Chapters 13.1 and 14

Basic assumptions describing the behavior of gases:

1.

2.

3.

Gas consist of small particles that are very far apart   Most of the volume of a gas is empty space Individual particles have essentially zero volume compared to the total volume of the gas Gases particles are in constant, random, straight line motion Attractive and repulsive forces between particles are negligible

Basic assumptions:

4.

5.

Random motion of gas particles is constantly interrupted by collisions between particles and between particles and surfaces   Such collisions are perfectly elastic Kinetic energy may be transferred between particles but is not lost (conserved)  The average kinetic energy of gas particles is directly proportional to absolute temperature (kelvin) The rate of motion of particles is related to temperature

Characteristics of gases:



Gases readily change

volume  Container containing gas is always full     Molecules separate and spread out uniformly until fill available space Compressibility : can be compressed or reduced to small fractions of original volume Expandability : can expand to fill virtually any volume Gases typically have low densities (mass to volume ratio)

Characteristics of gases:



Gases exert

pressure   Due to constant random motion, molecules collide with each other and with the walls of their container Collisions exert pressure on surroundings

Characteristics of gases:



Gas

Temperature 

and Kinetic Energy

Relationship between average kinetic energy of group of gas molecules and the temperature measured in degrees Kelvin is directly proportional    As temperature decreases, average kinetic energy of molecules ____________ Relationship only exists when temperature is expressed in Kelvin scale  As temperature increases, average kinetic energy of molecules ____________ Molecules have 0 kinetic energy when temp is 0 K  Only have kinetic energy at temps above 0 K

   Describes movement of gases through other materials Gas molecules move from areas of high concentration to low concentration Examples:  Perfume molecules spreading across a room  Aromas of cooking foods waft through the house

DIFFUSION

Animation by Caitlin Claunch

Animation by Caitlin Claunch    Gas molecules escape through tiny openings Gas molecules move from an area of high pressure to low pressure Examples:   Tire flattens due to a puncture Balloon deflates when stuck with a pin

EFFUSION

 Gas particles are in constant, ________ motion. Particles travel in ____________.

 The size of the particles is ___________ compared to empty  Attraction and repulsion between particles is __________.

 Collisions are ______ because particles transfer constant  ____________ is the measure of average kinetic average kinetic all  In diffusion, gas particles move from high concentration effusion, particles move from high ________ to low pressure

 Pressure (P): force per unit area  Volume (V): amount of space taken up  Temperature (T): measure of average kinetic energy  Number of Moles (n): amount of substance

 Results from billions of gas particles in motion and colliding against their container  Barometer : instrument used to measure atmospheric pressure  Manometer : instrument used to measure gas pressure in a closed container

 UNITS:

atm mm Hg

1 760

torr

760

psi

14.7

kPa

101.325

Pa

101,325

SI UNIT

 The relationships among the different units of pressure can be used as conversion factors  Values expressed in table above describe the same amount of pressure

The total pressure of a mixture of gases is equal to the sum of the pressures of all the gases in the mixture

P

total

= P

1

+ P

2

+ P

3

+ … P

n

kPa kPa kPa kPa kPa 101.3

kPa

A mixture of oxygen (O 2 ), carbon dioxide (CO 2 ), and nitrogen (N 2 ) has a total pressure of 0.97 atm. What is the partial pressure of O 2 , if the partial pressure of CO 2 partial pressure of N 2 is 0.70 atm and the is 0.12 atm?

Known P N 2 = 0.12 atm P CO 2 = 0.70 atm P total = 0.97 atm Unknown P O 2 = ? atm P total P O2 P O2 = P = P O2 total + P CO2 – P CO2 + P – P N2 N2 = 0.97 atm – 0.70 atm – 0.12 atm =

0.15 atm

PRACTICE/EXAMPLES

What is the partial pressure of hydrogen gas in a mixture of hydrogen and helium if the total pressure is 600 mm Hg and the partial pressure of helium is 439 mm Hg?

