Honors Chemistry-B

Chapter 14 - Physical Characteristics of GasesAnd Molecular Composition of Gases

Chapter Objectives

After students have completed Chapters 14, they will be able to:

The Kinetic-Molecular Theory of Matter

-The kinetic Molecular Theory can help us understand the behavior of gas molecules and the physical properties of gases.

-It is based on a model called an IDEAL GAS.

-Ideal gas is an imaginary gas that perfectly fits all the assumption of the kinetic-molecular theory.

The Kinetic-Molecular Theory of gases is based on the following five assumptions:

1. Gases consist of large numbers of particles that are far apart relative to their sizes.2. Collisions between gas particles and between particles and container walls are elastic

collisions.+Elastic Collisions: there is no net loss of kinetic energy-energy is just transferred between the particles but is not converted to any other type of energy.

3. Gas particles are in continuous, rapid, random motion4. There no forces of attraction or repulsion between gas particles5. The average kinetic energy of gas particles depends on the temperature of gases. Ideal gases:

- follow the kinetic molecular theory of gases+ no intermolecular forces+ elastic collisions

- molecules of an ideal gas occupy no volume.+ this means they are infinitely compressible, will never become a liquid under high pressures.

No real gas behaves exactly like an ideal gas, but some are more ideal than others.

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Objectives:

State the Kinetic Molecular theory of matter. And describe how it explains certain properties of matter.

Describe the conditions under which a real gas deviates from “ideal” behavior.

Reading 413-418 428-429

Conceptual problems-1) Are the collisions of pool balls on a pool table elastic?

If the collisions were elastic the kinetic energy must be conserved. Therefore, kinetic energy being the energy of motion, the pool balls would never stop moving. We know this doesn’t happen so the collisions of pool balls must not be elastic collisions. These collisions would be referred to as inelastic collisions.

2) Which of these real gases would be the most like an ideal gas.

Gas Boiling Point (oC)Water 100

Carbon Dioxide -78Xenon -108.12Helium -268.93

To be like an ideal gas a gas would have to:- have no attraction to other molecules, which means it would unlikely to be a liquid or

solid at any temperature.- The gas with lowest B.P. has the least attraction for other molecules. Therefore,

Helium will be most like an ideal gas and water will be the least ideal of the molecules.

Development of the Kinetic Theory of Gases

Focus on three measurable properties of gases.

V olume - amount of space taken up by a substance.should be expressed in Liters (L). The volume of 1 mole of a gas at STP is 22.4 dm3. STP mean 1 atm of pressure and 0 oC

P ressure is defined as a force applied over an area.should be expressed in atm, kPa, mmHg, or torr. kPa is the SI unit of Pressure.

Standard pressure is:1.0 atm = 101.3 kPa = 760 mmHg = 760 torr

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In a gas pressure is caused by collisions.

Each time a gas hits the walls of the container a force is applied to the area of the wall. This is pressure. Atmospheric pressure is the pressure being exerted by all of the air pushing down from above a location. More collisions = more pressure. Less collisions = less pressure.

Problems: 560 mm Hg to atm

560 mm Hg x 1 atm / 760 mm Hg = 0.739 atm

3.45 atm to kPa

3.45 atm x 101.3 kPa / 1 atm = 349.5 kPa

Measuring Pressure

A barometer measures the pressure of the outside air by comparing the force the air exerts on the mercury to the force exerted by the weight of column downwards ( weight = mass x acceleration due to gravity. Note that the space above the mercury in the tube is vacuum, essentially no matter is present in this space. When the pressure outside equals the downward force of the weight the barometer stabilizes. Thus the amount of pressure can be related to the height of a column of mercury and this is where the units of mm Hg for pressure come from. The higher the outside pressure, the more of the column will be filled. The lower the outside pressure the lower the height of the mercury column.

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Manometer

A manometer is a tool for the comparison of pressures of two different gases. Typically, the apparatus has one end open to the air, this side has atmospheric pressure pushing down on it. The other side has the pressure of the gas from the sample chamber. The level of liquid on each side relates to the pressure exerted on that side. (think see-saw) The side with more pressure will push downward with more force and thus the liquid on its side will be lower and the liquid on the lower pressure side will be higher in the U tube. (no, not that U tube the chemistry U tube). If the pressures are equal then the levels of mercury will be equal on both sides of the U tube.

