A Gas
Uniformly fills any container. GASES Mixes completely with any other gas
Exerts pressure on its surroundings.
Pressure Figure 5.2: A • Force exerted per unit area of surface by torricellian barometer. The molecules in motion. tube, completely P = Force/unit area filled with mercury, is • 1 atmosphere = 14.7 psi inverted in a dish of mercury. • 1 atmosphere = 760 mm Hg (see Fig. 5.2) • 1 atmosphere = 101,325 Pascals • 1 Pascal = 1 kg/m.s2
Pressure
is equal to force/unit area Figure 5.3: Atmospheric SI units = Newton/meter2 = 1 Pascal (Pa) pressure from air 1 standard atmosphere = 101,325 Pa mass. P = gdh 1 standard atmosphere = 1 atm = To compare the height of
Hg and H2O columns at 1 atm pressure: Both = 1 atm, so 760 mm Hg = 760 torr
gdwaterhwater = gdHghHg, or
(hwater/hHg ) = (dHg/dwater)
1 Boyle’s Law* Figure 5.3: A simple manometer.
Pressure × Volume = Constant (T = constant)
P1V1 = P2V2 (T = constant) V ∝ 1/P (T = constant) (*Holds precisely only at very low pressures.)
As pressure increases, the volume
of SO2 decreases.
Figure 5.5: Figure 5.4: Plotting Boyle's data A J-tube from Table 5.1. (a) A similar to plot of P versus V shows the one used that the volume doubles by Boyle. as the pressure is halved. (b) A plot of V versus 1/P gives a straight line. The slope of this line equals the value of the constant k.
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2 A gas that strictly obeys Boyle’s Law is called an ideal gas.
Figure 5.6: A plot of PV versus P for several gases at pressures below A Problem to Consider 1 atm. • A sample of chlorine gas has a volume of 1.8 L at 1.0 atm. If the pressure increases to 4.0 atm (at constant temperature), what would be the new volume? using Pf × Vf = Pi × Vi
Pi × Vi (1.0 atm)× (1.8 L) Vf = = Pf (4.0 atm)
Vf = 0.45 L
Figure 5.7: Effect of temperature on a volume of gas. (A)
Charles’s Law Photo courtesy of James Scherer.
The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin. V = bT (P = constant) b = a proportionality constant
3 Figure 5.7: Effect of temperature on a volume of gas. (B)
Photo courtesy of James Scherer. Charles’s Law
V V 1 ==2 (P constant) T1 T2
Figure 5.9: Plots of V versus T as Figure 5.8: Plots of V versus T in Fig. 5.8 except here the Kelvin (ºC) for several gases. scale is used for temperature.
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A Problem to Consider The Empirical Gas Laws • A sample of methane gas that has a volume of 3.8 L at 5.0 oC is heated to 86.0 oC at constant pressure. • Gay-Lussac’s Law: The pressure exerted Calculate its new volume. by a gas at constant volume is directly V V proportional to its absolute temperature. using f = i T T f i P α Tabs (constant moles and V) V ×T (3.8L)(359K) or P V = P V V = i f = i i f f f T (278K) Ti Tf i P P f = i At constant n Vf = 4.9 L Tf Ti
4 A Problem to Consider • An aerosol can has a pressure of 1.4 atm at 25 oC. What pressure would it attain at 1200 oC, assuming the volume remained constant? P P using f = i Tf Ti P ×T (1.4atm)(1473K ) P = i f = f Ti (298K ) A. Pressure will decrease and will be lower that Pi B. Pressure will increase and will be higher that P Pf = 6.9atm i
Avogadro’s Law Figure 5.10: For a gas at constant temperature and These balloons pressure, the volume is directly proportional each hold 1.0L to the number of moles of gas (at low of gas at 25ºC pressures). and 1 atm. Each balloon contains V = an 0.041 mol of a = proportionality constant gas, or 2.5 x 22 V = volume of the gas 10 molecules. n = number of moles of gas
Standard Temperature The Empirical Gas Laws and Pressure
• Avogadro’s Law: Equal volumes of any “STP” two gases at the same temperature and P = 1 atmosphere pressure contain the same number of molecules. T = 0°C – The volume of one mole of gas is called the The molar volume of an ideal gas is 22.414 liters at STP molar gas volume, Vm. (See figure 5.10) – Volumes of gases are often compared at standard temperature and pressure (STP), chosen to be 0 oC and 1 atm pressure.
