12 Gases and Kinetic Molecular Theory -...

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12 Gases and Kinetic Molecular Theory

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CHAPTER GOALS 1. Comparison of Solids, Liquids, and Gases 2. Composition of the Atmosphere and Some

Common Properties of Gases 3. Pressure 4. Boyle’s Law: The Volume-Pressure Relationship 5. Charles’ Law: The Volume-Temperature

Relationship; The Absolute Temperature Scale 6. Standard Temperature and Pressure 7. The Combined Gas Law Equation 8. Avogadro’s Law and the Standard Molar Volume

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CHAPTER GOALS 9. Summary of Gas Laws: The Ideal Gas Equation 10. Determination of Molecular Weights and Molecular

Formulas of Gaseous Substances 11. Dalton’s Law of Partial Pressures 12. Mass-Volume Relationships in Reactions Involving

Gases 13. The Kinetic-Molecular Theory 14. Diffusion and Effusion of Gases 15. Real Gases: Deviations from Ideality

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Comparison of Solids, Liquids, and Gases

• The density of gases is much less than that of solids or liquids.

Densities (g/mL)

Solid Liquid Gas

H2O 0.917 0.998 0.000588

CCl4 1.70 1.59 0.00503

Gas molecules must be very far apart compared to liquids and solids.

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Composition of the Atmosphere and Some Common Properties of Gases

Gas % by Volume N2 78.09 O2 20.94 Ar 0.93

CO2 0.03 He, Ne, Kr, Xe 0.002

CH4 0.00015 H2 0.00005

Composition of Dry Air

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Pressure

• Pressure is force per unit area. – lb/in2 – N/m2

• Gas pressure as most people think of it.

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Pressure

• Atmospheric pressure is measured using a barometer.

• Definitions of standard pressure – 76 cm Hg – 760 mm Hg – 760 torr – 1 atmosphere – 101.3 kPa

Hg density = 13.6 g/mL

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Boyle’s Law: The Volume-Pressure Relationship

• V ∝ 1/P or • V= k (1/P) or PV = k • P1V1 = k1 for one sample of a gas. • P2V2 = k2 for a second sample of a gas. • k1 = k2 for the same sample of a gas at

the same T. • Thus we can write Boyle’s Law

mathematically as P1V1 = P2V2

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Boyle’s Law: The Volume-Pressure Relationship

• Example 12-1: At 25oC a sample of He has a volume of 4.00 x 102 mL under a pressure of 7.60 x 102 torr. What volume would it occupy under a pressure of 2.00 atm at the same T?

( )( ) torr1520

mL 400 torr760PV PV

V PV P

2

112

2211

=

=

= 2211 V PV P =

( )( )

mL 1000.2 torr1520

mL 400 torr760PV PV

V PV P

2

2

112

2211

×=

=

=

=

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Boyle’s Law: The Volume-Pressure Relationship • Notice that in Boyle’s law we can use any

pressure or volume units as long as we consistently use the same units for both P1 and P2 or V1 and V2.

• Use your intuition to help you decide if the volume will go up or down as the pressure is changed and vice versa.

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Charles’ Law: The Volume-Temperature Relationship

0

5

10

15

20

25

30

35

0 50 100 150 200 250 300 350 400

Volume (L) vs.

Temperature (K)Gases liquefy before reaching 0K

absolute zero = -273.15 0C

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• Charles’s law states that the volume of a gas is directly proportional to the absolute temperature at constant pressure. – Gas laws must use the Kelvin scale to be

correct. • Relationship between Kelvin and

centigrade.

K = C + 273o

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• Mathematical form of Charles’ law.

kTVor kT = Vor TV =∝

form usefulmost in the TV

TV

so equal are sk' ehowever thk TV andk

TV

kTVor kT = Vor TV

2

2

1

1

2

2

1

1

=

==

=∝

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• Example 12-2: A sample of hydrogen, H2, occupies 1.00 x 102 mL at 25.0oC and 1.00 atm. What volume would it occupy at 50.0oC under the same pressure?

