Gases Properties of gases and gas laws.

We’re going to talk about the behavior of gases, but first…what is a gas???? There is a lot of “free” space in a gas. Gases can be expanded infinitely. Gases fill containers uniformly and completely. Gases diffuse and mix rapidly.

Around 1827, dust particles were seen to move in a random, zig-zag pattern under a microscope, Brownian Movement “I’ve been behind this guy in the hall!” From the idea of Brownian Movement came the explanation for the behavior of gases and, later, for other particles of matter.

I. Properties of gases:

1. Gases have mass. Gases seem to be weightless, but they are classified as matter, which means they have mass. The density of a gas – the number of particles per unit of volume – is much less than the density of a liquid or solid, however.

GAS DENSITY GAS DENSITY Higherdensity Lowerdensity

2 nd – Gases are Compressible If you squeeze a gas, its volume can be reduced considerably The low density of a gas means there is a lot of empty space between gas molecules.

3 rd – Gases fill their containers Gases spread out to fill containers until the concentration of gases is uniform throughout the entire space. This is why there is never an absence of air around you!

4 th – Gases diffuse Because of all of the empty space between gas molecules, another gas molecule can pass between them until each gas is spread out evenly throughout the entire container. This is called diffusion.

5 th – Gases exert pressure The sum of all of the collisions makes up the pressure the gas exerts. Gas particles exert pressure by colliding with objects in their path (like sides of a container).

6 th – Pressure depends on Temp The higher the temperature of a gas the faster the gas molecules move Results in more collisions -the higher the pressure that the gas exerts The reverse of that is true as well, as the temperature of a gas decreases – the pressure decreases. How are temperature & pressure related? DIRECTLY

II. Kinetic Molecular Theory 1. All matter is composed of tiny, discrete particles (atoms, ions, or molecules). 2. These particles are in rapid, random, constant straight line motion. This motion can be described by well-defined and established laws of motion.

3. All collisions between particles are perfectly elastic, meaning that there is no change in the total kinetic energy of two particles before and after their collision (they don’t gain or lose energy when they collide). II. Kinetic Molecular Theory

Gas variables In order to describe a gas sample completely and then make predictions about its behavior under changed conditions, it is important to deal with the values of: 4) amount of gas 3) temperature 2) volume 1) pressure

Standard Temperature and Atmospheric Pressure (STP) When we do calculations involving gases, we usually assume the gases are at “Standard Temperature and Pressure” unless otherwise noted.

Standard Temperature and Pressure of Gases The volume of a gas( V ) the number of gas particles in that volume( n ) the pressure of the gas( P ), Pressure is the force of the collisions between the gas particles and the sides of the container. the temperature of the gas( T ) The average kinetic energy of all the molecules is proportional to the temperature. These variables depend on one another.

STP Standard atmospheric pressure = 1 atmosphere = kilopascals = 760 mmHg Standard temperature = 0°Celsius = 273 K

Normal Air (atmospheric pressure) Pressure is the average pressure of the air at sea level under normal conditions. It will support a column of mercury 760 mm high

Converting Between Units of Pressure There are three different units of pressure used in chemistry. This is an unfortunate situation, but we cannot change it. You must be able to use all three. Here they are: 1. atmospheres (symbol = atm) 2. millimeters of mercury (symbol = mmHg), also referred to as (torr) 3. kiloPascals (symbol = kPa)

Converting Between Units of Pressure Conversion bridges are: 1 atm = kPa = 760 mmHg(torr) EX: Convert atm to mmHg atm ( 760 mmHg ) = 665mmHg 1 ( 1 atm )

EX: Convert kPa to mmHg kPa ( 760 mmHg ) = 1901 mmHg 1 ( kPa ) Converting Between Units of Pressure

1 atm = kPa = 760 mmHg(torr) kPa  atm mmHg  kPa mmHg  atm kPa  mmHg

Converting Between Units of Temperature – EVERY TEMPERATURE USED IN A CALCULATION MUST BE IN KELVINS, NOT DEGREES CELSIUS. K = °C EX: Convert 25°C to K K = K = 298