Conceptual Problem:a) Where is there greater pressure at sea level or on top of Pikes Peak, Colorado? (elevation 4302 m above sea level) WHY?

At sea level there are more molecules of air above the surface of the earth so there are more collisions and there is more pressure pushing down. At Pikes Peak there are less molecules of air = less collisions so there is less pressure.

b) When you blow up a balloon it achieves a certain size and then remains constant. What pressure forces dictate the size of a balloon?

- There are two sources of pressure acting on a balloon; the pressure outside the balloon pushing inwards (smaller volume) and the pressure inside the balloon pushing out (larger volume.) The balloon stabilizes at the size where Pinside = Poutside.

T emperature is the average kinetic energy of the molecules in a sample.Should always be expressed in Kelvin (K). Standard temperature is 273 K.oC + 273 = KK - 273 = oC

Problems:443 K is how many oC?

443 K = 273 + oC170 oC

-15 oC is how many K?

K = 273 + (-15 oC)K = 258 K

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The Gas Laws3 Scientists that studied the behavior of gases

1. Boyle - Related pressure and volume2. Charles - Related volume and temperature3. Gay-Lussac - Related pressure and temperature

Boyle’s Law: Pressure and the Volume of a Gas

Volume and pressure of a gas are inversely related if temperature remains constant.

Notice as the piston descends the gas particles get closer and closer, this results is more collisions which means more pressure as volume decrease.

P1V1 = P2V2

Conceptual problem:

A) If we take a balloon from sea level to pikes peak (2403 m above sea level) what will happen to the size of the balloon?

The amount of gas in the balloon is constant. The amount of gas outside the balloon is decreasing. As the amount of gas outside the balloon decreases there is less collisions on the outside of the balloon. On the inside of the balloon there is no change in collisions initially. The balloon will expand until the pressure inside and outside are again equal. The volume of the balloon will increase.Math Problem:

If 1.5 L of CO2 gas at STP is added to a 0.750 L flask. What is the new pressure?

P1 = 1.0 atmV1 = 1.5 L

P2 = ?V2 = 0.750 L

P1V1 = P2V2

1.0 atm X 1.5 L = P2 X 0.750 L

P2 = 2.0 atm

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Objectives:

Understand the ideal gas laws formulated by Boyle, Charles, Gay-Lussac, Dalton, and Graham and solve problems based on these laws.

Use the kinetic theory to explain the theoretical basis for the gas laws.

Show how the absolute temperature (Kelvin) scale is derived.

Reading: 418-425

Charles’s Law: Volume-Temperature Relationship

Volume and temperature of a gas are directly related if pressure remains constant.

In both parts of this diagram the gas is at the same pressure, but you can see that the temperature has risen from –65°C to 250°C. The gas particles move faster at higher temperatures and collide more often, but the gas particles now take up a larger volume. This results in no change in pressure. According to Charles's law, when the temperature of a gas is increased at constant pressure, its volume also increases.

V1 V2

---- = ----T1 T2

Conceptual Problem

A balloon is filled with very hot air and then allowed to cool. What will happen to the volume of the balloon if the pressure in the balloon stays constant?

As the temperature of the hot air in the balloon decreases the rate of collisions between the gas molecules and the walls decrease. In order for the pressure to stay constant the volume must also decrease so that the fewer collision can still cause the same pressure.

If a balloon, with a volume of 1.0 L, is heated from 15oC to 25oC, what is the new volume?

V1 = 1.0 LT1 = 15oC = 288 K

V2 = ?T2 = 25oC = 298 K

V1 V2

---- = ----T1 T2

1.0 L V2

--------- = ---------288 K 298 K

V2 = 1.03 L

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Development of the Kelvin Scale

The Kelvin scale was developed based on Charles law. According to this law as the temperature decreases so does the volume. What Lord Kelvin (real name William Thomson) did was he took many data points at different temperatures and volumes for the same system. Then he graphed his data points and calculated the x intercept (temperature) to determine what would be the temperature if the volume was 0. The answer was -273.15 oC, which became the basis of the Kelvin scale.