Known P He P total = 439 mm Hg = 600 mm Hg Unknown P H 2 = ? mm Hg P total P H2 P H2 = P H2 = P total + P – P He He = 600 mm Hg – 439 mm Hg =

161 mm Hg

Find the total pressure for a mixture that contains four gases with partial pressures of 5.00 kPa, 4.56 kPa, 3.02 kPa, and 1.20 kPa. Known P 1 = 5.00 kPa P 2 = 4.56 kPa P 3 P 4 = 3.02 kPa = 1.20 kPa Unknown P total = ? kPa P total P total = P 1 + P 2 + P 3 + P 4 = 5.00 kPa + 4.56 kPa + 3.02 kPa + 1.20 kPa =

13.78 kPa

Find the partial pressure of carbon dioxide in a gas mixture with a total pressure of 30.4 kPa if the partial pressures of the other two gases in the mixture are 16.5 kPa and 3.7 kPa.

Known P 1 = 16.5 kPa P 2 = 3.7 kPa P total = 30.4 kPa Unknown P CO 2 = ? kPa P total P CO2 P CO2 = P = P 1 + P total 2 + P CO2 – P 1 – P 2 = 30.4 kPa – 16.5 kPa – 3.7 kPa

= 10.2 kPa

p. 415: #68, #69, #70, #71

Pressure Unit Conversions Worksheet

 Gases expand to fill their container  That is, they do not have a definite volume  Gas laws explain how and when gas volume changes  UNITS:

L

1

mL

1000

cm 3

1000

quart

1.057

SI UNIT

 Measure of average kinetic energy of particles  Recall KMT: At any given temperature, all gases have same average kinetic energy  UNITS:    Measured using the Kelvin scale (absolute temperature scale) developed by Lord Kelvin 0 degrees kelvin is defined as absolute zero Absolute zero is the hypothetical temperature at which all motion stops

K = °C + 273



S

tandard

T

emperature and

P

ressure



0°C or 273 K



1 atm or 101.325 kPa

States the relationship among pressure (P), volume (V), and temperature (T) when amount

of gas (n) remains constant

Pressure and Volume are proportional Pressure and Temperature are proportional Volume and Temperature are proportional

A gas occupies 7.84 cm 3 at 71.8 kPa and 25°C. What volume will it occupy at STP?

Rearrange to solve for unknown variable, V 2 VARIABLES P V 1 T 1 P 2 V 2 T 2 1 = = = = = unknown = 71.8 kPa 7.84 cm 273 K 3 25°C + 273 K = 298 K 101.325 kPa

A gas occupies 7.84 cm 3 at 71.8 kPa and 25°C. What volume will it occupy at STP?

P 1 V 1 T 1 P 2 V 2 T 2 VARIABLES = = = V 2 71.8 kPa 7.84 cm 3 25°C + 273 K = 298 K

V 2

= (71.8 kPa)(7.84 cm 3 )(273K) (101.325 kPa)(298K)

= 5.1 cm 3

= 101.325 kPa = unknown = 273 K

p. 449: #92, #93, #94

PRACTICE/EXAMPLES

A helium-filled balloon at sea level has a volume of 2.1 L at 0.998 atm and 36°C. If it is released and rises to an elevation at which the pressure is 0.900 atm and the temperature is 28°C, what will be the new volume of the balloon?

VARIABLES P 1 V 1 T 1 P 2 V 2 T 2 = = = = = 0.998 atm 2.1 L 36°C + 273 K = 303 K 0.900 atm = unknown 28°C + 273 K = 301 K

V

2

= 2.3 L

At 0.00°C and 1.00 atm pressure, a sample of gas occupies 30.0 mL. If the temperature is increased to 30.0°C and the entire gas sample is transferred to a 20.0-mL container, what will be the gas pressure inside the container?