5 Figure 5.10: The molar volume of a gas. Photo courtesy of James Scherer.
Return to Slide 12
The Ideal Gas Law The Ideal Gas Law
• This implies that there must exist a • The numerical value of R can be derived using proportionality constant governing these Avogadro’s law, which states that one mole of any relationships. gas at STP will occupy 22.4 liters. – Combining the three proportionalities, we can VP obtain the following relationship. R = nT
nTabs (22.4 L)(1.00 atm) V ="R" ( P ) R = (1.00 mol)(273 K) where “R” is the proportionality constant L⋅atm referred to as the ideal gas constant. = 0.08206 mol⋅K
Ideal Gas Law Ideal Gas Law
PV = nRT R = proportionality constant An equation of state for a gas. = 0.08206 L atm Κ−1 mol−1 “state” is the condition of the gas at a given P = pressure in atm time. V = volume in liters PV = nRT n = moles T = temperature in Kelvins Holds closely at P < 1 atm
6 Molecular Weight Determination A Problem to Consider – An experiment calls for 3.50 moles of • In Chapter 3 we showed the relationship between moles and mass. chlorine, Cl2. What volume would this be if the gas volume is measured at 34 oC and 2.45 atm? mass nRT moles = molecular mass since V = P (3.50 mol)(0.08206 L⋅atm )(307 K) or then V = mol⋅K 2.45 atm n = m then V = 36.0 L Mm
Molecular Weight Figure 5.15: Finding the vapor density of a substance. Determination • If we substitute this in the ideal gas equation, we obtain PV = ( mass )RT Mm If we solve this equation for the molecular mass, we obtain mRT M = m PV
Gas Density Determination A Problem to Consider
• A 15.5 gram sample of an unknown gas occupied • If we look again at our derivation of the a volume of 5.75 L at 25 oC and a pressure of 1.08 molecular mass equation, atm. Calculate its molecular mass. PV = ( m )RT mRT Mm Since M = m PV we can solve for m/V, which represents (15.5 g)(0.08206 L⋅atm )(298 K) the gas density. then M = mol⋅K m m PM (1.08 atm)(5.75 L) = D = m V RT Mm = 61.0 g/mol
7 Figure 5.14: (a) One mole of N2(l) has a volume of approximately 35 mL and density of 0.81 g/mL. b) One mole of N2(g) has a volume of 22.4 L (STP) and a density of 1.2 x 10-3 g/mL. A Problem to Consider
Thus the ratio of the volumes of gaseous N2 and liquid N2 is 22.4/0.035 = 640 and the spacing of the molecules is 9 times farther • Calculate the density of ozone, O3 (Mm = 48.0g/mol), at o apart in N2(g). 50 C and 1.75 atm of pressure. PM Since D = m RT (1.75 atm)(48.0 g/mol) then D = L⋅atm (0.08206 mol⋅K )(323 K) D = 3.17 g/L
Dalton’s Law of Partial Pressures
For a mixture of gases in a container, A. All have the same number of atoms B. The flask with Xe because its M is greatest, so mass/V is the m P = P + P + P + . . . largest Total 1 2 3 C. The He flask - (PV/T) = constant If T increases, so must P D. All have the same number of moles - we can’t create or destroy mass!
Figure 5.17: An illustration of Dalton’s law of partial pressures. Collecting Gases “Over Water”
• A useful application of partial pressures arises when you collect gases over water.
• As gas bubbles through the water, the gas becomes saturated with water vapor. • The partial pressure of the water in this “mixture” depends only on the temperature.
