2012 general chemistry i 1 chapter 4. the properties of gases 2012 general chemistry i the nature of...
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2012 General Chemistry I 1
Chapter 4.THE PROPERTIES OF GASES
2012 General Chemistry I
THE NATURE OF GASES
THE GAS LAWS
4.1 Observing Gases4.2 Pressure4.3 Alternative Units of Pressure
4.4 The Experimental Observations4.5 Applications of the Ideal Gas Law4.6 Gas Density4.7 The Stoichiometry of Reacting Gases4.8 Mixtures of Gases
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THE NATURE OF GASES (Sections 4.1-4.3)
4.1 Observing Gases4.1 Observing Gases
Many of physical properties of gases are very similar, regardless of the identity of the gas. Therefore, they can all be described simultaneously.Samples of gases large enough to study are examples of bulk matter – forms of matter that consist of large numbers of molecules
Compressibility – the act of reducing the volume of a sample of a gasExpansivity - the ability of a gas to fill the space available to it rapidly
Two major properties of gases:
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4.2 Pressure4.2 Pressure
-
- SI unit of pressure, the pascal (Pa)
- Pressure arises from the collisions of gas molecules on the walls of the container.
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Barometer – A glass tube, sealed at one end, filled with liquid mercury, and inverted into a beaker also containing liquid mercury (Torricelli)
where h = the height of a column, d = density of liquid, and g = acceleration of gravity (9.80665 ms-2)
Measurement of Pressure
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Manometer
-Two types of Hg manometer:(a) open-tube and (b) closedtube system
This is a U-shaped tube filled with liquid and connected to an experimental system, whose pressure is being monitored.
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Self-Test 4.1B
The density of water at 20 oC is 0.998 g.cm-3. What heightwould the column of liquid be in a water barometer at 20 oCwhen the atmospheric pressure corresponds to 760. mm ofmercury?
Solution
P = dwhwg = dHghHgg
Hence hw =dHghHg
dw
=(13.595 g cm-3)(760. mm)
(0.998 g cm-3)
= 1.04 x 104 mm (10.4 m)
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Self-Test 4.2A
What is the pressure in kilopascals in a system when themercury level in the system-side column in an open-tubemercury manometer is 25 mm lower than the mercury levelin the atmosphere-side column and the atmospheric pressurecorresponds to 760. mm of mercury at 15 oC?
P = dhg = (13595 kg m-3)(0.785 m)(9.80665 m s-1)
= 104657 Pa = 105 kPa
Solution
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4.3 Alternative Units of Pressure4.3 Alternative Units of Pressure
- 1 bar = 105 Pa = 100 kPa- 1 atm = 760 Torr = 1.01325×105 Pa (101.325 kPa)- 1 Torr ~ 1 mmHg
mbar
Weather map
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Self-Test 4.3A
The US National Hurricane Center reported that the eyeof Hurricane Katrina (2005) fell as low as 902 mbar. Whatis the pressure in atmospheres?
1 atm = 1.01325 bar, hence 0.920 bar =
Solution
0.920 bar x 1 atm
1.01325 bar= 0.890 atm
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THE GAS LAWS (Sections 4.4-4.6)
4.4 The Experimental Observations4.4 The Experimental Observations
Boyle’s law: For a fixed amount of gas at constant temperature, volume is inversely proportional to pressure.
This applies to an isothermal system (constant T) with a fixed amount of gas (constant n).
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- For isothermal changes between two states (1 and 2),
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Self-Test 4.4B
In a petroleum refinery a 750.-L container containing ethylenegas at 1.00 bar was compressed isothermally to 5.00 bar. Whatwas the final volume of the container?
Solution
Isothermally means at constant temperature, hence Boyle'slaw can be used.
P1V1 = P2V2, (1.00 bar)(750. L) = (5.00 bar)V2
V2 = 150. L
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Charles’s law: For a fixed amount of gas under constant pressure, the volume varies linearly with the temperature.
This applies to an isobaric system (constant P) with a fixed amount of gas (constant n).
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- Kelvin temperature scale
T = 0 K = -273.15 oC,
when V → 0.