VARIABLES P 1 V 1 T 1 P 2 V 2 T 2 = = = = = 1.00 atm 30.0 mL 0.00°C + 273 K = 273 K unknown = 20.0 mL 30.0°C + 273 K = 303 K P 2

P

2

= P 1 V 1 T 2 V 2 T 1

= 2.3 L

States the relationship between pressure (P) and volume (V), when temperature (T) and amount

of substance (n) remain constant

Boyle’s Law

• Pressure and Volume are __________ proportional • As pressure increases, volume __________ • As pressure decreases, volume __________

A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa.

Rearrange to solve for unknown variable, V 2 VARIABLES P V 1 T 1 P 2 V 2 T 2 1 = = = = = 150. kPa 100. mL remains constant 200. kPa = unknown remains constant

A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa.

VARIABLES P V 1 T 1 P 2 V 2 T 2 1 = = = = = 150. kPa 100. mL remains constant 200. kPa = unknown remains constant V 2 V 2 = (150. kPa)(100. mL) (200. kPa) = 75.0 mL

States the relationship between volume (V) and temperature (T), when pressure (P) and amount

of substance (n) remain constant

Charles’ Law

• Volume and Temperature are __________ proportional • As volume increases, temperature __________ • As volume decreases, temperature __________

A gas occupies 473 cm 3 volume at 94° C .

at 36° C . Find its Rearrange to solve for unknown variable, V 2 VARIABLES P V 1 T 1 P 2 V 2 T 2 1 = = = = = unknown = remains constant 473 cm 3 36°C + 273 K = 309 K remains constant 94°C + 273 K = 367 K

A gas occupies 473 cm 3 volume at 94° C .

at 36° C . Find its VARIABLES P V 1 T 1 P 2 V 2 T 2 1 = = = = = unknown = remains constant 473 cm 3 36°C + 273 K = 309 K remains constant 94°C + 273 K = 367 K V 2 V 2 = (473 cm 3 )(367 K) (309 K) = 560 cm

3

States the relationship between pressure (P) and temperature (T), when volume (V) and amount

of substance (n) remain constant

Gay-Lussac’s Law

• Pressure and Temperature are __________ proportional • As pressure increases, temperature __________ • As temperature decreases, pressure __________

The pressure of a gas in a tank is 3.20 atm at 22.0° C . If the temperature rises to 60.0° C, what will be the gas pressure in the tank?

Rearrange to solve for unknown variable, P 2 VARIABLES P V 1 T 1 P 2 V 2 T 2 1 = = = = = 3.20 atm remains constant 22.0°C+ 273 K = 295 K unknown = remains constant 60.0°C+ 273 K = 333 K

The pressure of a gas in a tank is 3.20 atm at 22.0° C . If the temperature rises to 60.0° C, what will be the gas pressure in the tank?

VARIABLES P V 1 T 1 P 2 V 2 T 2 1 = = = = = 3.20 atm remains constant 22.0°C+ 273 K = 295 K unknown = remains constant 60.0°C+ 273 K = 333 K P 2 P 2 = (3.20 atm)(333.0K) (295.0K) = 3.61 atm

p. 448: #88, #89, #90, #91

PRACTICE/EXAMPLES

A sample of air in a syringe exerts a pressure of 1.02 atm at a temperature of 22.0° at 100.0

of air?

° C. The syringe is placed in a boiling water bath C. The pressure of the air is increased to 1.23 atm by pushing the plunger in, which reduces the volume to 0.224 mL. What was the original volume VARIABLES P V 1 T 1 P 2 V 2 T 2 1 = = = = = =

The volume of a gas at 99.0 kPa is 300.0 mL. If the pressure is increased to 188 kPa, what will be the new volume?

VARIABLES P V 1 T 1 P 2 V 2 T 2 1 = = = = = =

A gas at 89° C occupies a volume of 0.67 L. At what Celsius temperature will the volume increase to 1.12 L?

