Chapter 10 Gases

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Chapter 10

Gases

Lecture Presentation

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Characteristics of Gases• Unlike liquids and solids, gases

– Expand to fill their containers.– Are highly compressible.– Have extremely low densities.

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• Pressure is the amount of force applied to an area:

Pressure

P = FA

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Common Units of Pressure

Atmospheric PressureUnit Scientific Field

chemistryatmosphere(atm) 1 atm*

pascal(Pa); kilopascal(kPa)

1.01325x105Pa; 101.325 kPa

SI unit; physics, chemistry

millimeters of mercury(Hg)

760 mm Hg* chemistry, medicine, biology

torr 760 torr* chemistry

pounds per square inch (psi or lb/in2)

14.7lb/in2 engineering

bar 1.01325 bar meteorology, chemistry, physics

*This is an exact quantity; in calculations, we use as many significant figures as necessary.

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Sample Problem 5.1 Converting Units of Pressure

PROBLEM: A geochemist heats a limestone (CaCO3) sample and collects the CO2 released in an evacuated flask attached to a closed-end manometer. After the system comes to room temperature, Dh = 291.4 mm Hg. Calculate the CO2 pressure in torrs, atmospheres, and kilopascals.

SOLUTION:

PLAN: Construct conversion factors to find the other units of pressure.

291.4 mmHg 1torr

1 mmHg

= 291.4 torr

291.4 torr 1 atm

760 torr= 0.3834 atm

0.3834 atm 101.325 kPa

1 atm= 38.85 kPa

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Standard Pressure

• Normal atmospheric pressure at sea level is referred to as standard pressure.

• It is equal to– 1.00 atm

–760 torr (760 mmHg)–101.325 kPa

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The Gas Laws

• Gas experiments revealed that four variables affect the state of a gas:

• Temperature, T• Volume, V• Pressure, P• Quantity of gas present, n (moles)

• These variables are related through equations know as the gas laws.

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Boyle’s Law 1662

• The volume occupied by a gas is inversely related to its pressure

Boyle’s Law: ), (constant 1 TnP

V

P1 x V1 = P2 x V2

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A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?

P1 x V1 = P2 x V2

P1 = 726 mmHg

V1 = 946 mL

P2 = ?

V2 = 154 mL

P2 = P1 x V1

V2

726 mmHg x 946 mL154 mL= = 4460 mmHg

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P x V = constant

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Charles’ Law 1787

• At constant pressure, the volume occupied by a fixed amount of gas directly proportional to its absolute temperature

• First conceived by Guillaume Amontons in 1702 and published by Joseph Gay-Lussac 1802.

Charles’s Law: ), (constant PnTV

V1 /T1 = V2 /T2

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A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?

V1 = 3.20 L

T1 = 398.15 K

V2 = 1.54 L

T2 = ?

T2 = V2 x T1

V1

1.54 L x 398.15 K3.20 L= = 192 K

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V1 /T1 = V2 /T2

T1 = 125 (0C) + 273.15 (K) = 398.15 K

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Avogadro’s Law

• At fixed temperature and pressure, equal volumes of any ideal gas contain equal number of particles (or moles)

Avogadro’s Law: ),(constant TPnV

V1 /n1 = V2 /n2

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Figure 5.7 Standard molar volume.

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Ideal-Gas Equation

V 1/P (Boyle’s law)V T (Charles’s law)V n (Avogadro’s law)

• So far we’ve seen that

• Combining these, we get

V nTP

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Ideal-Gas Equation

The constant of proportionality is known as R, the gas constant.

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Ideal-Gas Equation

The relationship

then becomes

nTPV

nTPV = R

or

PV = nRT

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Molecular Mass

We can manipulate the density equation to enable us to find the molecular mass of a gas:

Becomes molar mass of a gaseous substance

PRTd =

dRTP =

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A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 0C. What is the molar mass of the gas?

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dRTPM = d = m

V4.65 g2.10 L

= = 2.21 gL

M =2.21 g

L

1 atm

x 0.0821 x 300.15 KL•atmmol•K

M = 54.6 g/mol

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1)What is the volume (in liters) occupied by 49.8 g of HCl at STP?

2) Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?

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What is the volume (in liters) occupied by 49.8 g of HCl at STP?

PV = nRT

V = nRTP

T = 0 0C = 273.15 K

P = 1 atm

n = 49.8 g x 1 mol HCl36.45 g HCl

= 1.37 mol

V =1 atm

1.37 mol x 0.0821 x 273.15 KL•atmmol•K

V = 30.6 L

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Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?

PV = nRT n, V and R are constantnRV = P

T = constant

P1

T1

P2

T2=

P1 = 1.20 atmT1 = 291 K

P2 = ?T2 = 358 K

P2 = P1 x T2

T1

= 1.20 atm x 358 K291 K

= 1.48 atm

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Dalton’s Law of Partial Pressures• Formulated in 1801• The total pressure of a mixture of gases is just the sum of the pressures that each gas would exert if it was alone.

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Dalton’s Law of Partial Pressures

V and T are

constant

P1 P2 Ptotal = P1 + P2

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Consider a case in which two gases, A and B, are in a container of volume V.

PA = nARTV

PB = nBRTV

nA is the number of moles of A

nB is the number of moles of B

PT = PA + PB XA = nA

nA + nBXB =

nB

nA + nB

PA = XA PT PB = XB PT

Pi = Xi PT

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mole fraction (Xi) = ni

nT

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A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the

partial pressure of propane (C3H8)?

Pi = Xi PT

Xpropane = 0.116

8.24 + 0.421 + 0.116

PT = 1.37 atm

= 0.0132

Ppropane = 0.0132 x 1.37 atm = 0.0181 atm

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In Groups

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Gas Stoichiometry

What is the volume of CO2 produced at 37 0C and 1.00 atm when 5.60 g of glucose are used up in the reaction:

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)

g C6H12O6 mol C6H12O6 mol CO2 V CO2

5.60 g C6H12O6

1 mol C6H12O6

180 g C6H12O6

x6 mol CO2

1 mol C6H12O6

x = 0.187 mol CO2

V = nRT

P

0.187 mol x 0.0821 x 310.15 KL•atmmol•K

1.00 atm= = 4.76 L

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Chapter 10 Gases

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