Practice °C  K °C  K °C  K °C  K °C  K °C  K

Volume (V) The volume of a gas is simply the volume of the container it is contained in. The metric unit of volume is the liter (L) 1 L = 1 dm 3 = 1000 mL = 1000 cm 3

Amount (n) The quantity of gas in a given sample is expressed in terms of moles of gas (n). This of course is in terms of 6.02 x molecules of the gas. Don’t forget to convert mass to moles you just divide by the molar mass of the gas.

III. Gas Laws

1. Boyle’s Law Robert Boyle was among the first to note the relationship between pressure and volume of a gas. He measured the volume of air at different pressures, and observed a pattern of behavior which led to his mathematical law. During his experiments Temperature and amount of gas weren’t allowed to change (remained constant)

Boyle’s Law He found that at a constant temperature and number of particles, pressure and volume are inversely related P 1 = V 2 P 2 V 1 As one goes up the other goes down P 1 V 1 = P 2 V 2 Robert Boyle ( ). Son of Earl of Cork, Ireland.

As the pressure increases Volume decreases Volume decreases

Applying Boyle’s Law: P 1 V 1 = P 2 V 2 Ex: A gas has a volume of 3.0 L at 2.5 atm. What is its volume at 4.3 atm? What if we had a change in conditions?

1)determine which variables you have: P and V = Boyle’s Law 2)determine which law is being represented: P 1 V 1 = P 2 V 2  P 1 = 2.5 atm  V 1 = 3.0 L  P 2 = 4.3 atm  V 2 = ?

3) Rearrange the equation for the variable you don’t know: V 2 = P 1 V 1 P 2 4 ) Plug in the variables and solve: (2.5 atm x 3.0L ) = (4.3 atm) V 2 = 1.74L

Applying Boyle’s Law EX: A gas is collected and found to fill 2.85 L at 245 kPa. What will be its volume at standard pressure? Step 1: Determine your variables P 1 (245kPa), P 2 (Standard is kPa), V 1 (2.85L), V 2 (?) Step 2: Isolate your unknown variable (V 2 ). Step 3: Plug in your known values. Step 4: Solve for your unknown.

How to Isolate a Variable (V 2 ): P 1 V 1 = P 2 V 2 P 2 P 2 P 1 V 1 = V 2 P 2

Now plug in your numeric values: 245 kPa x 2.85L = V kPa Solve: 6.89 L

More Boyle’s Problems 1. A sample of Neon gas has a volume of 250. mL at a pressure of 1.25 atm, what is the new volume if the pressure increases to 1.55 atm?

More Boyle’s Problems 2. A 5.00 L canister of O 2 gas has a pressure of 125 kPa. What volume would this gas occupy at 1435 mmHg?

More Boyle’s Problems L of Helium gas has a pressure of mmHg, what is the pressure of 7500mL?

2. Charles’ Law

Jacques Charles determined the relationship between temperature and volume of a gas. He measured the volume of air at different temperatures, and observed a pattern of behavior which led to his mathematical law. During his experiments pressure of the system and amount of gas were held constant.

Charles’s Law The volume of a gas varies directly with the absolute temperature, if pressure remains constant. Volume & Kelvin Temp. are directly proportional! V 1 = V 2 Jacques Charles ( ). Isolated boron and studied gases. Balloonist. T 1 T 2 V 1 T 2 = V 2 T 1

Volume of balloon at room Temperature Volume of balloon at 5 °C

If we place a balloon in liquid nitrogen it shrinks: How Volume Varies With Temperature So, gases shrink if cooled. Conversely, if we heat a gas it expands (as in a hot air balloon). Let’s take a closer look at temperature before we try to find the exact relationship of V vs. T.