Gay-Lussac’s Law: Gay Lussac’s Relationship

Pressure and temperature of a gas are directly related if volume remains constant.

P1 P2

---- = ----T1 T2

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If a fixed volume of a gas at STP is heated to 100oC, what is the new pressure?

P1 = 1.0 atmT1 = 273 K

P2 = ?T2 = 100oC=373 K

P1 P2

---- = ----T1 T2

1.0 atm P2 --------- = ---------273 K 373 K

P2 = 1.37 atm

The Combined Gas Law

Combined Gas Law

Combines Boyle’s LawCharles’s LawGay-Lussac’s Law

Allows for the calculation of systems where one of the three variables cannot be held constant.This is much more useful for real systems as it is often impossible to hold one of the variables constant.

P1V1 P2V2

-------- = -------- T1 T2

Given 500.0 mL of a gas at 15oC and 98 Kpa, what would be the new volume at STP?

P1 = 98 KpaV1 = 500.0 mLT1 = 15oC = 288 K

P2 = 101.3 KpaV2 = ?T2 = 273 K

P1V1 P2V2

------- = -------T1 T2

98 Kpa*500.0 mL 101.3 Kpa*V2

------------------------ = -------------------288 K 273 K

V2 = 458.5 mL

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Mass-Volume Problems at Non-Standard Conditions

Example-1:

Calculate the volume of H2 gas at STP that can be produced by 4.5g of Mg reacting with HCl.(Step 1 – Stoichiometry – Convert grams to volume at STP)(STP: Standard Temperature and pressure: 273K, 1atm and at 1mol volume is 22.4dm3)

Mg + 2 HCl H2 + MgCl2

4.5g Mg 1 mol Mg 1 mol H2 22.4 dm3 H2

x ------------- x ------------- x ---------------- = 4.2 dm3 H2 at STP 24g Mg 1 mol Mg 1 mol H2

Calculate the volume of the H2 gas if the temperature is 25oC and the pressure is 97 Kpa.(Step 2 - Combined gas law - convert volume at STP to non-STP)

P1 = 101.3 KpaV1 = 4.2 dm3

T1 = 273 K

P2 = 97 KpaV2 = ?T2 = 25oC = 298 K

P1V1 P2V2

------- = -------T1 T2

101.3 Kpa * 4.2 dm3 97 Kpa * V2

-------------------------- = ---------------273 K 298 K

V2 = 4.8 dm3

at 25oC and 97 Kpa

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Objectives:

Solve mixed mass-volume problems when there are non-standard conditions of temperature and pressure.

Use the ideal gas law

to solve gas-law problems.

Reading: 418-425

Example 2:

If 10.0 dm3 of H2 at 10.0oC and 10.0 Kpa reacts with excess N2 how many grams of NH3 (ammonia) can be produced?

Step 1 – Combined Gas Law – Convert volume to STPP1 = 10.0 KpaV1 = 10.0 dm3

T1 = 10.0oC = 283 K

P2 = 101.3 KpaV2 = ?T2 = 273 K

P1V1 P2V2 10.0 Kpa * 10.0 dm3 101.3 Kpa * V2

------- = ------- --------------------------- = --------------------T1 T2 283 K 273 K

V2 = 0.95 dm3 at STP

Step 2 – Stoichiometry – Convert Volume at STP to grams

3 H2 + N2 2 NH3

0.95 dm3 H2 1 mol H2 2 mol NH3 17g NH3

x ---------------- x -------------- x ------------ = 0.48g NH3 22.4 dm3 H2 3 mol H2 1mol NH3

Example 3:

Hints:

If given grams and want volume at non-STP: 1. Stoichiometry (Convert grams to volume at STP)2. Combined gas law (Convert volume at STP to non-STP)

If given volume at non-STP and want grams: 1. Combined gas law (Convert volume at non-STP to STP)2. Stoichiometry (Convert volume at STP to grams)

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Dalton’s Law of Partial Pressure

A partial pressure is the pressure an individual gas exerts in a mixture.Total pressure of a mixture of gas is equal to the sum of the partial pressures.