- Celsius temperature scale
t (oC) = T (K) - 273.15
0 oC = 273.15 K
The Kelvin Scale of Temperature
If a Charles’ plot of V versus T (at constant P and n) is extrapolated to V = 0, the intercept on the T axis is ~-273 oC.
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Another aspect of gas behavior (Gay-Lussac’s Law)
This applies to an isochoric system (constant V) with a fixed amount of gas (constant n).
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Self-Test 4.5A
A rigid oxygen tank stored outside a building has a pressureof 20.00 atm at 6:00 am when the temperature is 10. oC. Whatwill be the pressure in the tank at 6:00 pm, when the temperatureis 30. oC?
Solution
Volume is constant, hence Gay-Lussac's Law can be used.
P1
T1
= P2
T2
(20.00 atm)
(283.15 K)
= P2
(303.15 K)
P2 = 21.41 atm
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Avogadro’s Principle
Under the same conditions of temperature and pressure, a given number of gas molecules occupy the same volume regardless of their chemical identity.
- This defines molar volume
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Self-Test 4.6A
A helium weather balloon was filled at -20. oC and acertain pressure to a volume of 2.5 x 104 L with 1.2 x 103
mol He. What is the molar volume of helium under theseconditions?
Solution
Molar volume =Volume
No. moles= (2.5 x 104 L)
(1.2 x 103 mol)
= 21 L mol-1
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This is formed by combining the laws of Boyle, Charles, Gay-Lussac and Avogadro.
The ideal gas law:
Gas constant, R = PV/nT.It is sometimes called a “universal constant” andhas the value 8.314 J K-1 mol-1 in SI units, althoughother units are often used (Table 4.2).
The Ideal Gas Law
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-The ideal gas law, PV = nRT, is an equation of state that summarizes the relations describing the response of an ideal gas to changes in pressure, volume, temperature, and amount of molecules; it is an example of a limiting law.(it is strictly valid only in some limit: here, as P 0.)
Table 4.2. The Gas Constant, R
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- Standard ambient temperature and pressure (SATP)
298.15 K and 1 bar, molar volume at SATP = 24.79 L·mol-1
- Standard temperature and pressure (STP)
0 oC and 1 atm (273.15 K and 1.01325 bar)
- Molar volume at STP
- For conditions 1 and 2,
- Molar volume
4.5 Applications of the Ideal Gas Law4.5 Applications of the Ideal Gas Law
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EXAMPLE 4.4
In an investigation of the properties of the coolant gas used in an air-conditioning system, a sample of volume 500 mL at 28.0 oC wasfound to exert a pressure of 92.0 kPa. What pressure will the sampleexert when it is compressed to 30 mL and cooled to -5.0 oC?
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Self-Test 4.7B
A idling, badly tuned automobile engine can release asmuch as 1.00 mol of CO per minute into the atmosphere.At 27 oC, what volume of CO, adjusted to 1.00 atm, isemitted per minute?Solution
This question requires the use of PV = nRT, whereP = 1.00 atm, n = 1.00 mol (min-1), and T = 300.15 K.(R = 8.206 x 10-2 L atm K-1 mol-1)
V = (1.00 mol (min-1)(0.08206 L atm K-1 mol-1)(300.15 K)
(1.00 atm)= 24.6 L (min-1)
Calculating the pressure of a given sample
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Self-Test 4.8A
A sample of argon gas of volume 10.0 mL at 200. Torr isallowed to expand isothermally into an evacuated tubewith a volume of 0.200 L. What is the final pressure of theargon in the tube?
P1V1
T1
P2V2
T2
Solution
The volume is increased by a factor of 20, so we expect adecrease in pressure by the same factor, under isothermalconditions.
n1 n2
= , where T1 = T2 and n1 = n2(reduces to Boyle's Law)
(200. Torr)(10.0 mL) = P2(Torr)(200 mL)
P2 = 10.0 Torr
Using the combined gas law when one variable is changed
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Self-Test 4.9A
A parcel of air (the technical term in metereology for asmall region of the atmosphere) of volume 1.00 x 103 Lat 20. oC and 1.00 atm rises up the side of a mountainrange. At the summit, where the pressure is 0.750 atm,the parcel of air has cooled to -10. oC. What is the volumeof the parcel of air at that point?