VARIABLES P V 1 T 1 P 2 V 2 T 2 1 = = = = = =

A gas in a sealed container has a pressure of 125 kPa at a temperature of 30.0° C. If the pressure in the container is increased to 201 kPa, what is the new temperature?

VARIABLES P V 1 T 1 P 2 V 2 T 2 1 = = = = = =

The pressure of a sample of helium in a 1.00-L container is 0.988 atm. What is the new pressure if the sample is placed in a 2.00-L container?

VARIABLES P V 1 T 1 P 2 V 2 T 2 1 = = = = = =

An unopened, cold 2.00-L bottle of soda contains 46.0 mL of gas confined at a pressure of 1.30 atm at a temperature of 5.0° and the temperature is 2.09

° C. If the bottle is dropped into a lake and sinks to a depth at which the pressure is 2.85 atm C, what will be the volume of gas in the bottle?

VARIABLES P V 1 T 1 P 2 V 2 T 2 1 = = = = = =

Air trapped in a cylinder fitted with a piston occupies 145.7 mL at 1.08 atm pressure. What is the new volume of air when the pressure is increased to 1.43 atm by applying force to the piston?

VARIABLES P V 1 T 1 P 2 V 2 T 2 1 = = = = = =

A sample of gas of unknown pressure occupies 0.766 L at a temperature of 298 K. The same sample of gas is then tested under known conditions and has a pressure of 32.6 kPa and occupies 0.644 L at 303 K. What was the original pressure of the gas?

VARIABLES P V 1 T 1 P 2 V 2 T 2 1 = = = = = =

If a gas sample has a pressure of 30.7 kPa at 0.00° C, by how much does the temperature have to decrease to lower the pressure to 28.4 kPa?

VARIABLES P V 1 T 1 P 2 V 2 T 2 1 = = = = = =

 Addresses variable n (number of moles)  States that equal volumes of gases at the same temperature and pressure contain the same numbers of particles  Molar volume of gases: 22.4 L/mol at standard temperature and pressure ( STP )  Standard temperature: 273 K or 0°C  Standard pressure: 1.00 atm, 101.325 kPa, 760 torr, 760 mm Hg 

Conversion factor 22.4 L of any gas = 1 mol

Calculate the volume that 0.881 mol of gas at STP will occupy.

Is amount of substance (n) involved?

Yes Is gas at STP?

Yes Use relationship between volume and number of moles ( Avogadro’s Law )

Calculate the volume that 0.881 mol of gas at STP will occupy.

Conversion factor: 22.4 L = 1 mol Unit for volume?

Liter (L)

L = 0.881 mol

Calculate the volume that 0.881 mol of gas at STP will occupy.

Conversion factor: 22.4 L = 1 mol L = 0.881 mol mol

Calculate the volume that 0.881 mol of gas at STP will occupy.

Conversion factor: 22.4 L = 1 mol L = 0.881 mol mol

Calculate the volume that 0.881 mol of gas at STP will occupy.

Conversion factor: 22.4 L = 1 mol L = 0.881 mol L mol

Calculate the volume that 0.881 mol of gas at STP will occupy.

Conversion factor: 22.4 L = 1 mol L = 0.881 mol 22.4 L 1 mol

Calculate the volume that 0.881 mol of gas at STP will occupy.

L = 0.881 mol 22.4 L 1 mol L = 19.7 L

Calculate the volume that 2.0 kg of methane gas (CH

4

) will occupy at STP.

L = 2.0 kg 1000 g 1 kg 1 mol 16.0 g 22.4 L 1 mol L = 2800 L

p. 415: #66, #67

Express answers in atm , kPa , and torr .

p. 449: #95, #96

PRACTICE/EXAMPLES

Determine the volume of a container that holds 2.4 mol of gas at STP.

What size container do you need to hold 0.0459 mol N 2 gas at STP?

