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Chem 101 week 12 ch 5

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Gases Chapter 5 Gases
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Page 1: Chem 101 week 12 ch 5

Gases

Chapter 5Gases

Page 2: Chem 101 week 12 ch 5

Gases

Characteristics of Gases

• Unlike liquids and solids, they Expand to fill their containers. Are highly compressible. Have extremely low densities.

Page 3: Chem 101 week 12 ch 5

Gases

Measurements

• We have learned in our early chapters and experienced in the lab, the means of measurement for solids and liquids

• We will begin our study of gases by looking at the means we employ to make measurements on gases

Page 4: Chem 101 week 12 ch 5

Gases

Measurements involving Gases

• To describe a gas we must specify the following:

1.Volume

2. Amount

3.Temperature

4. Pressure

Page 5: Chem 101 week 12 ch 5

Gases

Volume of Gases

• Gases expand uniformly to fill any container they are placed in

• Therefore, the volume of a gas is the volume of its container

• Volumes are expressed in L, cm3 or m3

Page 6: Chem 101 week 12 ch 5

Gases

Amount

• The amount of matter in a sample of gas is expressed in terms of the number of moles(n)

• Sometimes the mas in grams is given

• n = g/mol mass(M)

• n(M)= g

Page 7: Chem 101 week 12 ch 5

Gases

Temperature

• Temperature of a gas is measured in Celsius

• Calculations will require conversion to the Kelvin scale

• TK = Tcelsius + 273.15

• Temperature is expressed to the nearest degree (simply add 273 to celsius temperature)

Page 8: Chem 101 week 12 ch 5

Gases

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

Pressure

• Atmospheric pressure is the weight of air per unit of area.

P =FA

Page 9: Chem 101 week 12 ch 5

Gases

Units of Pressure

• English System pounds per square inch (psi)

• Pascals (SI unit) 1 Pa = 1 N/m2

• Bar 1 bar = 105 Pa = 100 kPa

Page 10: Chem 101 week 12 ch 5

Gases

Units of Pressure

• mm Hg or torrThese units are literally the difference in the heights measured in mm (h) of two connected columns of mercury.

• Atmosphere1.00 atm = 760 torr

Page 11: Chem 101 week 12 ch 5

Gases

Manometer

Used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.

Page 12: Chem 101 week 12 ch 5

Gases

Standard Pressure

• Normal atmospheric pressure at sea level.

• It is equal to1.00 atm760 torr (760 mm Hg)101.325 kPa

Page 13: Chem 101 week 12 ch 5

Gases

Boyle’s Law

The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.

Page 14: Chem 101 week 12 ch 5

Gases

Boyle’s Law

Page 15: Chem 101 week 12 ch 5

Gases

As P and V areinversely proportional

A plot of V versus P results in a curve.

Since

V = k (1/P)This means a plot of V versus 1/P will be a straight line.

PV = k

Page 16: Chem 101 week 12 ch 5

Gases

Charles’s Law

• The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature.

A plot of V versus T will be a straight line.

• i.e.,VT

= k

Page 17: Chem 101 week 12 ch 5

Gases

Avogadro’s Law

• The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas.

• Mathematically, this means V = kn

Page 18: Chem 101 week 12 ch 5

Gases

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

Page 19: Chem 101 week 12 ch 5

Gases

Ideal-Gas Equation

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

Page 20: Chem 101 week 12 ch 5

Gases

Ideal-Gas Equation

The relationship

then becomes

nTP

V

nTP

V = R

or

PV = nRT

Page 21: Chem 101 week 12 ch 5

Gases

Ideal-Gas Equation

Page 22: Chem 101 week 12 ch 5

Gases

Densities of Gases

If we divide both sides of the ideal-gas equation by V and by RT, we get

nV

PRT

=

Page 23: Chem 101 week 12 ch 5

Gases

• We know thatmoles molecular mass = mass

Densities of Gases

• So multiplying both sides by the molecular mass () gives

n = m

PRT

mV

=

Page 24: Chem 101 week 12 ch 5

Gases

Densities of Gases

• Mass volume = density

• So,

• Note: One only needs to know the molecular mass, the pressure, and the temperature to calculate the density of a gas.

PRT

mV

=d =

Page 25: Chem 101 week 12 ch 5

Gases

Molecular Mass

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

Becomes

PRT

d =

dRTP =

Page 26: Chem 101 week 12 ch 5

Gases

The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).

PV = nRT

R = PVnT

=(1 atm)(22.414L)

(1 mol)(273.15 K)

R = 0.082057 L • atm / (mol • K)

5.4

Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.

Page 27: Chem 101 week 12 ch 5

Gases

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

5.4

Page 28: Chem 101 week 12 ch 5

Gases

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 constant

nRV

= PT

= constant

P1

T1

P2

T2

=

P1 = 1.20 atm

T1 = 291 K

P2 = ?

T2 = 358 K

P2 = P1 x T2

T1

= 1.20 atm x 358 K291 K

= 1.48 atm

5.4

Page 29: Chem 101 week 12 ch 5

Gases

Density (d) Calculations

d = mV =

PMRT

m is the mass of the gas in g

M is the molar mass of the gas

Molar Mass (M ) of a Gaseous Substance

dRTP

M = d is the density of the gas in g/L

5.4

Page 30: Chem 101 week 12 ch 5

Gases

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?

5.4

dRTP

M = d = mV

4.65 g2.10 L

= = 2.21 g

L

M =2.21

g

L

1 atm

x 0.0821 x 300.15 KL•atmmol•K

M = 54.6 g/mol

Page 31: Chem 101 week 12 ch 5

Gases

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

5.5

Page 32: Chem 101 week 12 ch 5

Gases

Dalton’s Law ofPartial Pressures

• The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone.