PT = P1 + P2 + P3 + …

This idea is often related with mole fraction because the fraction of moles of any one gas to the total moles of gas must be the same ratio as the partial pressure of a gas to the total pressure of the gas mixture.

xi = ni / ntotal must equal % gas = Pi / Ptotal or % gas = Vi / Vtotal

This is because we assume that all these molecules behave as ideal gases. Therefore, regardless of type, each molecules of gas behaves in an identical fashion. Through substitution we can see that

Pi = Ptotal xi Pressure of the individual gas = total pressure x mole fraction

Collecting a gas over water (bubbling it through water)Anytime a gas is collected over water a quantity of water is evaporated in the air and collected with the gas. The pressure caused by the evaporated water vapor is only dependent on the temperature.

PT = Pgas + PH2O

See water vapor pressure: Reference table in the back of your homework packet.

Problems:

1. If a mixture of two gases has a total pressure of 200 Kpa and a partial pressure of O2 of 150 Kpa.

a. Calculate the partial pressure of the 2nd gas in the mixture.

200 kPa = 150 kPa + P2

50 kPa = P2

b. Calculate the % O2 by volume in the mixture.

150 kPa x 100 / 200 kPa = 75 % O2 by volume

c. What is the mole fraction of the O2 ?Must be the same ratio as the ratio between the POxygen and the Ptotal.

Objectives:

Use Dalton’s law

Reading: 432 - 434

Therefore the mole fraction must be 0.75 = x oxygen

2. A mixture of gases has a total pressure of 120.0 Kpa. The gas is composed of 3.2 x 1023 molecules of He, 16g of Ne, and 0.70 moles of CO. Calculate the partial pressure of CO.

3.2 x 1023 molecules He x 1 mol He / 6.02 x 1023 molecules He = 0.532 mol He

16 g Ne x 1 mol Ne / 20.18 g Ne = 0.793 mol Ne

mole A + mole B + mole C = mole total0.70 mol CO + 0.793 Ne + 0.532 mol He = 2.03 mol = xtotal

PCO = 120.0 kPa (0.70 mol CO / 2.03 mol total) = 41.4 kPa

Chapter 11: Molecular Composition of Gases

Sec. 11.1: Measuring and Comparing the Volumes of Reacting Gases

Molar Volume of Gases: Volume occupied by one mole of a gas as STP is 22.4liters

The Ideal Gas Law

Allows us to calculate pressure, volume, number of moles of gas and temperature for any condition, not just STP.

PV = nRT

P = pressure (Kpa)V = volume (L)n = # of moles (mol)

R = Gas constantT = Temperature (K)

R = 8.31 Kpa L mol K

R= 0.0821 L atm mol K

Conceptual Problem

When will a gas deviate most from the behavior of an ideal gas? At low temperatures and high pressures gases deviate from ideal gas

because as the temperature goes down molecules move slower and the attractions between molecules become more relevant. As well, at high pressures the distance between the molecules must be lower. This means that the volume of the gas molecules is not as negligible causing the gas to behave least like an ideal gas.

Objectives:

Use the ideal law to solve gas law problems.

Reading: 426-427

Notice that this happens as a gas gets close to its boiling/condensing point.

1. Calculate the volume of 1.0 mole of a gas at STP.

STP means P = 1 atm, T = 273 K, n = 1 mol, R = 0.0821 L atm/ mol K

PV = nRT 1 atm ( V ) = 1 mol ( 0.0821 L atm mol-1 K-1) (273 K)

V = 22.4 L

Now we know where the molar volume number comes from !!

2. What would be the volume of 15.0g of CO2 at 100oC and 150.0 Kpa?

15.0 g CO2 x 1 mol CO2 / 44 g CO2 = 0.341 mol CO2

100 oC + 273 = 373 K 150.0 kPa x ( 1 atm/ 101.3 kPa) = 1.48 atm

1.48 atm ( V ) = 0.341 mol ( 0.0821 L atm/ mol K) (373 K)

V = 7.06 L

3. How many moles of a gas will a 1250 cm3 flask hold at 35oC and 95.4 Kpa?

1250 cm3 x 1 L / 1000 cm3 = 1.250 L 35 oC + 273 K = K 308 K = K

95.4 kPa x 1 atm / 101.3 kPa = 0.942 atm

1.250 L ( 0.942 atm ) = n (0.0821 L atm / mol K ) ( 298 K)

0.048 mol = n

Finding Molar Mass or Density from the Ideal Gas Law

Other versions of the Ideal Gas Law

PV = nRT n = mass in grams (m) molecular mass (M.M.)

