Chapter 5 Gases - MRS. KALMER'S WEBSITE -...

Chapter 5

Gases

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Chapter 5

A Gas Uniformly fills any container.

Easily compressed.

Mixes completely with any other gas.

Exerts pressure on its surroundings.

Copyright © Cengage Learning. All rights reserved 2

Chapter 5

Variables used to describe a gas

Pressure (P)

• collision of gas molecules with wall of container.

Volume (V)

Temperature (T)

related to average speed of gas molecules.

Must be in Kelvin

Number of moles (n)

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Chapter 5

Pressure

SI units = Newton/meter2 = 1 Pascal (Pa)

Standard Pressure = 1 atm

= 101,325 Pa = 101.3 kPa

= 760 mm Hg

= 760 torr Copyright © Cengage Learning. All rights reserved 4

forcePressure =

area

Chapter 5

Barometer

Device used to measure atmospheric pressure.

Hg flows out of the tube until the pressure of the column of Hg standing on the surface of the Hg in the dish is equal to the pressure of the air on the rest of the surface of the Hg in the dish.


Chapter 5

Manometer

Device used for measuring the pressure of a gas in a container.


Chapter 5

Pressure Conversions

Example: The pressure of a gas is measured as 2.5 atm. Represent this pressure in both torr and pascals.


3760 torr2.5 atm = 1.9 10 torr

1 atm

5101,325 Pa2.5 atm = 2.5 10 Pa

1 atm

Chapter 5

The Kinetic Molecular Theory provides a model for gases. Postulates of the Kinetic Molecular Theory:

1) The particles are so small compared with the distances between them that the volume of the individual particles can be assumed to be negligible (zero).


Chapter 5

Postulates of the Kinetic Molecular Theory

2) The particles are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas.


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Chapter 5


3) The particles are assumed to exert no forces on each other; they are assumed neither to attract nor to repel each other.


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Chapter 5


4) The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas.


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Chapter 5

Kinetic Molecular Theory


Chapter 5

You are holding two balloons of the same volume. One contains He, the other H . Complete each of the following statements with “different” or “the same” and be prepared to justify your answer.


Ne

H2

CONCEPT CHECK!

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Chapter 5

The numbers of moles of the gas in the two balloons are _______.

The temperatures of the gas in the two balloons are__________.

The pressures of the gas in the two balloons are __________.

The densities of the gas in the two balloons are __________.


He

H2

CONCEPT CHECK!

Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro

Concept Check: Liquid Nitrogen and a Balloon: Why does this happen?



Liquid Nitrogen and a Balloon What happened to the gas in the

balloon?

A decrease in temperature was followed by a decrease in the pressure and volume of the gas in the balloon.

Mathematical relationships among the properties of a gas (Pressure, Volume, Temperature and Moles) were discovered from observations like these.



Boyle’s Law Pressure and volume are inversely related

(constant T and n).

The formula: P1V1 = P2V2


(Plotting 1/V gives a

straight line)


Molecular View of Boyle’s Law



Boyle’s law


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A sample of helium gas occupies 12.4 L at 23°C and 0.956 atm. What volume will it occupy at 1.20 atm assuming that the temperature stays constant?

P1V1 = P2V2

(0.956 atm)(12.4 L) = (1.20 atm)(V2)

9.88 L Copyright © Cengage Learning. All rights reserved 20

EXERCISE!


Charles’s Law

Volume and Temperature (in Kelvin) are directly related (constant P and n).

K = °C + 273

0 K is absolute zero.


1 2

1 2

= V V

T T


Molecular View of Charles’s Law



Charles’s Law



Suppose a balloon containing 1.30 L of air at 24.7°C is placed into a beaker containing liquid nitrogen at –78.5°C. What will the volume of the sample of air become (at constant pressure)?

Remember to convert the temperatures to Kelvin.

1.30 L = V2

(24.7+273.15) (-78.5 + 273.15)

0.849 L


EXERCISE!

1 2

1 2

= V V

T T



• You have a sample of nitrogen gas (N2) in a container fitted with a piston that maintains a pressure of 6.00 atm. Initially, the gas is at 45°C in a volume of 6.00 L.

• You then cool the gas sample.

CONCEPT CHECK!


Which best explains the final result that occurs once the gas sample has cooled?

a) The pressure of the gas increases.

b) The volume of the gas increases.

c) The pressure of the gas decreases.

d) The volume of the gas decreases.

e) Both volume and pressure change.


CONCEPT CHECK!


The gas sample is then cooled to a temperature of 15°C. Solve for the new condition. (Hint: A moveable piston keeps the pressure constant overall, so what condition will change?)