Return to Slide 29
8 Figure 5.18: Collection of gas over water.
Zn (s) + 2 HCl (aq) º ZnCl2 (aq) + H2 (g)
A Problem to Consider A Problem to Consider
• Suppose a 156 mL sample of H gas was collected • Suppose a 156 mL sample of H gas was collected 2 2 over water at 19.5 oC and 769 mm Hg. What is the over water at 19 oC and 769 mm Hg. What is the mass of H2 collected? mass of H2 collected? – First, we must find the partial pressure of the – Table 5.6 lists the vapor pressure of water at 19 oC
dry H2. as 17.0 mm Hg. P = 769 mm Hg - 17.0 mm Hg P = P − P H2 H2 tot H2 0 P = 752 mm Hg H2
A Problem to Consider A Problem to Consider • Now we can use the ideal gas equation, along •From the ideal gas law, PV = nRT, you have with the partial pressure of the hydrogen, to PV (0.989 atm)(0.156 L) n = = determine its mass. L⋅atm RT (0.08206 mol⋅K )(292 K ) 1 atm PH = 752 mm Hg × 760 mm Hg = 0.989 atm 2 n = 0.00644 mol V = 156 mL = 0.156 L •Next,convert moles of H to grams of H . T = (19 + 273) = 292 K 2 2 2.02 g H2 n = ? 0.00644 mol H2 × = 0.0130 g H2 1 mol H2
9 P = P + P Figure 5.12: total A B Partial Pressures of Gas The partial Since P = (nRT/V) Mixtures pressure of we can substitute this into each gas in a Dalton’s Law. • The composition of a gas mixture is often described in terms of its mole fraction. mixture of ntotal(RT/V) = nA (RT/V) + n (RT/V) gases in a B •Themole fraction, χ , of a component gas is container the fraction of moles of that component in the Dividing by (RT/V): total moles of gas mixture. depends on the ntotal = nA + nB or 1 = (n /n ) + (n /n ) number of A total B total nA PA moles of that Finally, multiplying by Ptotal: χ A = Mole fraction of A = = P = (n /n )P + gas. total A total total ntot Ptot (nB/ntotal)Ptotal
Partial Pressures of Gas Mixtures • The partial pressure of a component gas, “A”, is then defined as
PA = χ A × Ptot A. No change to pressure of hydrogen gas • Applying this concept to the ideal gas equation, we find that each gas can be treated independently. B. They are the same (same V, T & P)
C. Ptotal = PH2 + PAr = 2 PH2 PAV = nART
A Problem to Consider
• Given a mixture of gases in the atmosphere at 760 torr, what is the partial pressure of o N2 (χ = 0 .7808) at 25 C? since P = χ × P N2 N2 tot then P = (0.7808) × (760 torr) N2 P = 593 torr N2 60
10 A. O 2 Practice problem 5.123
B. H2
C. Each has the same number of mole- cules
D. No change
E. Mole fraction = 1/4
so PH2= 0.25 of total P
2 C O (g) + O 2 (g) = 2 C O 2 (g) Kinetic Molecular Theory n CO = PV/RT = (0.500)(2.00)/(0.0821)(300) = 0.0406 m ol n O2 = PV/Rt = (1.00)(1.00)/(0.0821)(300) = 0.0406 m ol 1. Volume of individual particles is ≈ zero. C O is the lim iting rea ctant 2. Collisions of particles with container 0 .0406 m ol CO require 0.0203 m ole O 2 so walls cause pressure exerted by gas. t hat (0.0406 m ol - 0.0203 m ol) = 0.0203 m ol O 2 rem ain. 3. Particles exert no forces on each other.
P O2 = nRT/V = (0.0203)(0.0821)(300) = 0.167 4. Average kinetic energy ∝ Kelvin 3 .0 L atm temperature of a gas. P CO2 = nRT/V= (0.0406)().0821)(300) = 0.334 3 .0 L a tm
P to ta l = (0.167 + 0.334) atm = 0.501 atm
Figure 5.15: The effects of decreasing Figure 5.16: The effects of increasing the volume of a sample of gas at constant the temperature of a sample of gas at temperature and number of moles. constant volume and number of moles.
Boyle’s Law Gay-Lussac’s Law V 1/P % (Pi/Ti) = (Pf/Tf)
11 Figure 5.17: The effects of increasing Figure 5.18: The effects of increasing the temperature of a sample of gas at the number of moles of gas particles at constant pressure and number of moles. constant temperature and pressure.
