by 1oC. I.e., at -273oC , V=0 ???**All gases will solidify or liquefy before reaching zero**volume.**Vn**• V = an • (constant P& T) Avogadro’s Law**5.3 The Ideal Gas Law**Boyle’s law Charles's law Universal gas constant Avogadro’s law**The Ideal Gas Law**PV = nRT R = 0.0821 atm L mol-1 K-1 The Ideal Gas Law can be used to derive the gas laws as needed!**Molar Volume**At STP 4.0 g He 16.0 g CH4 44.0 g CO2 1 mole 1 mole 1mole (STP) (STP) (STP) V = 22.4 L V = 22.4 L V = 22.4 L**The value of R**• What is R for 1.00 mol of an ideal gas at STP (25 oC and 1.00 atm)?Given that V of 1 mol of gas at STOP= 22.4L**Example**• A reaction produces enough CO2(g) to fill a 500 mL flask to a pressure of 1.45 atm at a temperature of 22 oC. How many moles of CO2(g) are produced? PV = nRT**Ideal Gas Law**• The ideal gas law is an equation of state. • Independent of how you end up where you are at. Does not depend on the path. • The state of the gas is described by: P, V, T and n • Given 3 you can determine the fourth. • Ideal gas equation is an empirical equation - based on experimental evidence.**Example**A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?**Data Table**Set up Data Table P1 = 0.800 atm V1 = 0.180 L T1 = 302 K P2 = 3.20 atm V2= 90.0 mL T2 = ?? ??**Solution**Solve for T2 Enter data T2 = 302 K x atm x mL = K atm mL T2 = K - 273 = °C**Calculation**Solve for T2 T2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180.0 mL T2 = 604 K - 273 = 331 °C**Example**A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?**Solution**T1 = 308 K T2 = ? V1 = 675 mL V2 = 0.315 L = 315 mL P1 = 0.850 atm P2 = 802 mm Hg = 646 mm Hg T2 = 308 K x 802 mm Hg x 315 mL 646 mm Hg 675 mL = 178 K - 273 = - 95°C**5.4 Gas Stoichiometry**• Reactions happen in moles • At Standard Temperature and Pressure (STP, 0ºC and 1 atm) 1 mole of gas occuppies 22.4 L. • If not at STP, use the ideal gas law to calculate moles of reactant or volume of product.**Example**A.What is the volume at STP of 4.00 g of CH4? 4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L 16.0 g CH4 1 mole CH4 B. How many grams of He are present in 8.0 L of gas at STP? 8.00 L x 1 mole He x 4.00 g He = 1.43 g He 22.4 He 1 mole He**Example**A 12.25 L cylinder contains 75.5 g of neon at 24.5 oC. Determine the pressure in the cylinder. P = nRT V PV = nRT = (3.74 mol)(0.082L•atm)(297.5K) (12.25 L) mol•K ? P = V = n = R = T = 12.25 L mol = 1.009 atm 75.5 g = mol 3.74 20.18 g = 5670 torr 0.082 L•atm mol•K 24.5 + 273 = 297.5 K**Example**• 30.2 mL of 1.00 M HCl are reacted with excess FeS. What volume of gas is generated at STP? STP means standard temperature and pressure . . . 0 oC and 1 atm. 2 HCl + FeS FeCl2 + H2S HCl + FeS FeCl2 + H2S Now . . . Go for moles,**2 HCl + FeS FeCl2 + H2S**PV = nRT V = nRT/P**Example**The decomposition of sodium azide, NaN3, at high temperatures produces N2(g). What volume of N2(g), measured at 735 mm Hg and 26°C, is produced when 70.0 g NaN3 is decomposed. 2 NaN3(s) → 2 Na(l) + 3 N2(g)**nRT**(1.62 mol)(0.08206 L atm mol-1 K-1)(299 K) V= = P 1.00 atm (735 mm Hg) 760 mm Hg 2 NaN3(s) → 2 Na(l) + 3 N2(g) Determine moles of N2: 1 mol NaN3 3 mol N2 = 1.62 mol N2 nN2 = 70 g NaN3 X X 65.01 g N3/mol N3 2 mol NaN3 Determine volume of N2: = 41.1 L**Molar mass of a gas**P x V = n x R x T • P x V = m x R x T M • m = mass, in grams • M = molar mass, in g/mol • Molar mass = m R T P V**Density**• Density (d) is mass divided by volume • P x V = m x R x T M • P = m x R x T V x M • d = m V • P = d x R x T M **Example**A glass vessel weighs 40.1305 g when clean, dry and evacuated; it weighs 138.2410 when filled with water at 25°C (d=0.9970 g cm-3) and 40.2959 g when filled with propylene gas at 740.3 mm Hg and 24.0°C. What is the molar mass of polypropylene? Volume of the vessel**Vflask =**= 98.41 cm3 = 0.09841 L mgas = mfilled- mempty = (40.2959 g – 40.1305 g) = 0.1654 g**m**RT PV = M (0.6145 g)(0.08206 L atm mol-1 K-1)(297.2 K) M = (0.9741 atm)(0.09841 L) m RT M = PV PV = nRT M = 42.08 g/mol**Example**Calculate the density in g/L of O2 gas at STP. From STP, we know the P and T. P = 1.00 atm T = 273 K Rearrange the ideal gas equation for moles/L d = PXM R x T**The density of O2 gas at STP is**1.43 grams per liter**m**RT M = PV Example • 2.00 g sample of SX6(g) has a volume of 329.5 Cm3 at 1.00 atm and 20oC. Identify the element X. Name the compound • P= 1.00 atm • T = 273+20 = 293K**= 146 g SX6 /mol**f Molar mass of(X6 )= 146- 32 = 114 g/mol Molar mass of X = (114 g/mol X6)/6 = 19 X = with a molar mass of 19 = F The compound isSF6**5.5 Dalton’s Law of Partial Pressures**• For a mixture of gases in a container, the total pressure is the sum of the pressure each gas would exert if it were alone in the container. • The total pressure is the sum of the partial pressures. • PTotal = P1 + P2 + P3 + P4 + P5 ... • For each gas:**Partial pressure**• Each component of a gas mixture exerts a pressure that it would exert if it were in the container alone • PTotal = n1RT + n2RT + n3RT +... V V V • In the same container R, T and V are the same. • PTotal = (n1+ n2 + n3+...)RT V Thus,**A 250.0 mL flask contains 1.00 mg of He and 2.00 mg of H2 at**25.0oC. Calculate the total gas pressure in the flask in atmospheres. The total pressure is due to the partial pressures of each of these gases. so: For He: 1.00 x 10-3 g He mol _____________________ = mol He 2.50 x 10-4 4.00 g For H2: 2.00 x 10-3 g H2 mol ______________________ = mol H2 9.92 x 10-4 2.016 g**A 250.0 mL flask contains 1.00 mg of He and and 2.00 mg of**H2 at 25.0oC. Calculate the total gas pressure in the flask in atmospheres. so: 1.00 x 10-3 g He mol _____________________ = mol He For He: 2.50 x 10-4 4.00 g For H2: 2.00 x 10-3 g H2 mol ______________________ = mol H2 9.92 x 10-4 2.016 g And: Ptotal = (2.50 x 10-4 + 9.92 x 10-4)(RT/V) = (0.001242 mol)(0.0821 L•atm)(25 + 273)K mol•K (0.2500 L) Ptotal= 0.1216 atm