Use Charles’s Law to solve

6.00 L = V2

(45+273) (15+273)


CONCEPT CHECK!


Avogadro’s Law

Volume and number of moles are directly related (constant T and P).

The formula:


1 2

1 2

n n =

V V

Avogadro’s hypothesis states that at a fixed temperature and pressure, the volume of a gas is directly proportional to the amount of gas (that is, to the number of moles of gas, n, or to the number of molecules of gas).


If 2.45 mol of argon gas occupies a volume of 89.0 L, what volume will 2.10 mol of argon occupy under

the same conditions of temperature and pressure?

2.45 mol = 2.10 mol

89.0 L V2


EXERCISE!

1 2

1 2

n n =

V V


Lussac’s Law

The pressure of a gas is directly proportional to its Kelvin temperature when the volume is held constant.

P1 = P2

T1 T2

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A sample of nitrogen gas has a pressure of 6.58kPa at 539K. What will the pressure of the gas be when its temperature reaches 211K?

6.58kPa = P2

539K 211K

2.58 kPa


EXERCISE!

P1 = P2

T1 T2


The Combined Gas Law

The combined gas law combines Boyle Law, Charles’ Law, and Lussac’s Law.

The formula: Formula combining n:

P1V1 = P2V2

T1 T2

You can cancel any term (P, V, n, T) that is the same (or held constant) on both sides.

P1V1 P2V2 —— = —— n1T1 n2T2


At what temperature (in ⁰C) does 121 mL of CO2 at 27 ⁰ C and 1.05 atm occupy a volume of 293 mL at a pressure of 1.40 atm?

(1.05 atm)(121 mL) = (1.40 atm)(293 mL)

(27+273) (T2 + 273)

696°C


EXERCISE!

P1V1 = P2V2 T1 T2

Chapter 5

Section 5.3 The Ideal Gas Law

We can bring all of these laws together into one comprehensive law:

PV = nRT

Be careful of units!

R = The universal gas constant

= 0.08206 L·atm/mol·K

=8.314 L·kPa/mol·K

=62.36 L·mmHg/mol·K


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Molecular View of the Ideal Gas Law



An automobile tire at 23 ⁰C with an internal volume of 25.0 L is filled with air to a total pressure of 3.18 atm. Determine the number of moles of air in the tire.

PV = nRT

(3.18 atm)(25.0 L) = n(0.08206)(23+273K)

3.27 mol


EXERCISE!

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What is the pressure in atm in a 304.0 L tank that contains 5.670 kg of helium at 25 ⁰C?

First, convert kg of He into mol!

5.670 kg He ×1000 g He × 1mol He = 1416mol

1kg He 4.003 g He

PV = nRT

(P)(304.0L) = (1416 mol)(0.08206)(25+273K)

= 114 atm Copyright © Cengage Learning. All rights reserved 38

EXERCISE!

Section 5.4 Gas Stoichiometry

A sample of oxygen gas has a volume of 2.50 L at STP. How many grams of O2 are present?

PV = nRT

(1.00 atm)(2.50 L) = n(0.08206)(273)

n = 0.112 mol O2 × 32.00 g O2 = 3.57 g O2

1 mol O2

3.57 g


EXERCISE!

Section 5.3: The Ideal Gas Law


Molar Volume of an Ideal Gas at STP

For 1 mole of an ideal gas at 0 ⁰C and 1 atm, the volume of the gas is 22.42 L.

STP = standard temperature and pressure

0⁰C and 1 atm

Therefore, the molar volume is 22.42 L at STP.


1.000 mol 0.08206 L atm/K mol 273.2 KV = = = 22.42 L

1.000 atm

nRT

P

Section 5.3: Calculating 22.42L Using The Ideal Gas Law

Section 5.4 Gas Stoichiometry Applications of the Ideal Gas Law: Molecular Mass Determination

Alternative to equation: (A)

find n using the ideal gas

equation; (B) Divide m

(grams) by n (moles) to get

grams/mol.

