Gases. The Gas Laws Labs #18 Molar Mass of a Volatile Liquid #19 Calcium Carbonate Analysis: Molar Volume of Carbon Dioxide. Gases have general characteristics. Expansion : Expand indefinitely to fill the space available to them
The Gas Laws
#18 Molar Mass of a Volatile Liquid
#19 Calcium Carbonate Analysis: Molar Volume of Carbon Dioxide
Difference in heights of liquid levels inversely proportional to density of liquid and represents the pressure
Greater density of liquid, smaller difference in height
High density of mercury (13.6 g/mL) allows relatively small monometers to be built
Readings must be corrected for relative densities of fluid used and of mercury (mm Hg = mm fluid)
A sample of CH4 is confined in a water manometer. The temperature of the system is 30.0 °C and the atmospheric pressure is 98.70 kPa. What is the pressure of the methane gas, if the height of the water in the manometer is 30.0 mm higher on the confined gas side of the manometer than on the open to the atmosphere side. (Density of Hg is 13.534 g/mL).
X axis independent variable Y axis dependent variable
Temperature/# molecules (n) constant
A gas which has a pressure of 1.3 atm occupies a volume of 27 L. What volume will the gas occupy if the pressure is increased to 3.9 atm at constant pressure?
A gas at 30oC and 1.00 atm occupies a volume of 0.842 L. What volume will the gas occupy at 60.0oC and 1.00 atm?
When H burns in O, volume of H consumed is always exactly 2x volume of O
Direct relationship between pressure and temperature (P T)
P/T = constant
Pi/Ti = Pf/Tf
Avogadro's law predicts directly proportional relation between # moles of gas and its volume
Helped establish formulas of simple molecules when distinction between atoms and molecules was not clearly understood, particularly existence of diatomic molecules
Once shown that equal volumes of hydrogen and oxygen do not combine in manner depicted in (1), became clear that these elements exist as diatomic molecules and that formula of water must be H2O rather than HO as previously thought
A 5.20 L sample at 18.0oC and 2.00 atm pressure contains 0.436 moles of a gas. If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be?
Generalization applicable to most gases, at pressures up to about 10 atm, and at temperatures above 0°C.
Ideal gas’s behavior agrees with that predicted by ideal gas law.
Pressure (P) in atm
Volume (V) in L
Absolute temperature (T) in K (Charles’ Law)
Amount (number of moles, n)
R (universal ideal gas constant)
0.08206 liter ∙ atm/mole ∙ K or 0.08206 L·atm·K–1·mol–1
8.31 liter ∙ kPa/mole ∙ K
8.31 J/mole ∙ K
8.31 V ∙ C/mole ∙ K
8.31 x 10-7 g ∙ cm2/sec2 ∙ mole ∙ K
6.24 x 104 L ∙ mm Hg/mol ∙ K
1.99 cal/mol ∙ K
Given 3 of 4 variables and calculate 4th
Cancel all constants and R, make appropriate substitutions from given data to perform calculation
A sample containing 0.614 moles of a gas at 12.0oC occupies a volume of 12.9 L. What pressure does the gas exert?
Read 5.1-5.3, pp. 189-202
Q pp. 232-234, #29, 31-34, 44
A sample containing 15.0 g of dry ice, CO2(s), is put into a balloon and allowed to sublime according to the following equation: CO2(s) CO2(g) How big will the balloon be (what is the volume of the balloon) at 22oC and 1.04 atm after all of the dry ice has sublimed?
44.0 g CO2
0.500 L of H2(g) are reacted with 0.600 L of O2(g) according to the equation 2H2(g) + O2(g) 2H2O(g). What volume will the H2O occupy at 1.00 atm and 350oC?
A gas at 34.0oC and 1.75 atm has a density of 3.40 g/L. Calculate the molar mass (M.M.) of the gas.
Read 5.4, pp. 203-206
Q pp. 234-235, #52, 56, 58, 60
When two gases are mixed together, gas particles tend to act independently of each other
Pi/Pt = ni/nt
Pi = ni/nt x Pt
Pi = XiPt
Xi = mole fraction of gas component i
A volume of 2.0 L of He at 46oC and 1.2 atm pressure was added to a vessel that contained 4.5 L of N2 at STP. What is the total pressure and partial pressure of each gas at STP after the He is added?
0.293 mol 1.457 atm
0.293 mol 1.457 atm
Calculate the mole fraction of each gas.
nt = 3.00 mol + 4.00 mol + 1.00 mol = 8.00 mol
Xi = ni/nt
XHe = 3.00 mol/8.00 mol = 0.375
XAr = 4.00 mol/8.00 mol = 0.500
XNe = 1.00 mol/8.00 mol = 0.125
Calculate the partial pressure of each gas.