• In other words,

Ptotal = P1 + P2 + P3 + …

Page 33: Chem 101 week 12 ch 5

Gases

Dalton’s Law of Partial Pressures

V and T are constant

P1 P2 Ptotal = P1 + P2

5.6

Page 34: Chem 101 week 12 ch 5

Gases

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

PA = nART

V

PB = nBRT

V

nA is the number of moles of A

nB is the number of moles of B

PT = PA + PB XA = nA

nA + nB

XB = nB

nA + nB

PA = XA PT PB = XB PT

Pi = Xi PT

5.6

mole fraction (Xi) = ni

nT

Page 35: Chem 101 week 12 ch 5

Gases

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

5.6

Page 36: Chem 101 week 12 ch 5

Gases

Partial Pressures

• When one collects a gas over water, there is water vapor mixed in with the gas.

• To find only the pressure of the desired gas, one must subtract the vapor pressure of water from the total pressure.

Page 37: Chem 101 week 12 ch 5

Gases

2KClO3 (s) 2KCl (s) + 3O2 (g)

Bottle full of oxygen gas and water vapor

PT = PO + PH O2 2 5.6

Page 38: Chem 101 week 12 ch 5

Gases

5.6

Page 39: Chem 101 week 12 ch 5

Gases

Chemistry in Action:

Scuba Diving and the Gas Laws

P V

Depth (ft) Pressure (atm)

0 1

33 2

66 3

5.6

Page 40: Chem 101 week 12 ch 5

Gases

Kinetic-Molecular Theory

This is a model that aids in our understanding of what happens to gas particles as environmental conditions change.

Page 41: Chem 101 week 12 ch 5

Gases

Kinetic Molecular Theory of Gases1. A gas is composed of molecules that are separated from each

other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume.

2. Gas molecules are in constant motion in random directions, and they frequently collide with one another. Collisions among molecules are perfectly elastic.

3. Gas molecules exert neither attractive nor repulsive forces on one another.

4. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy

5.7

KE = ½ mu2

Page 42: Chem 101 week 12 ch 5

Gases

Kinetic theory of gases and …

• Compressibility of Gases

• Boyle’s Law

P collision rate with wall

Collision rate number densityNumber density 1/VP 1/V

• Charles’ LawP collision rate with wall

Collision rate average kinetic energy of gas molecules

Average kinetic energy T

P T

5.7

Page 43: Chem 101 week 12 ch 5

Gases

Kinetic theory of gases and …

• Avogadro’s Law

P collision rate with wall

Collision rate number densityNumber density nP n

• Dalton’s Law of Partial Pressures

Molecules do not attract or repel one another

P exerted by one type of molecule is unaffected by the presence of another gas

Ptotal = Pi

5.7

Page 44: Chem 101 week 12 ch 5

Gases

Main Tenets of Kinetic-Molecular Theory

The average kinetic energy of the molecules is proportional to the absolute temperature.

Page 45: Chem 101 week 12 ch 5

Gases

Apparatus for studying molecular speed distribution

5.7

Page 46: Chem 101 week 12 ch 5

Gases

The distribution of speedsfor nitrogen gas moleculesat three different temperatures

The distribution of speedsof three different gasesat the same temperature

5.7

urms = 3RTM

Page 47: Chem 101 week 12 ch 5

Gases

Diffusion

The spread of one substance throughout a space or throughout a second substance.

Page 48: Chem 101 week 12 ch 5

Gases

Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.

5.7

NH3

17 g/molHCl36 g/mol

NH4Cl

r1

r2

M2

M1=

Page 49: Chem 101 week 12 ch 5

Gases

Effusion

The escape of gas molecules through a tiny hole into an evacuated space.

Page 50: Chem 101 week 12 ch 5

Gases

Gas effusion is the is the process by which gas under pressure escapes from one compartment of a container to another by passing through a small opening.

5.7

r1

r2

t2

t1

M2

M1= =

Nickel forms a gaseous compound of the formula Ni(CO)x What is the value of x given that under the same conditions methane (CH4) effuses 3.3 times faster than the compound?

r1 = 3.3 x r2

M1 = 16 g/mol

M2 = r1

r2( )

2x M1 = (3.3)2 x 16 = 174.2

58.7 + x • 28 = 174.2 x = 4.1 ~ 4

Page 51: Chem 101 week 12 ch 5

Gases

Page 52: Chem 101 week 12 ch 5

Gases

Page 53: Chem 101 week 12 ch 5

Gases

Real Gases

In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure.

Page 54: Chem 101 week 12 ch 5

Gases

Deviations from Ideal Behavior

The assumptions made in the kinetic-molecular model break down at high pressure and/or low temperature.

Page 55: Chem 101 week 12 ch 5

Gases

Deviations from Ideal Behavior

1 mole of ideal gas

PV = nRT

n = PVRT

= 1.0

5.8

Repulsive Forces

Attractive Forces

Page 56: Chem 101 week 12 ch 5

Gases

Effect of intermolecular forces on the pressure exerted by a gas.

5.8

Page 57: Chem 101 week 12 ch 5

Gases

Corrections for Nonideal Behavior

• The ideal-gas equation can be adjusted to take these deviations from ideal behavior into account.

• The corrected ideal-gas equation is known as the van der Waals equation.

Page 58: Chem 101 week 12 ch 5

Gases

5.8

Van der Waals equationnonideal gas

P + (V – nb) = nRTan2

V2( )}

correctedpressure

}

correctedvolume

Page 59: Chem 101 week 12 ch 5

Gases

Page 60: Chem 101 week 12 ch 5

Gases


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