Applying Charles’ Law: EX: A gas has a volume of 3.0 L at 127.0°C. What is its volume at °C? What if we had a change in conditions? V 1 T 2 = V 2 T 1

1)Determine which variables you have: (ALWAYS CHANGE CELSIUS TO KELVIN!) 2)Determine which law is being represented: T and V = Charles’ Law V 1 T 2 = V 2 T 1  T 1 = 127.0°C = 400.0K  V 1 = 3.0 L  T 2 = 227.0°C = 500.0K  V 2 = ?

3) Rearrange to solve for unknown: (500.0K x 3.0L) = V 1 T 2 V 2 = 4) Plug in the variables and solve: T1T1 ( 400.0K V 2 = 3.8L

Applying Charles’s Law EX: A gas is collected and found to fill 2.85 L at 25.0°C. What will be its volume at standard temperature? Step 1: Determine your variables V 1 (2.85L), T 1 (25 O C = 298.0K), V 1 (?), T 2 (standard is 273K) Step 2: Isolate your unknown variable (V 2 ). Step 3: Plug in your known values. Step 4: Solve for your unknown.

Isolate unknown variable: V 1 T 2 = V 2 T 1 Solve: 2.85 L x 273 K = 2.61L K

More Charles’s Problems L of a gas is collected at 100. K and then allowed to expand to 20.0 L. What must the new temperature be in order to maintain the same pressure?

More Charles’s Problems 2. Calculate the decrease in temperature when 2.00 L at 20.0 °C is compressed to 1000 mL.

More Charles’s Problems mL of nitrogen gas is collected at 35.5 °C, calculate new volume if the gas is heated to 65.7 °C.

3. The Combined Gas Law

Combined Gas Law The good news is that you don’t have to remember each individual gas law! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION! P 1 V 1 T 2 = P 2 V 2 T 1 If you should only need one of the other gas laws, you can cover up the item that is constant and you will get that gas law!

Combined Gas Law P 1 V 1 T 2 = P 2 V 2 T 1 We use the combined gas law equation when pressure, volume, and temperature are all changing.

Using Combined Gas Law 2.00 L of a gas is collected at 25.0°C and mmHg. What is the volume at STP? Step 1: Make a table for known values: P 1 = mmHg P 2 = mmHg V 1 = 2.00 LV 2 = ? T 1 = KT 2 = 273 K

Isolate V 2 V 2 = P 1 V 1 T 2 P 2 T 1 V2 = mmHg x 2.00 L x 273 K mmHg x K V2 = 1.80 L

Combined Gas Law Problems A gas has a volume of mL at °C and torr. What would the volume of the gas be at °C and torr of pressure?

Combined Gas Law Problems The pressure of a gas is reduced from mm Hg to kPa as the volume of its container is increased by moving a piston from L to mL. What would the final temperature be if the original temperature was 90.0 °C?

4. Ideal Gas Law

Avogadro’s Hypothesis Equal volumes of gases at the same T and P have the same number of molecules. V and n are directly related. (twice as many molecules) 1 mole of gas (6.02 x atoms) at STP takes up 22.4 L of space (volume) 2 moles 2 x avogadro’s # 44.8 L at STP

4. The Ideal Gas Equation When we combine Boyle’s, Charles’s, and Avogadro’s Laws, we can derive the ideal gas equation: PV = nRT * P = pressure, V = volume, n = moles, T = temperature, and R is the gas constant L x atm mol x K

*Whenever you use the ideal gas constant (R), all units must agree when solving problems. *All volumes must be in liters. *All substances must be in moles. *All pressures must be in atm. *All temperatures must be in Kelvin. The Ideal Gas Equation

Solving Problems Using The Ideal Gas Equation EX: Calculate the volume of mol of an ideal gas at atm and 0.00°C. *all units must agree with the units of the gas constant. (L, atm, mol, and K) *convert 0.00°C to K = K

*isolate volume from PV=nRT V = nRT P *solve: V = mol x x K atm V = L

More Ideal Gas Problems #1. EX: A sample of CaCO 3 is decomposed, and the carbon dioxide is collected in a 250. mL flask. After the decomposition is complete, the gas has a pressure of mmHg at a temp of 31.0°C. How many moles of CO 2 were generated?

