GASES Chemistry I Honors – Chapter 13 1

Importance of Gases (don’t copy) Airbags fill with N 2 gas in an accident. Airbags fill with N 2 gas in an accident. Gas is generated by the decomposition of sodium azide, NaN 3. Gas is generated by the decomposition of sodium azide, NaN 3. 2 NaN 3 --->2 Na + 3 N 2 2 NaN 3 --->2 Na + 3 N 2 2

General Properties of Gases 1. There is a lot of “free” space in a gas. 2. Gases can be expanded infinitely. (they will fill whatever “container” they are in.) 3. Gases fill containers uniformly and completely. 4. Gases diffuse and mix rapidly. 3

Pressure Pressure is the amount of force per unit area. 1. Gas Pressure is the pressure caused by particles of gas striking an object. 2. Atmospheric pressure is the pressure of the column of atmosphere above you, pressing down on you. 4

Atmospheric Pressure 5 (COPY) Measured with a BAROMETER (developed by Torricelli in 1643)Measured with a BAROMETER (developed by Torricelli in 1643) Hg rises in tube until force of Hg (UP) balances the force of atmosphere (DOWN). (Just like a straw in a soft drink)Hg rises in tube until force of Hg (UP) balances the force of atmosphere (DOWN). (Just like a straw in a soft drink) (DON’T COPY) Q: Why is Hg so good for use in barometer? A: If you tried to use water, it would rise about 34 feet high!

Atmospheric. Pressure COPY The column height (of the barometer) measures atmospheric pressure 1 standard atmosphere (atm) * = kPa (SI unit is PASCAL) = 760 mm Hg = 14.7 pounds/in2 (psi) * These are the 2 units we will use most of the time From the height of mercury in the barometer

Pressure Conversions A. What is 2.71 atm expressed in kPa? B. The pressure in a tire reads 262 kPa. What is this pressure in atm? 7

Properties of Gases Gas properties can be modeled using math. The factors that affect gases are: V = volume of the gas (L) V = volume of the gas (L) T = temperature (K) T = temperature (K) n = amount (moles) n = amount (moles) P = pressure (atm, kPa, mm Hg, or psi) P = pressure (atm, kPa, mm Hg, or psi) 8 All temperatures must be in Kelvins! No exceptions!

Converting Temperatures to Kelvin Kelvin = ˚Celsius OR K = ˚C Copy onto P. Table Example: convert 25 ˚C to Kelvins. K = 25 ˚C = 298 K

Properties of Gases, cont. We can study the relationship between 2 variables if we keep the others the same. 10 VARIABLES: Moles, temperature, pressure, and volume

The relationship between: n (moles) & P (pressure) Pressure is caused by the particles striking the walls of the container. If the gas is in a rigid container, the volume is constant (not changing) If you increase the number of gas particles (n), you increase the pressure. AnimationAnimation 11

SAMPLE PROBLEM n (moles) & P (pressure) Question: A container containing 2 moles of helium gas exerts a pressure of 0.75 atm on the container walls. Another 2 moles of helium are pumped into the container. What is the new pressure inside the container? Analysis: The number of (moles of) gas particles doubles, so the pressure doubles. 12

Boyle’s Law: Boyle’s Law: The relationship between: P (pressure) & V (volume) All other variables kept the same (T, n, etc.) This relationship is seen in a flexible/adjustable container Ex: a rising balloon or a cylinder with a piston Animation 13

Boyle’s Law P α 1/V P α 1/V This means Pressure and Volume are INVERSELY PROPORTIONAL This means Pressure and Volume are INVERSELY PROPORTIONAL P goes up as V goes down. P goes up as V goes down. P 1 V 1 = P 2 V 2 P 1 V 1 = P 2 V 2 14 Robert Boyle ( ). Son of Earl of Cork, Ireland. Copy onto P. Table

Boyle’s Law A bicycle pump is a good example of Boyle’s law. As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire. 15

Sample Problem 14.1 (p 419) A balloon contains 30.0 L of helium gas at 103kPa. What is the volume of the helium when the balloon rises to an altitude where the pressure is only 25.0 kPa? 16 Givens: P 1 = 103 kPaP 2 = 25.0kPa V 1 = 30.0L Unknown: V 2 = ? SOLUTION STEP 1: Identify & label all your variables (given & unknown) SOLUTION STEP 2: Find the formula that has all the variables you are using. P 1 V 1 = P 2 V 2 P 2 P 2 P 2 P 2 SOLUTION STEP 3: Rearrange the formula to solve for the unknown. P 1 V 1 = P 2 V 2

SOLUTION, cont. SOLUTION STEP 4: substitute values 17 P 1 V 1 = P 2 V 2 P 2 P 2 P 2 P 2 SOLUTION STEP 5: Calculate an answer (103kPa)(30.0L) = V kPa 25.0kPa V 2 = V 2 = 124 L SOLUTION STEP 6: Evaluate your answer. Does it make sense?