T1 = 25 + 273 = 298 T2 = 50 + 273 = 323

mL 108K 298

K 323mL 101.00=V

TTV=V

TV

TV

2

2

1

212

2

2

1

1

=

××

∴=

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Standard Temperature and Pressure

• Standard temperature and pressure is given the symbol STP. – It is a reference point for some gas

calculations. • Standard P ≡ 1.00000 atm or 101.3 kPa • Standard T ≡ 273.15 K or 0.00oC

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The Combined Gas Law Equation • Boyle’s and Charles’ Laws combined into one

statement is called the combined gas law equation. – Useful when the V, T, and P of a gas are changing.

2

2

1

12211 T

VTV VPVP

Law Charles' Law sBoyle'

==

2

22

1

11

2

2

1

12211

TV P

TV P k

TV P

:is law gas combined The :gas of samplegiven aFor TV

TV VPVP

Law Charles' Law sBoyle'

==

==

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The Combined Gas Law Equation

• Example 12-3: A sample of nitrogen gas, N2, occupies 7.50 x 102 mL at 75.00C under a pressure of 8.10 x 102 torr. What volume would it occupy at STP?

12

2112

21

21

21

T PT V P=Vfor Solve

torr760=P torr 810=PK 273=TK 348=T

?=V mL 750=V

( )( )( )( )( )

mL 627K 348 torr760

K 273mL 750 torr810T PT V P=Vfor Solve

torr760=P torr 810=PK 273=TK 348=T

?=V mL 750=V

12

2112

21

21

21

=

=

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• Example 12-4 : A sample of methane, CH4, occupies 2.60 x 102 mL at 32oC under a pressure of 0.500 atm. At what temperature would it occupy 5.00 x 102 mL under a pressure of 1.20 x 103 torr?

You do it!

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( )( )( )( )( )C1580 K 1852=

mL 260 torr380mL 500 torr1200K 305

V PV P T= T

? = T K 305 = T torr 380 =

torr1200 = P atm 0.500 = PmL 500 = V mL 260 = V

o11

2212

21

21

21

≈

=

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Avogadro’s Law and the Standard Molar Volume

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Avogadro’s Law and the Standard Molar Volume • Avogadro’s Law states that at the same temperature and

pressure, equal volumes of two gases contain the same number of molecules (or moles) of gas.

• If we set the temperature and pressure for any gas to be STP, then one mole of that gas has a volume called the standard molar volume.

• The standard molar volume is 22.4 L at STP. – This is another way to measure moles. – For gases, the volume is proportional to the number of moles.

• 11.2 L of a gas at STP = 0.500 mole – 44.8 L = ? moles

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Avogadro’s Law and the Standard Molar Volume • Example 12-5: One mole of a gas occupies 36.5

L and its density is 1.36 g/L at a given temperature and pressure. (a) What is its molar mass? (b) What is its density at STP?

? . . . gmol

Lmol

gL

g / mol= × =36 5 136 49 6

g/L 21.2L 4.22

mol 1mol

g 6.49L

g ?

STP

=×=

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Summary of Gas Laws: The Ideal Gas Law

• Boyle’s Law - V ∝ 1/P (at constant T & n) • Charles’ Law – V ∝ T (at constant P & n) • Avogadro’s Law – V ∝ n (at constant T & P) • Combine these three laws into one statement

V ∝ nT/P • Convert the proportionality into an equality.

V = nRT/P

• This provides the Ideal Gas Law. PV = nRT

• R is a proportionality constant called the universal gas constant.

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Summary of Gas Laws: The Ideal Gas Law • We must determine the value of R.

– Recognize that for one mole of a gas at 1.00 atm, and 273 K (STP), the volume is 22.4 L.

– Use these values in the ideal gas law.

( )( )( )( )

R = PVnT

1.00 atm L1.00 mol K

L atmmol K

=

=

22 4273

0 0821

.

.

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Summary of Gas Laws: The Ideal Gas Law • R has other values if the units are changed. • R = 8.314 J/mol K

– Use this value in thermodynamics. • R = 8.314 kg m2/s2 K mol

– Use this later in this chapter for gas velocities. • R = 8.314 dm3 kPa/K mol

– This is R in all metric units. • R = 1.987 cal/K mol

– This the value of R in calories rather than J.

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• Example 12-6: What volume would 50.0 g of ethane, C2H6, occupy at 1.40 x 102 oC under a pressure of 1.82 x 103 torr?

– To use the ideal gas law correctly, it is very important that all of your values be in the correct units!