More Ideal Gas Problems #2. EX: Dinitrogen monoxide (N 2 O), laughing gas, is used by dentists as an anesthetic. If 126 grams of gas occupies a 20.0 L tank at 23.0°C, what is the pressure in the tank in the dentist office?

5. Law of Partial Pressures

5. Dalton’s Law of Partial Pressure Our calculations so far have been for pure gases. John Dalton formed a hypothesis about pressure exerted by a mixture of gases. Dalton’s Law of Partial Pressure: The total pressure in a container is the sum of the partial pressures of all the gases in the container. P total = P 1 + P 2 + P 3 …

Dalton’s Law of Partial Pressure

Partial Pressures The total pressure of a gas mixture depends on the total number of gas particles, not on the types of particles. P = 1.00 atm: mole O mole He mole Ar 1 mole H 2

We can find out the pressure in the fourth container By adding up the pressure in the first 3 0.3kPa + 0.2kPa + 0.5kPa = 1.00kPa

Gases in a single container are all the same temperature and have the same volume, therefore, the difference in their partial pressures is due only to the difference in the numbers of molecules present.

Partial Pressure of Air Air is an example of a mixture of gases. -nitrogen is % -oxygen is % -argon is % -carbon dioxide is % -neon, helium, krypton, and xenon are among the other trace gases.

The total pressure of the atmosphere at STP is kPa. If 78% of air is nitrogen, then 78% of pressure is due to nitrogen molecules x = 79 kPa 21% of air is oxygen so 21% of pressure is due to oxygen molecules x = 22 kPa Partial Pressure of Air

Partial Pressure Practice Problems 1. N 2 with a pressure of 220 mmHg, H 2 with a pressure of 176 mmHg, NH 3 at 300. mmHg, and SO 3 at a pressure of 101 mmHg are mixed. What is the total pressure of the container?

Partial Pressure Practice Problems 2. If a container of a mixture of gases has a total pressure of 150 kPa, what is the partial pressure of argon gas if the other gases are I 2 with a pressure of 35 kPa, CO 2 at 64 kPa, and Br 2 is at 24 kPa?

Collecting Gases by Water Displacement One method to collect gases is by water displacement. Gases must be insoluble in water. When collection is complete, water vapor is present in the collection container and must be accounted for in the partial pressures of gases.

Collecting Gases by Water Displacement

How to find the pressure of a dry gas? P dry gas = P total from problem – P water (table)

Example EX: A quantity of gas is collected over water at 8°C in a L vessel at 84.5 kPa.What volume would the dry gas occupy at standard atmospheric pressure and 10°C?

Example Find the pressure of the dry gas: P gas = P total – P water Obtain P water from table. P gas = 84.5 kPa – 1.23 kPa = kPa The remainder of this problem is a pressure and volume comparison. Boyle’s equation will now be used.

Example P 1 V 1 = P 2 V 2 P 1 V 1 = V 2 P kPa x L =.291 L kPa *Hint: adjust initial pressure, P 1 or P only! Look for dry gas, gas by water displacement.

Partial Pressure Practice Problems 1. A gas is collected over water and occupies a volume of 596 cm 3 at 43°C. The atmospheric pressure is kPa. What volume will the dry gas occupy at 50.0°C and standard atmospheric pressure?

Partial Pressure Practice Problems #2) 250. mL of nitrogen gas was collected over water at 24.0°C and mmHg. What is the new temperature of the dry gas if 350. mL is at a new pressure of mmHg?

Partial Pressure Practice Problems #3.) 1.50 moles of oxygen gas was collected over water at 25°C and 95.5 kPa of pressure. Calculate the volume of the dry gas.

Gases THE END!!!