PV = m RT M.M.

Solve for M.M.

M.M. = mRT

Objectives:

Solve mixed mass-volume problems when there are non-standard conditions of temperature and pressure.

Use the ideal gas law

to solve gas-law problems.

Reading: 418-425

PV

Gas Density Calculations

M.M. = m RT and D = m _ V P v

Notice D can replace m/v in the equation giving the following equation

M.M. = DRT P

Solved for D:

D = P (M.M.)

RT

1. Calculate the molecular mass of a gas of 500.0 cm3 has a mass of 1.00g at -23.0oC and 105 Kpa.

23.0 oC + 273 = K296 = K

M.M. = 1.0 g ( 8.31 L kPa/mol K) 296 K = 46.7 g/ mol105 kPa ( 0.500 L)

2. Calculate the density of 1.50 mol of CO2 gas at STP.

1atm, 273 K, 1.50 mol, MMco2 = 44.0 g/mol

D = 1 atm ( 44.0 g/mol) mol K0.0821 L atm ( 273 K)

D = 1.96 g/L

3. Calculate the molecular mass of a gas if 1500 cm3 has a mass of 4.00 g at -23.0oC and 204 Kpa.

1500 cm3 x 1 L / 1000 cm3 = 1.5 L 23.0 oC + 273 = 296 K

MM = mRT/ PV

MM = 4.00 g (8.31L kPa / mol K) 296 K = 32.24 g/mol1.5 L ( 204 kPa)

Effusion and Diffusion

Graham’s Law of Diffusion

Diffusion: Spontaneous spreading of a gas. Diffusion continues until gases are evenly spread throughout the space.

Effusion: diffusion of a gas through a small hole. (think pin hole leak)

Gases spread with relation to their velocity as described by their kinetic energy. But kinetic energy is calculated with KE = ½ m v2. This means two different gases with the same kinetic energies are not traveling at the same rate, the gas with more mass moves at a slower rate and the gas with the smaller mass moves at a faster rate.

Results: gases at lower masses diffuse faster.

Graham’s Law: Gases diffuse at a rate inversely proportional to the square roots of their densities (or molar masses) at constant temperature and pressure.

1 √D2 1 √M.M.2

---- = ---- ---- = --------2 √D1 2 √M.M.1

Conceptual: Two bottles of perfume A and B are opened in a room (A has a molar mass twice that of B), the perfumes immediately go into the gas state, your on the opposite side of the room which will you smell first?

The lighter gas diffuses faster, therefore B should diffuse faster and be smelled before A.

Problems:1. Calculate the ratio of rates of diffusion of nitrogen gas and carbon dioxide gas.

a. If the time of diffusion of N2 is 5 min., what would the time of diffusion be for CO2?

(28 g/mol)1/2 = 0.798 is the rate of CO2 to rate of N2 with CO2 (44 g/mol)1/2

Objectives:

Use graham’s law to discuss effusion and diffusion of gases

Discuss the effect of mass of a gas on effusion.

Reading: 435-436

N2 diffuses faster than CO2 so 0.798 X = 5min 6.27 min = X

Ideal gas law stoichiometry: (Return of the Ice box)

Two gases are mixed, hydrogen and oxygen, in a 1:1 ratio in a 500 mL container. The resulting mixture is sealed in a container that has 2.00 mL of H2O (l) at the bottom. If the temperature is 23 oC and the total pressure is 550 mm Hg before a spark is initiated calculate the limiting reactant for this problem and state the pressure exerted of the excess reactant under the same temperature.