Solution
P1V1
n1T1=
P2V2
n2T2
where n1 = n2
(1.00 atm)(1.00 x 103 L)
(293.15 K)= (0.750 atm)V2(L)
(263.15 K)
V2 = 1.20 x 103 L
Using the combined gas law when two variables are changed
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Self-Test 4.10A
Calculate the volume occupied by 1.0 kg of hydrogen at 25 oCand 1.0 atm.
Solution
We can use the ideal gas equation PV = nRT, after first findingthe number of moles of H2 in 1.0 kg.
n = Mass
Molar mass
1.0 x 103 g
2.016 g/mol= = 496 mol
(1.0 atm)V(L) = (496 mol)(8.206 x 10-2 L atm K-1mol-1)(298.15 K)
V = 1.2 x 104 L
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4.6 Gas Density4.6 Gas Density
Molar concentration of a gas at STP (where molar volume is 22.4141 L):
Molar concentration of a gas is the number moles divided by the volumeoccupied by the gas.
Density, however, does depend on the identity of the gas.
This value is the same for all gases, assuming ideal behavior.
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Density at STP
• For a given P and T, the greater the molar mass, the greater its density.
• At constant T, the density increases with P. In this case, P is increased either by adding more material or by compression (reduction of V).
• Raising T allows a gas to expand at constant P, increases V and therefore reduces its density.
Gas Density Relationships
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Self-Test 4.11A
The oil produced from eucalyptus leaves contains the volatileorganic compound eucalyptol. At 190. oC and 60 Torr, a sampleof eucalyptol vapor had a density of 0.320 g L-1. Calculate themolar mass of eucalyptol.
Solution
Density =MP
RTwhere M is molar mass
0.320 g L-1 = M(g mol-1)(60 Torr)
(62.324 L Torr K-1 mol-1)(463.15 K)
M = 154 g mol-1
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-Molar volumes of gases are generally > 1000 times those of liquids and solids. e.g. Vm (gases) = ~ 25 L mol-1; Vm (liquid water) = 18 mL mol-1
-Reactions that produce gases from condensed phases can be explosive.
e.g. sodium azide (NaN3) for air bags
4.7 The Stoichiometry of Reacting Gases4.7 The Stoichiometry of Reacting Gases
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EXAMPLE 4.6
The carbon dioxide generated by the personnel in the artificial atmosphere of submarines and spacecraft must be removed form the air and theoxygen recovered. Submarine design teams haveinvestigated the use of potassium superoxide, KO2,as an air purifier because this compound reacts withcarbon dioxide and releases oxygen:
4 KO2 (s) + 2 CO2(g) → 2 K2CO3(s) + 3 O2(g)
Calculate the mass of KO2 needed to react with 50 Lof CO2 at 25 oC and 1.0 atm.
Vm = 24.47 Lmol-1; 1 mol CO2 -> 2 mol KO2; MKO2 = 71.10 gmol-1
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Solution
6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g)
Self-Test 4.12A
Calculate the volume of carbon dioxide, adjusted to 25 oC and 1.0 atm,that plants need to make 1.00 g of glucose, C6H12O6, by photosynthesisin the reaction
From the equation, the stoichiometry of CO2:glucose is 6:1.The molar mass of glucose is 180 g/mol.The molar volume of CO2 at 25 oC and 1 atm is 24.47 L mol-1
Volume of CO2 =24.47 L
1 mol CO2
x6 mol CO2
1 mol glucosex 1.00 g glucose
180 g glucose
1 mol glucosex = 0.82 L
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4.8 Mixtures of Gases4.8 Mixtures of Gases
- A mixture of gases that do not react with one another behaves like a single pure gas.
Partial pressure: The total pressure of a mixture of gases is the sum of the partial pressures of its components. (John Dalton).