M = molar mass and m = mass in grams

PV n = —— RT

m (grams) m M (molar mass)= ————— so n = — n (moles) M

The ideal gas equation

rearranges to:

Setting the equations equal to one another:

m PV — = —— M RT

Rearranges to:

mRT M = ——— PV


Example

If 0.550 g of a gas occupies 0.200 L at 0.968 atm and

289 K, what is the molecular mass of the gas?

Applications of the Ideal Gas Law: Molecular Mass Determination

(0.550g) (0.08206 L-atm/mol-K))(289K) M = ———————————— (0.968atm)(0.200L) M = 67.4 g/mol

mRT M = ——— PV


Alternative: find volume of one mole (n = 1) or other

fixed quantity of gas. Divide mass of that quantity

by volume to find g/L.

mRT m MP M = ——— rearranges to — = —— PV V RT

m MP and density = — so d = —— V RT

Applications of the Ideal Gas Law: Molecular Mass / Density Determination

dRT Rearranges to M = —— P


Molar Mass of a Gas

d = density of gas

T = temperature in Kelvin

P = pressure of gas

R = universal gas constant


g L atmK

gL mol K = =

atm molMolar mass =

dRT

P

Applications of the Ideal Gas Law: Molecular Mass /Density Determination


What is the density of F2 at STP (in g/L)?

OR, since it is at STP:

d = mass = molar mass = 38.00g vol molar vol 22.42L

OR, use PV=nRT:


EXERCISE!

Applications of the Ideal Gas Law: Molecular Mass / Density Determination

dRT M = —— P

MP (19.00g/mol × 2)(1.00 atm) d = —— = RT (0.08206 L-atm/mol-K)(273 K)


The Law of Combining Volumes When gases measured at the same temperature and pressure react,

the volumes of gaseous reactants and products are in small whole-number ratios.

Example: At a given temperature and pressure, 2.00 L of H2 will react with 1.00 L of O2

Each flask contains the same number of molecules. Therefore, the

ratio of volumes is the same as the mole ratio from the balanced

equation: 2 H2 + O2 2 H2O


We can use the law of combining volumes for

stoichiometry only for gases and only if the

gases are at the same temperature and

pressure.

Otherwise, we must use stoichiometric

methods from Chapter 3 – combined with the

ideal gas equation.


Example

How many liters of O2(g) are consumed for every

10.0 L of CO2(g) produced in the combustion of

liquid pentane, C5H12, if all volumes are measured

at the same temperature and pressure?

C5H12(l) + 8O2(g) 5 CO2(g) + 6 H2O(l)

8 L O2 10.0 L CO2 x ———— = 16.0 L O2 5 L CO2


As in other stoichiometry calculations, the problem centers around the mole ratio:

• If A is a gas, we find moles of A first by using the ideal gas equation and P, V, and T.

• If B is a gas, we solve for moles of B (n), then use the ideal gas equation to find P, V, or T.


Example

In the chemical reaction used in automotive air-bag safety

systems, N2(g) is produced by the decomposition of sodium

azide, NaN3(s), at a somewhat elevated temperature:

2 NaN3(s) 2 Na(l) + 3 N2(g)

What volume of N2(g), measured at 25⁰C and 0.980 atm, is

produced by the decomposition of 62.5 g NaN3? 1 mol NaN3 3 mol N2 62.5 g NaN3 x ————— x ————— = 1.44 mol N2 65.01 g NaN3 2 mol NaN3

PV = nRT

(0.980 atm)(V) = (1.44 mol)(0.08206 L-atm/mol-K) (273+25K)

V = 35.9 L N2


For a mixture of gases in a container,

PTotal = P1 + P2 + P3 + . . .

The total pressure exerted is the sum of the partial pressures that each gas would exert if it were alone.

51

n1RT P1 = ——— V

n2RT P2 = ——— V

n3RT P3 = ——— V


27.4 L of oxygen gas at 25.0⁰C and 1.30 atm, and 8.50 L of helium gas at 25.0 ⁰ C and 2.00 atm were pumped into a tank with a volume of 5.81 L at 25 ⁰ C.

Calculate the new partial pressure of oxygen.

P1V1=P2V2

(1.30 atm)(27.4 L) = (P2)(5.81 L)

P2 = 6.13 atm

Calculate the new partial pressure of helium.

P1V1=P2V2

(2.00 atm)(8.50 L) = (P2)(5.81 L)

P2 = 2.93 atm

Calculate the new total pressure of both gases.

6.13 + 2.93 = 9.06 atm. 52

EXERCISE!


Which of the following best represents the mass ratio of Ne:Ar in the balloons?

1:1

1:2

2:1

1:3

3:1 Copyright © Cengage Learning. All rights reserved 53

Ne

Ar VNe = 2VAr

CONCEPT CHECK!

1 L Ne = 1/2 L Ar

1 mol Ne = 1/2 mol Ar

1 mol Ne = 20 g

1 mol Ar = 40 g; ½ mol Ar = 20g

Therefore: 1:1 Mass ratio


Example

A 1.00-L sample of dry air at 25 ⁰C contains 0.0319

mol N2, 0.00856 mol O2, 0.000381 mol Ar, and

0.00002 mol CO2. Calculate the partial pressure of

N2(g) in the mixture.

