Chapter 13 Gases - Lecture notes 3 PDF

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Bussey - Gases...

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Chapter 13: Properties of Gases • Kinetic energy • Pressure: the measurement of relative collisions of gas particles (direct result of kinetic energy) • Volume: measurement of the space occupied by the gas • Temperature: relative measurement of the kinetic energy of particles (Kelvin) • Pressure and Volume are inversely proportional P ↑ V ↓ • Pressure and Temperature are directly proportional. P ↑ T ↑ • As temperature increases, volume increases in an elastic container o Temperature is a measure of relative kinetic energy of a substance. • Adding more gas to a system increases pressure and volume. o If we increase the moles of gas, there will be more collisions between gas molecules and they will occupy more space • The Ideal Gas Law o PV = nRT o Pressure * Volume = moles * R * Temp o What is an ideal gas? • Gases behave ideally at lower pressures • Characteristics of being ideal/assumptions made: ▪ Size of the gas molecules relative to the space they are occupying is negligible ▪ The gas molecules do not attract each other. o Units! • T = must be in Kelvin. (direct measure of kinetic energy) (+273) • V = Liters • P = atm, bar, Pa, mmHg, Torr [Pascal SI unit] [usually atm or bar] ▪ Units of pressure and conversion ▪ 1 atm = 1.01325 * 10^5 Pa ▪ 1 atm = 1.01325 bar ▪ 1 atm = 760 Torr ▪ 1 atm = 760 mmHg • R = multiple units ▪ Gas constant R • R = 8.314447 J/K*mol [pascals] • R = 0.0820575 L*atm/mol*K [atm] • R = 0.0831447 L*bar/mol*K [bar] o How do we calculate gas parameters under changing conditions? o State 1 vs State 2 • A balloon has a volume 1.2: on a warm 305K day. If the same balloon is put in a freezer at 255K, what will be the new volume ▪ PV = nRT ▪ V/T = V/T ▪

▪

1.2L

=

V

305K

=

255K

1.2*255/305

•

▪ V = 1.0L Nitrogen is often injected into cans and bottles to keep the contents from oxidizing. If the N2 is injected into a bottle w/ a volume of 10.0mL @ a pressure of 1.2 bar, what volume will it occupy @ a pressure of 0.95 bar? ▪ PV = nRT PV = PV ▪

10.0mL o

1.2 bar

==

V

.95 bar

▪ V = 13 mL Ideal Gas Calculations • PV = nRT • Calculate the number of grams of propane, C3H8 in a 50.0L container at a pressure of 7.50atm and 25deg C. ▪

50.0L

o

==

n

0.0820575 L*atm/mol*K

25+273

N = 15.335 mol C3H8 15.335 mol * 44.095g/mol = 676g Gas Mixtures • When 2 gases are mixed, it will affect the pressure, volume, and mole amounts of the mixture. ▪ When adding gases together: • Moles and pressure are additive [P1 + P2 = Ptotal] ▪ We can figure out the fraction of the moles present based on the pressure contributed by each gas • X, mole fraction / % of moles of gases in a mixture • Mole fractions and Partial pressures ▪ A gaseous mixture contains 431.0 torr H2, 337.1 torr N2, and 77.7 torr Ar. What is the mole fraction of each gas? • PV = nRT • N total = nH2 + nN2 + nAr • Ptotal = PH2 + PN2 + Par • 431 + 337.1 + 77.7 = 845.8 torr • PH2 = 431.0 / 845.8 = 0.510 (mole fraction) • PN2 = 337.1 / 845.8 = 0.3986 (mole fraction) • PAr = 77.7/845.8 = 0.0919 (mole fraction) ▪ A mixture of He and Ne gases at 245K contains 3x the # of He atoms as Ne atoms. The concentration of the mixture is 0.150 mol/L. What is the partial pressure of He, in bar? • P(1L) = (0.150)(0.0820575)(245) • Ptotal = 3.70 bar • 3.7/4 = 0.925*3 = 2.775 • Partial Pressure He = 2.775 bar Gas Stoichiometry • Cellular respiration occurs: ▪ C6H12O6 + 6O2 --> 6CO2 + 6H2O ▪ ▪

o

7.50atm

•

Calculate the volume of CO2 gas produced at 37deg C and 1.00bar when 1.00g of glucose is metabolized

•

1.00g glucose

1 mol glucose

6 mol CO2

180.15g

1 mol glucose

=

0.0333 mol CO2

V = nRT/P ▪ 0.0333 mol CO2 * 0.08206 (310K) / 1bar ▪ V = 0.858L Effusion of Gases • Rate of effusion: (how quickly a gas moves from high pressure to low pressure) ▪ Affected by size of gas and temperature ▪ Root mean square speed (looking at how fast a gas particle can move (effuse)) •

o

▪ ▪ ▪

•

M = kg/mol R = J/mol*K [pascal value] AVERAGE KINETIC ENERGY = (3/2) RT

Calculate the Vrms of an N2 gas molecule at 20deg C ▪

Vrms =

3(0.3145J/mol*K)(293)

^1/2

(0.280kg/mol) •

▪ = 511 m/s How to assess the effusion of two gases ▪ Rate A / Rate B = (MWb / MWa)1/2 ▪ Two balloons, one with He and one with N. If N2 leaks at a rate of 75ml/hr, what will be the rate for He? ▪ N2:A H2:B ▪

=

4.0028 g/mol H2

X H2

=

28.013 g/mol N2

75/x = (4.0028/28.013)^1/2 75/x = 0.378 = 198.413 mL/h = 2.0 x 10^2 mL/h Deviations from Ideality • Not all gases are ideal! • How do we know? Dry ice, liquid oxygen. ▪ ▪ ▪

o

75ml/h N2

^1/2

• • •

• •

Ideal definition: gas molecules are not attracted to one another, and is infinitely compressible Van Der Waal's Equation: (P + a (N^2/v^2) ( V - nb) = nRT

A and B are correctional values for non-ideal gases. Use the van der waals equation to calculate the pressure in bars exerted by 24.5g of NH3 gas in a 2.15L container at 300K. A = 4.3044, B = 0.037847 ▪

24.5g

1 mol

= 1.438

17.04g NH3 ▪ ▪ ▪

(P + (4.30)(1.438^2 / 2.15^2)) ( 2.15 - (1.438)(0.037847)) = 1.438(0.0831447)(300) P = 15.2 bar If we use PV = nRT, Pideal = 16.7 bar...