What volume will 1.02 mol of carbon monoxide gas occupy at STP?

How many moles of nitrogen gas will be contained in a 2.00-L flask at STP?

If a balloon will rise off the ground when it contains 0.0226 mole of helium in a volume of 0.460 L, how many moles of helium are needed to make the balloon rise when its volume is 0.865 L? Assume that temperature and pressure stay constant.

How many grams of carbon dioxide gas are in a 1.0-L balloon at STP?

What volume in milliliters will 0.00922 g H 2 occupy at STP?

gas

 Describes the physical behavior of an ideal gas in terms of all four variables     Pressure (P) Volume (V) Temperature (T) Amount of gas (n)

 Described by Kinetic Molecular Theory  Particles have no volume, no intermolecular attractive forces, and perfectly elastic collisions  Follow all gas laws under all conditions of temperature and pressure  Behave less like ideal gases at extremely high pressures and low temperatures  Polar gases do not behave as ideal gases  Larger gas molecules behave less like ideal gases than smaller gas molecules

IDEAL GASES REAL GASES

 Combines:     Boyle’s Law Charles’ Law Gay-Lussac’s Law Avogadro’s Law Combined Gas Law If number of moles (n) is constant If number of moles (n) is not constant, and At STP → Not at STP Avogadro’s Law → Ideal Gas Law → Combined Gas Law

 Experimentally determined  Three possible numerical values

How will you know which value to use?

A sample of nitrogen gas (N 2 ) has a volume of 5.56 liters at 0.00°C and 1.50 atm pressure. How many moles of nitrogen are present?

VARIABLES P = 1.50 atm Rearrange to solve for unknown variable, n V = 5.56 L n = ? mol R = T = 273 K

A sample of nitrogen gas (N 2 ) has a volume of 5.56 liters at 0.00°C and 1.50 atm pressure. How many moles of nitrogen are present?

VARIABLES P = 1.50 atm V = 5.56 L n = ? mol R = T = 273 K n = ( (1.50 atm)(5.56 L) ( )(273 K) n = 0.372 mol

PRACTICE/EXAMPLES

Avogadro’s Law

22.4 L = 1 mol

What volume does 9.45 g of C 2 H 2 occupy at STP?

L = 9.45 g C

2

H

2

1 mol C

2

H

2

26.0 g C

2

H

2

22.4 L C

2

H

2

1 mol C

2

H

2

L = 8.14 L

Ideal Gas Law

PV = nRT

What volume does 9.45 g of C 2 H 2 occupy at STP?

Rearrange to solve for unknown variable, V VARIABLES P = 101.325 kPa V = ? L n = 9.45 g 26.0 g/mol = 0.3634615385 mol R = T = 273 K

What volume does 9.45 g of C 2 H 2 occupy at STP?

VARIABLES P = 101.325 kPa V = ? L n = 9.45 g 26.0 g/mol = 0.3634615385 mol R = T = 273 K

What volume does 9.45 g of C 2 H 2 occupy at STP?

V = (0.3634615385 mol)( )(273 K) (101.325 kPa) V =

8.14 L

You decide to go on a long hot air balloon ride, and pack a bottle of shampoo for your trip. There is some gas inside the shampoo bottle when you climb into the basket at the beginning of your journey. Being the good scientist that you are, you decide to take constant measurements of your surroundings. The shampoo bottle contains 435 mL of gas, is under a pressure of 1.10 atm, and is at a temperature of 30.0°C. As you climb high into the air, the bottle starts to expand and eventually explodes, covering you and your companions with Pert Plus®. Eager to explain this phenomenon, you take some quick measurements, noting that the pressure has dropped to 0.734 atm and the temperature is now 5.00°C. To what new volume did the gas inside the shampoo bottle expand?

VARIABLES P V 1 T 1 1 = = = 1.10 atm 435 mL 30.0°C = 303 K P 2 V 2 T 2 = = 0.734 atm = ?