1. T = 140 + 273 = 413 K 2. P = 1820 torr (1 atm/760 torr) = 2.39 atm 3. 50 g (1 mol/30 g) = 1.67 mol

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( ) ( )

atm 39.2

K 413K mol

atm L 0821.0mol 67.1

PT Rn = V

=

PT Rn = V

( ) ( )

L 6.23atm 39.2

K 413K mol

atm L 0821.0mol 67.1

PT Rn = V

=

=

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Summary of Gas Laws: The Ideal Gas Law • Example 12-7: Calculate the number of

moles in, and the mass of, an 8.96 L sample of methane, CH4, measured at standard conditions.

You do it!

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( )( )

( )n = PV

RT1.00 atm L

0.0821 L atmmol K

K0.400 mol CH

g CH mol 16.0 gmol

6.40 g

4

4

=

=

= × =

8 96

273

0 400

.

? .

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Summary of Gas Laws: The Ideal Gas Law • Example 12-8: Calculate the pressure exerted

by 50.0 g of ethane, C2H6, in a 25.0 L container at 25.0oC.

You do it!

( ) ( )

atm 63.1PL 25.0

K 298K mol

atm L0.0821mol 1.67P

VT Rn = P

K 298 = T and mol 1.67 =n

=

=

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Determination of Molecular Weights and Molecular Formulas of Gaseous Substances

• Example 12-9: A compound that contains only carbon and hydrogen is 80.0% carbon and 20.0% hydrogen by mass. At STP, 546 mL of the gas has a mass of 0.732 g . What is the molecular (true) formula for the compound?

• 100 g of compound contains 80 g of C and 20 g of H.

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C mol 67.6C g 12.0C mol 1C g 80.0 = atoms C mol ? =×

15 = mass with CH is formula empirical the36.6719.8

ratio.number holesmallest w theDetermine

H mol 8.19H g 1.01H mol 1H g 20.0 = atoms H mol ?

C mol 67.6C g 12.0C mol 1C g 80.0 = atoms C mol ?

3∴=

=×

=×



=×

=×

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( ) 6223 HCCHdoubled. formula empirical the

is formulamolecular theThus

20.150.30

mass empiricalmass actual

molg0.30

mol 0.0244g 0.732 =n

moles. ofnumber by the divided mass theis massmolar theRemember,

=

==

=

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• Example 12-10: A 1.74 g sample of a compound that contains only carbon and hydrogen contains 1.44 g of carbon and 0.300 g of hydrogen. At STP 101 mL of the gas has a mass of 0.262 gram. What is its molecular formula?

You do it!

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( )( )( )

mol 00451.0K 273

K molatm L0.0821

L 0.101atm 00.1RTPV=n

29 = mass with HC5.20.1200.297



52

=

=

∴=

=×

=×

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( ) 104252 HCHC229

58.1

g/mol 1.58mol 0.00451g 262.0

molg ?

=∴=

==

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Dalton’s Law of Partial Pressures • Dalton’s law states that the pressure

exerted by a mixture of gases is the sum of the partial pressures of the individual gases.

Ptotal = PA + PB + PC + .....

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Dalton’s Law of Partial Pressures • Example 12-11: If 1.00 x 102 mL of hydrogen,

measured at 25.0 oC and 3.00 atm pressure, and 1.00 x 102 mL of oxygen, measured at 25.0 oC and 2.00 atm pressure, were forced into one of the containers at 25.0 oC, what would be the pressure of the mixture of gases?

P P P

3.00 atm + 2.00 atm = 5.00 atm

Total H O2 2= +

=

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Dalton’s Law of Partial Pressures • Vapor Pressure is the pressure exerted by

a substance’s vapor over the substance’s liquid at equilibrium.

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Dalton’s Law of Partial Pressures • Example 12-12: A sample of hydrogen was

collected by displacement of water at 25.0 oC. The atmospheric pressure was 748 torr. What pressure would the dry hydrogen exert in the same container?

( )P P P P P P

P torr = 724 torr

P from table

Total H H O H Total H O

H

H O

2 2 2 2

2

2

= + ∴ = −

= −748 24

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Dalton’s Law of Partial Pressures

• Example 12-13: A sample of oxygen was collected by displacement of water. The oxygen occupied 742 mL at 27.0 oC. The barometric pressure was 753 torr. What volume would the dry oxygen occupy at STP?