Step 1: finding moles of reactants and product:

Ptotal = PO2 + PH2 + PH2O (vapor) Since it’s a 1 to 1 ratio PO2 = PH2

Ptotal = x + PH2O (vapor) 550 mm Hg = x + 21.1 mm Hg

X = 521 mm Hg (The pressure of each gas is 521 mm Hg / 2 )

521 mm Hg = x atm x= 0.686 atm760 mm Hg 1 atm

Once we have the pressure of the dry gas in, atm, we can solve for the moles using PV = nRT

0.686 atm ( 0.500 L) = n ( 0.0821 L atm mol-1 K-1) ( 296 K)

ntotal = 0.0141 mol gas Which means ntotal / 2 = nO2 ( and nH2)

7.05 x 10-3 mol O2 and 7.05 x 10-3 mol H2

We also have water in our starting mix so we should find moles of that, too.

2 mL H2O = 2 g H2O based on density of water is 1 g/mL

2 g H2O x 1 mol H2O / 18.0 g H2O = 0.1111 mol H2O

Now we can ICE BOX2 H2 (g) + O2 (g) 2 H2O (l)

Initial 7.05 x 10-3 7.05 x 10-3 0.1111

Change - 7.05 x 10-3 -3.525 x 10-3 + 7.05 x 10-3

Final 0 mol 3.525 x 10-3 mol .1182 mol

Therfore: H2 is limiting and O2 is in excess by 3.25 x 10-3 mol.

To get P of O2 remaining use PV= nRT

0.500 L ( P ) = 3.525 x 10-3 mol ( 0.0821 L atm mol-1 K-1) (296 K)

PO2 = 0.171 atm = 130 mm Hg

What is the total pressure of the container? (Hint it is not the pressure of the O2)

Ptotal = PO2 + PH2O Ptotal = 130 mm Hg + 21.1 mm Hg = 151 mm Hg

Is there liquid water remaining at the bottom of the container when the reaction is complete or would it all evaporate at this temperature?

If all the water evaporated there would be:

0.500 L ( P ) = 0.1182 mol ( 0.0821 L atm mol-1 K-1) 296 K = 5.74 atm of pressure from the water alone!!! This is 4366 mm Hg, clearly all of the water did not evaporate so the air in our container must be saturated with water vapor and there is a significant portion of liquid water remaining in the container.

Boyle’s Law P1V1 = P2V2

Charles’s LawV1 V2

---- = ----T1 T2

Gay-Lussac’s Law

P1 P2

---- = ----T1 T2

Combined Gas Law

P1V1 P2V2

-------- = --------T1 T2

Ideal Gas LawPV = nRT

Density

D = P (M.M.) RT

Graham’s Law

1 √MM2

---- = ----2 √MM1

Vapor Pressure Data for H2OTemperature kPa Temperature kPa0 0.61129 50 12.3441 0.65716 51 12.972 0.70605 52 13.6233 0.75813 53 14.3034 0.81359 54 15.0125 0.8726 55 15.7526 0.93537 56 16.5227 1.0021 57 17.3248 1.073 58 18.1599 1.1482 59 19.02810 1.2281 60 19.93211 1.3129 61 20.87312 1.4027 62 21.85113 1.4979 63 22.86814 1.5988 64 23.92515 1.7056 65 25.02216 1.8185 66 26.16317 1.938 67 27.34718 2.0644 68 28.57619 2.1978 69 29.85220 2.3388 70 31.17621 2.4877 71 32.54922 2.6447 72 33.97223 2.8104 73 35.44824 2.985 74 36.97825 3.169 75 38.56326 3.3629 76 40.20527 3.567 77 41.90528 3.7818 78 43.66529 4.0078 79 45.48730 4.2455 80 47.37331 4.4953 81 49.32432 4.7578 82 51.34233 5.0335 83 53.42834 5.3229 84 55.58535 5.6267 85 57.81536 5.9453 86 60.11937 6.2795 87 62.49938 6.6398 88 64.95839 6.9969 89 67.49640 7.3814 90 70.11741 7.784 91 72.82342 8.2054 92 75.61443 8.6463 93 78.49444 9.1075 94 81.46545 9.5895 95 84.52946 10.094 96 87.68847 10.62 97 90.94548 11.171 98 94.30149 11.745 99 97.759

100 _______ = _______ mm Hg

Source: Handbook of Chemistry and Physics: 73rd Edition (1992-93)