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TRANSCRIPT
Chapter 5: Gasses
9/16
1. Gas volume changes greatly with pressure2. Gas volume changes greatly with temperature3. Gases have relatively low viscosity4. Most gases have relatively low densities under normal
conditions5. Gases are miscible
Elements as gases:Noble gases, Hydrogen, Nitrogen, Oxygen, Fluorine, Chlorine
Pressure= force
areaForce=Mass x Acceleration
Units of pressure1 Pascal (Pa) = 1 N/m^2 1 atm = 760 mmHg =760 torr1 atm= 101kPa (101,000)
Sea level= 1 atm4 miles= 0.5 atm10 miles=0.2 atm
*Table 5.1
Boyle’s law: Constant temperature=constant amount of gas
Linear, inverse P σ1/V PxV=k1 P1xV1=P2xV2 Inverse
Predicts how:Volume changes with pressurePressure changes with volume
A 946 mL sample of Cl2 is at 736mmHg. What is the pressure if the volume is reduced to 154 mL?
P1V1=P2V2
P1=736mmHg V1=946mL V2=154mL
P2= P1xV1 736mmHg x 946 mL P2 = = 4460mmHg V2 154 mL
P2=4460 mmHg
9/20
Temperature and volume*As T increases:V increases=direct relationship
Charles’s law: change of volume with temperature (constant P)Temperature must be in Kelvin V σ T V=k2 x T
V1/T1=V2/T2Predict:How volume changes with temperatureHow temperature changes with volume
A 3.20 L sample of CO is at 125°C. At what temperature will the CO occupy at 154 L?
125C=398.15K
V1/T1=V2/T2 V1=3.20 L T1=398.15K V2=154 L
T2= V2 x T1 = 1.54L x 398.15L =192K
V1 3.20 L
Amonton’s law: directly related P1 P2 x T1 T2
Predict:How temperature changes with pressureHow pressure changes with temperature
A sample of Co is at 25°C and 1atm. What will the pressure be at 100°C?
P1=1atm T1= 298.15K T2=373.15K
T2P1 373.15K x 1atmP2 = = = 1.25atm T1 298.15 K
Avogadro’s law: V σ n (n=#of moles)V=constant x nV1/n1=V2/n2
Predict:Constant temperatureConstant pressure
What volume of NO can be made from combusting 1L of NH3 in excess O2 (constant T and P)?
4NH3 + 5O2 à 4NO + 6H2O4 mole NH3à4 mole NO1 mole NH3à1 mole NOAt constant T and P1L NH3à1L NO *Only works with gases
Standard molar volume: Figure 5.7
Standard temperature and pressure (STP)
At STP (1atm, 0°C)1 mole of an ideal gas à 22.4L
Ideal gas law (combined gas law) (PV=nRT)P and V=inversely relatedP and T=directly relatedP and n=directly related
P1V1 P2V2 = n1T1 n2T2Boyle’s lawV= k1/P
Charles’s law
V=k2T
Avogadro’s lawV=k3n
A light bulb has Ar at 1.20atm and 18°C is heated to 85°C. What is the final pressure of the Ar?
P1V1 P2V2 = n1T1 n2T2
P1=1.20atm P2=? T1=291.15K T2=358.15K
1.20atm x 358.15 K = 1.48atm291.15 K
Relation to volume:We can change…Pressure: V=k1/PTemperature: V=Tk2Moles: V=nk3V=nTR/P or PV=nRT
*R is the gas constant
R=0.0821 Latm/Kmol
What is the volume of 49.8g of HCL acid at STP?PV=nRT
n=49.8g 1 mol HCL x = 1.37 mol
36.45 g HCl
(1atm)(?L)=(1.37 mol)(0.0821Latm/Kmol)(273.15K)
V= 30.6L
The density of a gas: density=m/v
Gas density is: Directly proportional to m (also n, P, and M)Inversely proportional to V (also T)
Density (d) Calculationsm=mass (g)d=m/vV=m/dSince V=nRT/P m/d=nRT/Pd=mP/nRTFINAL EQUATIONL à d= MP/RT
What is the density of CO2 at 0.990 atm and 55°C
D= MP/RT
d= (44.01g/mol)(0.990atm) d=1.62 g/L (0.8021 Latm/Kmol) (328K)
9/23
Molar mass (M) of a gaseous substance density=g/Ld= MP RT
M= dRT P
A 2.10 L jar contains 4.65 g of a gas at 1 atm and 27°C. What is the molar mass?
D=m/v 4.65g/2.10L=2.21g/L
M= 2.21g/L X 0.0821 Latm/Kmol X 300.15 K = 54.6 g/mol
1 atm
Mixture of gases:Gases mix homogenouslyEach behaves as if it were the only gas present
Consider a case in which two gases, A and B, are in a container of volume V.