You do it!

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Dalton’s Law of Partial Pressures

( )

V 742 mL V ?T K T KP = 726 torr P torr

V mL 273 K300 K

726 torr torr

645 mL @ STP

1 2

1 2

1 2

2

= == =

= − =

= × × =

300 273753 27 760

742760

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Mass-Volume Relationships in Reactions Involving Gases

2 mol KClO3 yields 2 mol KCl and 3 mol O2 2(122.6g) yields 2 (74.6g) and 3 (32.0g) Those 3 moles of O2 can also be thought of as:

3(22.4L) or 67.2 L at STP

g)(2(s)&MnO

(s)3 O 3 + KCl 2KClO 2 2 → ∆

•In this section we are looking at reaction stoichiometry, like in Chapter 3, just including gases in the calculations.

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• Example 12-14: What volume of oxygen measured at STP, can be produced by the thermal decomposition of 120.0 g of KClO3?

You do it!

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2STP2STP

2

2STP

3

2

3

332STP

O L 9.32O L ?O mol 1

O L 4.22KClO mol 2

O mol 3 KClO g 122.6

KClO mol 1 KClO g 120.0O L ?

=

×××=

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The Kinetic-Molecular Theory • The basic assumptions of kinetic-

molecular theory are: • Postulate 1

– Gases consist of discrete molecules that are relatively far apart.

– Gases have few intermolecular attractions. – The volume of individual molecules is very

small compared to the gas’s volume. • Proof - Gases are easily

compressible.

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The Kinetic-Molecular Theory

• Postulate 2 – Gas molecules are in constant, random,

straight line motion with varying velocities. • Proof - Brownian motion displays

molecular motion.

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• Postulate 3 – Gas molecules have elastic collisions with

themselves and the container. – Total energy is conserved during a collision.

• Proof - A sealed, confined gas exhibits no pressure drop over time.

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• Postulate 4 – The kinetic energy of the molecules is

proportional to the absolute temperature. – The average kinetic energies of molecules

of different gases are equal at a given temperature.

• Proof - Brownian motion increases as temperature increases.

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• The kinetic energy of the molecules is proportional to the absolute temperature. The kinetic energy of the molecules is proportional to the absolute temperature. • Displayed in a Maxwellian distribution.

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The Kinetic-Molecular Theory • The gas laws that we have looked at earlier in this chapter are

proofs that kinetic-molecular theory is the basis of gaseous behavior.

• Boyle’s Law – P ∝ 1/V – As the V increases the molecular collisions with container

walls decrease and the P decreases. • Dalton’s Law

– Ptotal = PA + PB + PC + ..... – Because gases have few intermolecular attractions, their

pressures are independent of other gases in the container. • Charles’ Law

– V ∝ T – An increase in temperature raises the molecular velocities,

thus the V increases to keep the P constant.

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53


u RTMrms

m=

3

• The root-mean square velocity of gases is a very close approximation to the average gas velocity.

• Calculating the root-mean square velocity is simple:

• To calculate this correctly: – The value of R = 8.314 kg m2/s2 K mol – And M must be in kg/mol.

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• Example 12-17: What is the root mean square velocity of N2 molecules at room T, 25.0oC?

( )

kg/mol 028.0

K 298molK sec

m kg8.3143u

2

2

rms

=

( )u

3 8.314 kg msec K mol

K

kg / mol m / s = 1159 mi / hr

rms

2

2

=

=

298

0 028515

.

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• Example 12-18: What is the root mean square velocity of He atoms at room T, 25.0oC?

You do it!

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( )u

3 8.314 kg msec K mol

K

kg / mol m / s = 3067 mi / hr

rms

2

2

=

=

298

0 0041363

.

• Can you think of a physical situation that proves He molecules have a velocity that is so much greater than N2 molecules?

• What happens to your voice when you breathe He?

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Diffusion and Effusion of Gases

• Diffusion is the intermingling of gases.

• Effusion is the escape of gases through tiny holes.

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• This is a demonstration of diffusion.

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• The rate of effusion is inversely proportional to the square roots of the molecular weights or densities.

1

2

2

1

1

2

2

1

DD

RR

or

MM

RR

=

=

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• Example 12-15: Calculate the ratio of the rate of effusion of He to that of sulfur dioxide, SO2, at the same temperature and pressure.