? =

598 mL

5.00°C = 278 K

Gas Laws Notes

   Stoichiometric calculations always require a balanced equation (1) N 2 3 2 (g) → 2 NH 3 (g) Coefficients indicate ratios of reactants and products  1 mole of nitrogen gas reacts with 3 of hydrogen gas to produce 2 ammonia gas OR moles moles of  1 volume volumes volumes of nitrogen gas reacts with 3 of hydrogen gas to produce 2 of ammonia gas Volume ratios only work when all reactants and products are gases

N 2 (g) + 3 H 2 (g) → 2 NH 3 (g) How many liters of gaseous ammonia will be made from 5.00 L of hydrogen gas reacting with excess nitrogen gas?

L NH 3 = 5.00 L H 2

N 2 (g) + 3 H 2 (g) → 2 NH 3 (g) How many liters of gaseous ammonia will be made from 5.00 L of hydrogen gas reacting with excess nitrogen gas?

L NH 3 = 5.00 L H 2 L H 2

N 2 (g) + 3 H 2 (g) → 2 NH 3 (g) How many liters of gaseous ammonia will be made from 5.00 L of hydrogen gas reacting with excess nitrogen gas?

L NH 3 = 5.00 L H 2 2 3 L NH 3 L H 2 = 3.33 L NH 3

Volume-Volume WS

Gas Laws Notes

Gas stoichiometry problems can involve mass and volume 2 → H 2 (g) + ZnCl 2 (aq) What volume (in L) of hydrogen can be produced when 6.54 g of zinc reacts with hydrochloric acid at STP?

L H 2 = 6.54 g Zn

Gas stoichiometry problems can involve mass and volume Zn(s) + 2 HCl(aq) → H 2 (g) + ZnCl hydrochloric acid at STP?

2 (aq) What volume (in L) of hydrogen can be produced when 6.54 g of zinc reacts with L H 2 = 6.54 g Zn 1 mol Zn 65.4 g Zn

Gas stoichiometry problems can involve mass and volume Zn(s) + 2 HCl(aq) → H 2 (g) + ZnCl hydrochloric acid at STP?

2 (aq) What volume (in L) of hydrogen can be produced when 6.54 g of zinc reacts with L H 2 = 6.54 g Zn 1 mol Zn 1 mol H 2 65.4 g Zn 1 mol Zn

Gas stoichiometry problems can involve mass and volume Zn(s) + 2 HCl(aq) → H 2 (g) + ZnCl hydrochloric acid at STP?

2 (aq) What volume (in L) of hydrogen can be produced when 6.54 g of zinc reacts with L H 2 = 6.54 g Zn 1 mol Zn 1 mol H 2 22.4 L H 2 65.4 g Zn 1 mol Zn 1 mol H 2 2.24 L H

2

Gas stoichiometric calculations only involving volume do not take the temperature and pressure conditions into consideration  After mixing, each gas is at the same temperature and pressure Gas stoichiometric calculations involving volume and mass must take temperature and pressure conditions into account  

volume-volume relationships from balanced chemical equation are combined with the Ideal Gas Law

When 3.00 L of propane gas is completely combusted to form water vapor and carbon dioxide at a temperature of 350°C and a pressure of 0.990 atm, what mass of water vapor will result? (

p. 449, #102

) C C 3 H 8 (g) + O 2 (g) → H 2 O (g) + CO 2 2 (g) 1.

Determine volume of water produced ( volume-volume relationships).

L H 2 O = 3.00 L C 3 H 8 4 L H 2 O 1 L C 3 H 8 = 12.0 L H 2 O

When 3.00 L of propane gas is completely combusted to form water vapor and carbon dioxide at a temperature of 350°C and a pressure of 0.990 atm, what mass of water vapor will result? (

p. 449, #102

) C 3 H 8 (g) + 5 O 2 (g) → 4 H 2 O (g) + 3 CO 2 (g) 2.