PA= nART nA is the number of moles of A V
PB= nBRT nB is the number of moles of B VPT=PA + PB XA= nA XB= nB
nA + nB nA + nB
Pi=XiPT
Mole fraction (Xi) = ni/nT
A gas sample has 8.24 mol CH4, 0.421 mol of C2H6, and 0.116 mol C3H8. If total P=1.37 atm, what is the P of C3H3?
Pi=XiPT PT= 1.37 atm
XC3H8 = 0.116 = 0.0132 8.24 + 0.421 + 0.116
PC3H8= 0.0132 x 1.37 atm = 0.0181 atm
Gas Stoichiometry:Amount of reactant in grams/Làmoles of reactantàmoles of productàamount of product in grams/L
1. Balance the chemical equation2. Use ideal gas equation: L knownàmol known or
Molar mass: g knownà mol known3. Use balanced equation
Mol knownà mol unknown4. Use ideal gas equation
Mol unknownà L unknown orMolar mass: mol unknownàg known
What is the volume of CO2 produced at 37°C and 1 atm when 5.60 g of glucose are used up in the reaction?
C6H12O6 (s) + 6O2 (g)à 6CO2 (g) + 6H2O (l)g C6H12O6àmol C6H12O6àmol CO2àV CO2
5.60 C6H12O6 x 1 mol C6H12O6 6 mol CO2 X = 0.187 mol
CO2
180 g C6H12O6 1 Mol C6H12O6
V=nRT 0.187 mol x 0.0821 Latm/Kmol x 310.15 K = = 4.76 L P 1 atm
Kinetic Molecular theory of gases
Gas molecules:1. Have mass but have negligible volume2. Are in constant random motion3. Collisions are perfectly elastic 4. Neither attract nor repel one another5. The average kinetic energy (KE) is proportional to the Kelvin
temperatureTwo gases @ same temp. have the same average KE
Kinetic theory of gases and…Boyle’s law:
Collisions with wallàPLess Vàless surfaceàmore collisionsà more PPressure indirectly related to volume
Charles’s law:Higher Tàmore velocityàfaster collisionsàmore PPressure directly related to temperature
Avogadro’s law:More atomsàmore collisionsà more PPressure is directly related to number of moles
Dalton’s law: of Partial Pressures:Molecules do not attract or repel one anotherP exerted by one type of molecule is unaffected by the presence of another gas Ptotal= PΣ i
Calculating rootmeansquare speed
R=8.314 J/(Kmol)M = molar mass in kg/mol
Urms= √3RT M
What is the rootmeansquare velocity of He at 25°C?
√398.314J/(Kmol)(298K) 4.003 x 10^3 kg/mol
√1.86 x 10^6 J/kg √1.86 x 10^6 m^2/s^2
=1.36 x 10^3 m/s
9/25
EffusionGas escapes into a vacuumGraham’s law of effusionRate of effusion σ 1/√M
rate A √ MB
= rate B MA
Calculate the ratio of effusion rates between nitrogen and water vapor.
rate H2O √28.02
= = 1.247rate N2 18.02
Calculating molar mass:
MHe (rateHe)^2
= Mx
rate X
If the effusion ratio of an unknown gas is 9.378 relative to He, what is the molar mass of the unknown gas?
4.003d/mL x (9.378)^2 = Mx =352.0g/mol
Diffusion:Distribution of molecular speedsMean free pathCollision frequency
*Mixing of one gas with another. Light gases diffuse faster.
rate A √ MB
= rate B MA
Deviations from ideal behavior
Affect the intermolecular forces on pressure.
Atoms have volume!
Real gases appropriate ideal at…Low pressureHigh temperature
4.2 g of an unknown gas occupies a volume of 45 cm^3 at 25°C. The molar mass of the gas is 26.04 g/mol. What is the pressure of the gas?
P=nRT/V
n=4.2g 1 mol x = 0.16 mol
26.04 g
P=(0.16mol)(0.0821Latm/Kmol)(298.15K)/(0.045L) = 87atm
A sample of a gas has a pressure of 750 torr and a volume of 2.74 L at 0°C. If the temperature is changed to 200°C and the pressure to 780 torr, what is the final volume of the gas?
(754torr)(2.74 L) (780 torr)(V2) = = 4.56 L (273.15K) (473.15K)Nitrogen and hydrogen react to form ammonia. How many grams of ammonia can be formed when 25.0L of nitrogen (at 760 torr and 25°C) react with 58.5L of hydrogen (at 850 torr and 20°C)?
N2 + 3H2 à 2NH3
PV (760 torr x 1atm/760torr)(25.0L) = = 1.0 mol N2
RT (0.0821Latm/Kmol)(298.15K)
1.0 mol N2 2 mol NH3 17.25g NH3 x x 34.5g NH3
1 mol N2 1 mol NH3
PV (850 torr x 1atm/760torr)(58.5L) = = 2.7 mol H2RT (0.0821Latm/Kmol)(293.15K)
2.7 mol H2 2 mol NH3 17.25gNH3 x x = 31.05g NH3 3 mol H2 1 mol NH3