RR

MM

g / mol4.0 g / mol

R R

He

SO

SO

He

He SO

2

2

2

=

=

= = ∴ =

641

16 4 4

.

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Diffusion and Effusion of Gases • Example 12-16: A sample of hydrogen, H2, was found to

effuse through a pinhole 5.2 times as rapidly as the same volume of unknown gas (at the same temperature and pressure). What is the molecular weight of the unknown gas?

You do it!

g/mol 54 = g/mol) 0.2(27Mg/mol 0.2

M27

g/mol 0.2M2.5

MM

RR

unk

unk

unk

H

unk

unk

H

2

2

=

=

=

=

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Real Gases: Deviations from Ideality

• Real gases behave ideally at ordinary temperatures and pressures.

• At low temperatures and high pressures real gases do not behave ideally.

• The reasons for the deviations from ideality are:

1. The molecules are very close to one another, thus their volume is important.

2. The molecular interactions also become important.

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• van der Waals’ equation accounts for the behavior of real gases at low temperatures and high pressures.

( )P + n aV

V nb nRT2

2

− =

• The van der Waals constants a and b take into account two things: 1. a accounts for intermolecular attraction 2. b accounts for volume of gas molecules

• At large volumes a and b are relatively small and van der Waal’s equation reduces to ideal gas law at high temperatures and low pressures.

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• What are the intermolecular forces in gases that cause them to deviate from ideality?

1. For nonpolar gases the attractive forces are London Forces.

2. For polar gases the attractive forces are dipole-dipole attractions or hydrogen bonds.

65

Real Gases: Deviations from Ideality • Example 12-19: Calculate the pressure exerted

by 84.0 g of ammonia, NH3, in a 5.00 L container at 200. oC using the ideal gas law.

You do it!

( ) ( )

atm 4.38PL 5.00

K 473K mol

atm L0821.0mol 94.4

VnRT = P

mol 94.4g 17.0

mol 1NH g 84.0 =n 3

=

=

=×

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Real Gases: Deviations from Ideality • Example 12-20: Solve Example 12-19

using the van der Waal’s equation.

( )

2

2

2

2

2

2

Van

nb-VnRT=P

nRTnb-VV

an + P

molL0.0371=b

molatm L 4.17 = a mol 4.94 =n

−

∴=

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( ) ( )( ) ( ) ( )( )

ideal from difference 7.6% a is which atm 7.35P

)atm 1.4atm 8.39(atm 07.4L 817.4atm L 8.191P

L 00.517.4mol 94.4

)0371.0)(mol 94.4(L 00.5K4730821.0 mol 94.4P 2

molatm L2

molL

K molatm L

2

2

=

−=−=

−−

=

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Synthesis Question

• The lethal dose for hydrogen sulfide is 6.0 ppm. In other words, if in 1 million molecules of air there are six hydrogen sulfide molecules then that air would be deadly to breathe. How many hydrogen sulfide molecules would be required to reach the lethal dose in a room that is 77 feet long, 62 feet wide and 50. feet tall at 1.0 atm and 25.0 oC?

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Synthesis Question

L106.76cm 1000

L1cm 106.76

cm 106.76cm 1524cm 1890cm 2347V

cm 1524in

cm 54.2ftin 12ft 05

cm 1890in

cm 54.2ftin 12ft 26

cm 2347in

cm 54.2ftin 12ft 77

63

39

39

×=

×

×=××=

=

=

=

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Synthesis Question

( )( )( )( )

SH of molecules 1096.9air of molecules 10

SH of molecule 6.0air of molecules 1066.1 dose Lethal

air of molecules 1066.1mol

molecules 106.022mol 276,000

air of mol 276,000K 2980.0821

L 106.76atm 1RTPVn

K 298 25 273 T

L 106.76cm 1000

L 1cm 106.76

223

6229

2923

K molatm L

6

63

39

×=

×=

×=

×

=×

==

=+=

×=

×

71

Group Question

• Tires on a car are typically filled to a pressure of 35 psi at 3.00 x 102 K. A tire is 16 inches in radius and 8.0 inches in thickness. The wheel that the tire is mounted on is 6.0 inches in radius. What is the mass of the air in the tire?

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12 Gases and Kinetic Molecular Theory

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