=0.0319 mol (0.08206 L- atm/mol-K) 298 K

1.00 L

= 0.780 atm

nN2RT PN2 = ——— V


n1 x1 = —— ntotal

The mole fraction (x) is the ratio of

the number of moles of a given

component in a mixture to the total

number of moles in the mixture

P1 = —— Ptotal

P1 = x1 Ptotal

Partial Pressure = Mole fraction x Total Pressure

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Ex. The partial pressure of oxygen gas was

observed to be 156 torr in air with an

atmospheric pressure of 743 torr. Calculate

the mole fraction of O2 present.

X1 = P1 = 156torr

Ptotal 743torr

= 2.10x10-1 (no unit)

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Ex. The mole fraction of nitrogen in the air is

0.7808. Calculate the partial pressure (in torr)

of N2 at STP.

.7808 = P1

760 torr

P1 = 593torr

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Collection of Gases over Water As (essentially insoluble) gas is bubbled

into the container for collection, the water is displaced.

The gas collected is usually saturated with water vapor.

… what TWO gases are

present here?

Assuming the gas is saturated with

water vapor, the partial pressure of

the water vapor is the vapor

pressure of the water.

Pgas = Ptotal – PH2O(g) = Pbar – PH2O(g)

Pbar = barometric pressure

If O2 is being

generated …


Vapor Pressure of Water as a Function of Temp

Temp.(C) Pressure (mmHg) 15 12.8 16 13.6 17 14.5 18 15.5 19 16.5 20 17.5 21 18.7 22 19.8 23 21.1


Example: Hydrogen produced in the following reaction is

collected over water at 23⁰C when the barometric pressure is 742

Torr: 2 Al(s) + 6 HCl(aq) 2 AlCl3(aq) + 3 H2(g)

What volume of hydrogen gas will be collected in the reaction of

1.50 g Al(s) with excess HCl(aq)?

Ptotal = PH2 + PH2O 742 Torr = PH2 + 21.1 Torr PH2 = 721 Torr

=.949atm

1.50 g Al x 1 mol Al x 3 mol H2 = 0.0834 mol H2

26.98 g Al 2 mol Al

V = nRT =0.0834 mol (0.08206 L-atm/mol-K)(273 + 23 K)

P .949 atm

= 2.14 L

Section 5.6 The Kinetic Molecular Theory of Gases

Root Mean Square Velocity

Gas molecules do not all move at the same speed.

The average velocity of the gas is known as the Root Mean Square Velocity.

R = 8.315 L-Kpa/mol-K **Must use this R so units cancel!

T = temperature of gas (in K)

M = mass of a mole of gas in kg

Final units are in m/s. Copyright © Cengage Learning. All rights reserved

3 =

RTurms M

Section 5.6 The Kinetic Molecular Theory of Gases

Example: Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 32°C. * M must be in Kg

32.00 g ×1 kg = .03200 kg

1000g

* T must be in K

T= 32 + 273 = 305 K

= 488m/s

3 =

RTurms M

3(8.315)(305 )

0.03200rms

Ku

kg

Section 5.7 Effusion and Diffusion

Diffusion:

The mixing of gases.

At a fixed temperature, molecules of higher mass (M) move more slowly than molecules of lower mass.

Effusion

Describes the passage of a gas through a tiny orifice into an evacuated chamber.

Rate of effusion measures the speed at which the gas is transferred into the

chamber. 64


Gases with lower molecular masses will diffuse and/or effuse faster than gases with higher molecular masses.


Effusion



Diffusion



Concept Check:

Arrange the following gases in order of increasing speed of the particles at the same temperature: SO2, NH4, Ne

SO2 = 64.1 g

NH4 = 18.0 g

Ne = 20.2 g

SO2, Ne, NH4


Graham’s Law of Effusion

M1 and M2 represent the molar masses of the gases.


gas a

gas b a

Rate

Rate

gas b

gas

Molarmass

Molarmass


Example:

Which gas effuses faster, Helium or Nitrogen? How much faster?

g

g

Rate

Rate

N

He

0.4

0.28

2

Helium is 2.6 times faster than N2;

Nitrogen is .38 times slower than He


Example:

Which gas effuses faster, Oxygen or Carbon Dioxide? How much faster?

g

g

Rate

Rate

CO

O

0.32

0.44

2

2

Oxygen effuses 1.17 times faster than CO2; CO2 effuses .855 times slower than O2


Since Rate is a unit per time (1/t), Graham’s law can be rearranged to solve for time instead of rate

assuming the same amount of each gas.