Determine number of moles under given conditions ( Ideal Gas Law ).

n = 0.232265784 mol L  atm mol  K 2 O)

When 3.00 L of propane gas is completely combusted to form water vapor and carbon dioxide at a temperature of 350°C and a pressure of 0.990 atm, what mass of water vapor will result? (

p. 449, #102

) C 3 H 8 (g) + 5 O 2 (g) → 4 H 2 O (g) + 3 CO 2 (g) g H 2 O = 0.232265784 mol H 2 O 18.0 g H 2 O 1 mol H 2 O g H 2 O = 4.180784112 g H 2 O

Gas Stoichiometry WS

p. 449 - #100, #103, #104

100.

4.15 g C

3

H

8

103. 2KClO

3

(s)

→

2KCl (s) + 3O

2

5.70 L O

2

104. a. 2CO (g) + 2NO (g)

→

N

2

(g) (g) + 2CO

2

b.

2 volumes CO 2 volumes CO

2

c. 17.1 L N

2

(g)

volume = L, mL or cm 3

PRACTICE/EXAMPLES

If the pressure exerted by a gas at 25°C in a volume of 0.044 L is 3.81 atm, how many moles of gas are present?

Variable(s) involved?

P, V, T, n Gas law/application?

Ideal Gas Law P: 3.81 atm V: n: R: T: ?

0.044 L 0.0821

298 K L  atm mol  K

n = 0.0069 mol

Calculate the volume that 4.5 kg of ethylene gas (C 2 H 4 ) will occupy at STP.

Variable(s) involved?

V, n (mass) @STP Gas law/application?

Avogadro’s Law

3600 L C

2

H

4

The Celsius temperature of a 3.00-L sample of gas is lowered from 80.0° C to 30.0

° C. What will be the resulting volume of this gas?

Variable(s) involved?

V, T Gas law/application?

Combined Gas Law VARIABLES P V 1 T 1 P 2 V 2 T 2 1 = = = = = =

V

2

= 2.58 L

Determine the volume of hydrogen gas needed to react completely with 5.00 L of oxygen gas to form water.

Variable(s) involved?

Gas law/application?

V volume-volume stoichiometry  Write the balanced chemical equation.

2H 2 (g) + O 2 (g) → 2H 2 O (g) Unknown: Given: L

H

5.00 L Answer: 10.0 L 2

O

2

H

2

Calculate the volume that a 0.323-mol sample of a gas will occupy at 265 K and a pressure of 0.900 atm.

Variable(s) involved?

P, V, T, n Gas law/application?

Ideal Gas Law P: 0.900 atm V: n: R: T: ?

0.323 mol 0.0821

265 K L  atm mol  K

V = 7.81 L

What is the pressure in atmospheres of a 0.108-mol sample of helium gas at a temperature of 20.0°C if its volume is 0.505 L?

Variable(s) involved?

P, V, T, n Gas law/application?

Ideal Gas Law P: ?

V: n: R: T: 0.505 L 0.108 mol 0.0821

293 K L  atm mol  K

P = 5.14 atm

A rigid plastic container holds 1.00 L methane gas at 660 torr pressure when the temperature is 22.0° to 44.6

C. How much more pressure will the gas exert if the temperature is raised ° C?

Variable(s) involved?

Gas law/application?

P, T Combined Gas Law VARIABLES P V 1 T 1 P 2 V 2 T 2 1 = = = = = =

P

2

= 711 torr Additional pressure of 51 torr will be exerted.

What volume of oxygen is needed to react with solid sulfur to form 3.5 L SO 2 ?

Variable(s) involved?

V Gas law/application?

volume-volume stoichiometry  Write the balanced chemical equation.

S (s) + O 2 (g) → SO 2 (g) Unknown: L

O

2 Given: 3.5 L Answer: 3.5 L

SO

2

O

2