P = PA + PB + … for the mixture containing A, B, …
- Humid gas: P = Pdry air + Pwater vapor (Pwater vapor = 47 Torr at 37 oC)
mole fraction: the number of moles of molecules of the gas expressed as a fraction of the total number of moles of molecules in the sample.
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EXAMPLE 4.7
Air is a source of reactants for many chemical processes. To determinehow much air is needed for these reactions, it is useful to know thepartial pressures of the components. A certain sample of dry air oftotal mass 1.00 g consists almost entirely of 0.76 g of nitrogen and0.24 g of oxygen. Calculate the partial pressures of these gases whenthe total pressure is 0.87 atm.
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Self-Test 4.14A
A baby with a severe bronchial infection is in respiratory distress.The anesthetist administers heliox, a mixture of helium and oxygenwith 92.3% by mass O2. What is the partial pressure of oxygen beingadministered to the baby if the atmospheric pressure is 730 Torr?
n(He)0.077 g
4.00 g mol-1n(O2) =
0.923 g
32.0 g mol-1=
Solution
= 0.0193 mol = 0.0288 mol
x(He) =0.0193 mol
0.0481 mol= 0.401
x(O2) =0.0288 mol
0.0481 mol= 0.599
[= 1.00 - 0.401 (only two components)]
P(O2) = x(O2)P = 0.599 x 730 Torr = 437 Torr
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Chapter 4.THE PROPERTIES OF GASES
2012 General Chemistry I
MOLECULAR MOTION
REAL GASES
4.9 Diffusion and Effusion4.10 The Kinetic Model of Gases4.11 The Maxwell Distribution of Speeds
4.12 Deviations from Ideality4.13 The Liquefaction of Gases4.14 Equations of State of Real Gases
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MOLECULAR MOTION (Sections 4.9-4.11)
4.9 Diffusion and Effusion4.9 Diffusion and Effusion
Diffusion: gradual dispersal of one substance through another substance
Effusion: escape of a gas through a small hole into a vacuum
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Graham’s law: At constant T, the rate of effusion of a gas is inversely proportional to the square root of its molar mass:
Strictly,Graham’s law relates to effusion, but it can also be used for diffusion.
- For two gases A and B with molar masses MA and MB,
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- Rate of effusion and average speed increase as the square root of the temperature:
Combined relationship: The average speed of molecules in a gas is directly proportional to the square root of the temperature and inversely proportional to the square root of the molar mass.
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Self-Test 4.15A
It takes 30. mL of argon 40. s to effuse through a porousbarrier. The same volume of vapor of a volatile compoundextracted from Caribbean sponges takes 120. s to effusethrough the same barrier under the same conditions. Whatis the molar mass of the compound?
Solution
Time for Ar to effuse
Time for unknown to effuse
M(Ar)
M(unknown)=
40 (s)
120 (s)
39.95 (g mol-1)
M(unknown)
M(unknown)
=
= 360 g mol-1
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4.10 The Kinetic Model of Gases4.10 The Kinetic Model of Gases
Kinetic molecular theory (KMT) of a gas makes four assumptions:
1. A gas consists of a collection of molecules in continuous random motion.
2. Gas molecules are infinitesimally small points.
3. The molecules move in straight lines until they collide.
4. The molecules do not influence one another except during collisions.
- Collision with walls: consider molecules traveling only in one dimensional x with a velocity of vx.
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The change in momentum (final – initial)of one molecule: 2mvx
All the molecules within a distance vxt of the walland traveling toward it will strike the wall during theInterval t.
If the wall has area A, all the particles in a volumeAvxt will reach the wall if they are traveling toward it.
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The number of molecules in the volume Avxt is thatfraction of the total volume V, multiplied by the totalnumber of molecules:
The average number of collisions with the wall duringthe interval t is half the number in the volume Avxt:
The total momentum change = number of collisions × individual molecule momentum change
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Force = rate of change of momentum = (total momentum change)/t
for the average value of <vx2>
Mean square speed:
Pressure on wall:
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where vrms is the root mean square speed,
or
- The temperature is proportional to the mean square speed of the molecules in a gas.- This was the first acceptable physical interpretation of temperature: a measure of molecular motion.