)(

)(

bmolarmass

amolarmass

Time

Time

b

a


Example: It takes 354 s for 1.00 mL of Xe to effuse through a small hole. Under the same conditions, how long will it take for 1.00 mL of nitrogen to effuse?

Solution A: x = 164 s

( )

( )

a

b

molarmass aTime

Time molarmass b

228.02

354 131.3

NTime

s


Example: It takes 354 s for 1.00 mL of Xe to effuse through a small hole. Under the same conditions, how long will it take for 1.00 mL of nitrogen to effuse?

Solution B:

x = 164 s

gas a

gas b a

Rate

Rate

gas b

gas

Molarmass

Molarmass

1.00mL131.3X

1.00mL 28.02

354 s

g

g

Section 5.8 Real Gases

Real Gases and van der Waals Equation

An ideal gas is a hypothetical concept. No gas exactly follows the ideal gas law.

Real gas molecules do have volume, and the molecules of real gases do experience attractive forces.

Gases behave most like ideal gases at low pressure, high

temperature.


Real Gases (van der Waals Equation)


n2 a (P + —— ) (V-nb) = nRT V2

‘a’ and ‘b’ are constants derived from experimental results a has units of L2atm/mol2

b has units of L/mol.

Corrected pressure

Corrected volume

Two corrective Factors:

1. n2a/V2 alters the pressure

•It accounts for the intermolecular attractive forces between gas molecules

•A low value for a reflects weak IM forces; a high value reflects stronger IM forces.

2. nb alters the volume

• It accounts for the volume occupied by the gas molecules

• The value of b is generally much lower than a

• The value of b generally increases with the size of the molecule

77

n2 a (P + —— )(V-nb) = nRT V2

Example: 1.00 mol of carbon dioxide gas at 373 K occupies 536 mL. What is the calculated value of the pressure in atm using: (a) Ideal gas equation (b) Van der Waals equation?

Van der Waals constants for carbon dioxide: a = 3.61 L2 atm mol-2; b = 0.0428 L mol-1 .

(a) Use the ideal gas law equation! PV=nRT (P) (.536L) = (1.00mol)(0.08206)(373K) P = 57.1atm

Example: 1.00 mol of carbon dioxide gas at 373 K occupies 536 mL. What is the calculated value of the pressure in atm using: (a) Ideal gas equation (b) Van der Waals equation?

Van der Waals constants for carbon dioxide: a = 3.61 L2 atm mol-2; b = 0.0428 L mol-1 .

(a) Use Van der Waals equation!

n2 a (P + —— )(V-nb) = nRT V2

(1.00)2(3.61) [P + -----------------][0.536–(1.00)(0.0428)]=(1.00)(0.08206)(373) (.536)2

(P + 12.6)(0.493) = 30.6 P + 12.6 = 62.1 P = 49.5 atm

Predict which substance has the largest "b" constant: NH3, N2, CH2Cl2, Cl2, CCl4

• The value of b constant is merely the actual volume of the gas.

• A larger molecular volume results in a larger b value.

• From the compound in the list, CCl4 is the largest so it will have the greatest b constant.

1.Using Van der Waals equation, calculate the temperature of 20.0 mole of helium in a 10.0 liter cylinder at 120 atm.

• Van der Waals constants for helium:

• a = 0.0341 L2 at mol-2; b = 0.0237 L mol-1

2. Compare this value with the temperature calculated from the ideal gas equation.

• (i) Using the Van der Waals equation: • Substituting in

(P + an2/V2)(V - nb) = nRT, P = 120 atm n = 20.0 mol V = 10.0 L

• (120 + 0.0341 x (20.0/10.0)2)(10.0 - 20.0 x 0.0237) = 20.0 x 0.0821 x T • [Note the value of R = 0.0821 because P is in atm and V in L]

(120 + 0.1364)(10.0 - 0.5) = 1.64 x T T = 696 K [Note that the correction to P is not significant as T is well above the temperature at which helium gas will liquefy while the volume correction is significant due to the high pressure.]

• (ii) Using the Ideal Gas Equation: Substituting in PV = nRT, 120 x 10.0 = 20.0 x 0.0821 x T T = 1200/1.642

• = 731 K

Section 5.10 Chemistry in the Atmosphere

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Section 5.10 Chemistry in the Atmosphere

Consider the following apparatus containing helium in both sides at 45°C. Initially the valve is closed.

After the valve is opened, what is the pressure of the helium gas?


3.00 atm 3.00 L

2.00 atm 9.00 L

EXERCISE!

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