Pi = XiPi
PHe = 0.375 x 1,200 torr = 450. torr
PAr = 0.500 x 1,200 torr = 600. torr
PNe = 0.125 x 1,200 torr = 150. torr
He-1.2 L, 0.63 atm, 16oC Ne-3.4 L, 2.8 atm, 16oC
Pt = Pgas + PH2O
Consequently, partial pressure of collected gas is Pgas = Pt – PH2O, where partial pressure of water depends on temperature and corresponds to vapor pressure of water
What is the partial pressure of oxygen? Given the vapor pressure of water at 30°C is 31.8 mm Hg.
Pt = PO2 + PH2O
PO2 = Pt – PH2O = 645 mm Hg – 31.8 mm HgPO2 = 613 mm Hg
What are the mole fractions of oxygen and water vapor?
Recall that the partial pressures of O2 and H2O are related to their mole fractions, PO2 = XO2Pt PH2O = XH2OPt
X O2 = 613/645 = 0.950, and XH2O = 31.8/645 = 0.049
Note also that the sum of the mole fractions is 1.0, within the number of significant figures, given: X O2 + XH2O = 0.950 + 0.049 = 0.999
The vapor pressure of water in air at 28oC is 28.3 torr. Calculate the mole fraction of water in a sample of air at 28oC and 1.03 atm pressure.
Pair 783 torr
The safety air bags in cars are inflated by nitrogen gas generated by the rapid decomposition of sodium azide. If an air bag has a volume of 36 L and is to be filled with nitrogen gas at a pressure of 1.15 atm at a temperature of 26.0oC, how many grams of NaN3 must be decomposed?
2NaN3(s) 2Na(s) + 3N2(g)
3 mol N2 1 mol NaN3
Read 5.5, pp. 206-211
Q pg. 235, #64, 66, 68, 70, 72
Describes the properties of gases
Average KE, e, related to root mean square (rms) speed u
4 molecules in gas sample have speeds of 3.0, 4.5, 5.2, and 8.3 m/s.
Because mass of molecules does not increase, rms speed of molecules must increase with increasing temperature
Experiments show that molecules of a gas do not all move at same speed but are distributed over a range.
Highest point on curve marks most probable speed-some move with speeds much less than most probable speed, while others move with speeds much greater than this speed
As temp. increases, curve flattens out (more molecules have higher kinetic energies and therefore higher average speeds)
Gas molecules with higher molar masses will have slower average speeds, while lighter molecules will have higher average speeds
Rates two gases effuse through a pinhole in a container are inversely related to the square roots of the molecular masses of gas particles
Use equation to find relative rates of diffusion of gases, whether evacuated or not)
All gases expand to fill container
Heavier gases diffuse more slowly than lighter ones
KE-average kinetic energy
c = constant that is same for all gases
T = temperature in Kelvin
M = mass of gas
v2 = average of the square of the velocities of the gas molecules
Since cT will be the same for all gases at the same temperature, the average KE of any two gases at the same T will be the same
KE1 = KE2
½ m1V21 = ½ m2v22
root mean square velocity = vrms = √v2
√m1/m2 = vrms2/vrms1
Higher the T, higher vrms
Cooling gas decreases average KE of its molecules
All gases when cooled enough will condense to liquid
Suggests that intermolecular forces of attraction exist between molecules of real gases
Condensation occurs when average KE is not great enough for molecules to break away from intermolecular attractive forces
When they stick together, there are fewer particles bouncing around and creating P, so real P in nonideal situation will be smaller than P predicted by ideal gas equation
Developed modification of ideal gas law to deal with nonideal behavior of real gases
Must correct ideal gas behavior when at high pressure(smaller volume) and low temperature(attractive forces become important)
correction for small/finite volume
Using the ideal gas law
Using van der Waal’s equation
PV = nRT
(X)(0.2000L) = (0.3000 mol)().08206 L atm/K mol)(248) = 30.55 atm
From table 5.3, a = 0.0341 atm L2/mol2 and b = 0.0237 L mol
(P + 0.0341 x 0.30002/0.20002)(0.2000 – 0.3000 x 0.0237) = 0.3000(0.08206)(248)= 31.60 atm
Read 5.6-5.9, pp. 212-229
Q pp. 235-236, #74, 76, 78, 80, 82, 84, 86, 93
Do one additional exercise and one challenge problem.
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