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EXAMPLE 4.7
What is the root mean square speed of nitrogenMolecules in air at 20 oC?
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Self-Test 4.16A
Solution
Estimate the root mean square of water molecules in the vaporabove boiling water at 100. oC.
Molar mass of water is 18.01 g mol-1 or 0.01801 kg mol-1.
From vrms = (3RT/M)1/2,
vrms =3 x (8.3145 J K-1 mol-1) x (373 K)
0.01801 kg mol-1
= 719 m s-1
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4.11 The Maxwell Distribution of Speeds4.11 The Maxwell Distribution of Speeds
v = a particle’s speed
N = the number of molecules with speeds in the range between v +v
N = total number of molecules; M = molar mass
f(v) = Maxwell distribution of speeds
For an infinitesimal range,
average speed
Maxwell derived equation 22, for calculating the fraction of gas moleculeshaving the speed v at any instant, from the kinetic model.
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- Molar mass (M) dependence:as M increases, the fraction of molecules withspeeds greater than a specific speed decreases.
- Temperature dependence:
as T increases, the fraction of molecules with speeds greater than a specific speed increases.
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REAL GASES (Sections 4.12-4.14)
4.12 Deviations from Ideality4.12 Deviations from Ideality
- Gases condense to liquids when cooled or compressed (attraction).- Liquids are difficult to compress (repulsion).
Deviation from ideal gases
- Deviations from the ideal gas law are significant at high pressures and low temperatures (where significant intermolecular interactions exist).
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Compression factor (Z): the ratio of the actual molar volume of the gas to the molar volume of an ideal gas under the same conditions.
For an ideal gas, Z = 1
Long range attractions; smaller Z, condensation of gases
Short range repulsions; larger Z, low compressibility of liquids and solids,finite molecular volume
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- For many gases, attractions dominate at low pressure (Z < 1), while repulsive interactions dominate at high pressure (Z > 1).
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4.13 The Liquefaction of Gases4.13 The Liquefaction of Gases
Joule-Thomson effect: when attractive forces dominate, a real gas cools as it expands.
-In this case expansion requires energy, which comes from the kinetic energy of the gas, lowering the temperature.
-The effect is used in some refrigerators and to effect the condensation of gases such as oxygen, nitrogen, and argon.
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- Linde refrigerator for the liquefaction of gases
i.e. Adiabatic cooling; temperature decrease under isentropic expansionof any gas (w 0)
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4.14 Equations of State of Real Gases4.14 Equations of State of Real Gases
Virial equation:
van der Waals equation:
or
pressure reduced due to attractions between pairs of molecules
–nb volume excluded since molecules cannot overlap
b volume excluded by 1 mol ~ molar volume in the liquid state
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Virial expansion of the van der Waals equation
At low particle densities
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Table 4.5 Van der Waals Parameters for some Common Gases
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Model of gas
1. A large number of gas molecules in ceaseless, random, and straight motion.
2. The average speed and the spread of speeds increase with T and decrease with m.
3. Molecules travel in straight lines until they collide with other molecules or the container wall.
4. Widely separated. Intermolecular forces have only a weak effect on the properties.
5. Repulsions increase the molar volume, whereas attractions decrease the molar volume.
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EXAMPLE 4.9 Refrigerant gas (a = 16.2 L2 atm mol–2, b = 0.084 L/mol), 1.50 mol in 5.00 L at 0 oC; Estimate the pressure.
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Self-Test 4.17A
Solution
A 10.0-L tank containing 25 mol of O2 is stored in a diving supplyshop at 25 oC. Use the data in table 4.5 and the van der Waalsequation to estimate the pressure in the tank.
From P = nRT/(V - nb) -an2/V2,
P =(25 mol) x (0.08206 L atm K-1 mol-1) x (298 K)
10.0 L (25 mol) x (3.19 x 10-2 L mol-1)_
_ (1.364 L2 atm mol-2) x(25 mol)2
(10.0 L)2
= 58 atm