Avogadro’s Law V % n Charles’ Law V % T
Figure 5.20: Figure 5.21: A plot of A plot of the relative number of the relative number of O molecules that 2 N2 molecules have a given velocity that have a given at STP. velocity at three temperatures.
Figure 5.19: Path of one particle in a gas. Any given particle The Meaning of Temperature will continuously change its course as a result of collisions with other particles, as well as with the walls of the container. 3 (KE) = RT avg 2
Kelvin temperature is an index of the random motions of gas particles (higher T means greater motion.)
12 Diffusion: describes the mixing of Figure 5.22: The effusion of a gases. The rate of diffusion is the gas into an evacuated chamber. rate of gas mixing.
Effusion: describes the passage of gas into an evacuated chamber.
Figure 5.23: Relative molecular Figure 5.24: (top) When HCl(g) and NH (g) meet in the tube, a speed distribution of H2 and UF6. 3 white ring of NH4Cl(s) forms. (bottom) A demonstration of the relative diffusion rates of NH3 and HCl molecules through air.
Molecular Speeds; Diffusion and Effusion • According to Graham’s law, the rate of effusion or diffusion is inversely proportional to the square root of its A. We use 1.0 mol of He for experiment X and 1.0 mol of Ar for molecular mass. (See Figure 5.22) experiment Y. If both valves are opened at the same time, which gas would you expect to reach the end of the long tube first? Rate of effusion of gas "A" Mm of Gas B Ans: He - it has a lower molar mass = B. If you wanted the Ar to reach the end of the tube at the same time Rate of effusion of gas "B" Mm of gas A as the He, what experimental condition could you change to make this happen? Ans: Heat the Ar
13 A Problem to Consider Effusion:
• How much faster would H gas effuse through an Rate of effusion for gas 1 M2 2 = opening than methane, CH ? 4 Rate of effusion for gas 2 M1 Rate of H M (CH ) 2 = m 4 Rate of CH4 Mm (H2 ) Diffusion: Rate of H 16.0 g/mol Distance traveled by gas 1 M 2 = = 2.8 = 2 Rate of CH4 2.0 g/mol Distance traveled by gas 2 M1
So hydrogen effuses 2.8 times faster than CH4
Figure 5.25: Plots of PV/nRT Figure 5.26: Plots of PV/nRT versus P for several gases (200 versus P for nitrogen gas at three K). temperatures.
Figure 5.27: (a) Gas at low concentration— relatively few interactions between particles. Figure 5.28: (b) Gas at high concentration—many more Illustration interactions between particles. of pairwise interactions among gas particles.
14 Figure 5.29: The volume taken up by the gas Real Gases particles themselves is less important at (a) large container volume (low pressure) than at (b) small container volume (high pressure).
2 []Paobs + (/nV ) ×−() VnbnRT= ↑ ↑ corrected pressure corrected volume
Pideal Videal
A Problem to Consider
• If sulfur dioxide were an “ideal” gas, the pressure at 0 oC exerted by 1.000 mol occupying 22.41 L would be 1.000 atm. Use the van der Waals equation to estimate the “real” pressure.
Given are the following values for SO2 a = 6.865 L2.atm/mol2 b = 0.05679 L/mol
87
A Problem to Consider A Problem to Consider
• First, let’s rearrange the van der Waals nRT n2a P = - equation to solve for pressure. V - nb V2 nRT n2a L⋅atm 2 L2 ⋅atm P = - (1.000 mol) (0.08206 )(273.2K) (1.000 mol) (6.865 2 ) 2 P = mol⋅K - mol V - nb V 22.41 L - (1.000 mol) (0.05679 L/mol) (22.41 L)2 . . R= 0.0821 L atm/mol K P = 0.989 atm T = 273.2 K a = 6.865 L2.atm/mol2 • The “real” pressure exerted by 1.00 mol of SO2 at V = 22.41 L b = 0.05679 L/mol STP is slightly less than the “ideal” pressure of 1.000 atm.
15 Real Gases
Must correct ideal gas behavior when at high pressure (smaller volume) and low temperature (attractive forces become important).
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