Charles’s Law Charles’s Law The relationship between: V (volume) & T (temperature) All other variables kept the same (P, n, etc.) Example: balloon 18

Charles’s Law V α T V α T V and T are directly proportional. V and T are directly proportional. V 1 V 2 = T 1 T 2 If one temperature goes up, the volume goes up! If one temperature goes up, the volume goes up! 19 Jacques Charles ( ). Isolated boron and studied gases. Balloonist.

Sample Problem 14.2 Using Charles Law (p 421) A balloon inflated in a room at 24˚C has a volume of 4.00L. The balloon is then heated to a temperature of 48˚C. What is the new volume if the pressure remains constant? 20 Givens: T 1 = 24˚C = 297KT 2 = =321K V 1 = 4.00 L Unknown: V 2

Solution 21 V1 = V 2  V 2 = V 1 T 2 T1 T2 T 1 V2 = 4.00L*321K 297K V2 = CW/HW Practice Problems, pp # 7-10 p 439 #47-49

Charles’s Law 22 What will happen if the syringe is put in a hot water bath? A cold water bath?

Gay-Lussac’s Law If n and V are constant, then P α T P and T are directly proportional. P 1 P 2 = T 1 T 2 If one temperature goes up, the pressure goes up! If one temperature goes up, the pressure goes up! 23 Joseph Louis Gay- Lussac ( )

Practice Problem (G-L) A 20 L cylinder containing 6 atm of gas at 27 °C. What would the pressure of the gas be if the gas was heated to 77 °C? 24 GIVENSUNKNOWN 20 L = V ? = P 6 atm = P 27 °C = 300.K = T 77 °C = 350.K= T

Section 14.1 Assessment p Why is a gas easy to compress? 2. List 3 factors that can affect gas pressure. 3. Why does a collision with an air bag cause less damage than a collision with a steering wheel? 4. How does a decrease in temp affect the pressure of a contained gas? 5. If the temp is constant, what change in volume would cause the pressure of an enclosed gas to be reduced to ¼ of its original value? 6. Assuming the gas in a container remains at a constant temp, how could you increase the gas pressure in a container a hundredfold? 25

Combined Gas Law Since all 3 gas laws are related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION! P 1 V 1 = P 2 V 2 T 1 T 2 26 No, it’s not related to R2D2

Combined Gas Law If you only need one of the other gas laws, you can cover up the item that is constant and you will get that gas law! = 27 P1P1 V1V1 T1T1 P2P2 V2V2 T2T2 Boyle’s Law Charles’ Law Gay-Lussac’s Law

Combined Gas Law Problem 28 A sample of helium gas has a volume of 180 mL, a pressure of atm and a temperature of 29°C. What is the new temperature(°K) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm? Set up Data Table P 1 = atm V 1 = 180 mL T 1 = 302 K P 2 = 3.20 atm V 2 = 90 mL T 2 = ??

Solution P 1 = atm V 1 = 180 mLT 1 = 302K P 2 = 3.20 atm V 2 = 90 mL T 2 = ?? P 1 V 1 P 2 V 2 = P 1 V 1 T 2 = P 2 V 2 T 1 T 1 T 2 T 2 = P 2 V 2 T 1 P 1 V 1 T 2 = 3.20 atm x 90.0 mL x 302 K atm x mL T 2 = 604 K = 331 °C 29 = 604 K

Learning Check A gas has a volume of 675 mL at 35°C and atm pressure. What is the temperature when the gas has a volume of L and a pressure of 802 mm Hg? 30

One More Practice Problem A balloon has a volume of 785 mL on a fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon? 31

And now, we pause for this commercial message from STP 32 OK, so it’s really not THIS kind of STP… STP in chemistry stands for Standard Temperature and Pressure Standard Pressure = 1 atm (or an equivalent) Standard Temperature = 0 deg C (273 K) STP allows us to compare amounts of gases between different pressures and temperatures

Try This One A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C? 33

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. 34 twice as many molecules

IDEAL GAS LAW Brings together gas properties. BE SURE YOU KNOW THIS EQUATION! 35 P V = n R T

Using PV = nRT P = Pressuren = number of moles V = VolumeT = Temperature R is a constant, called the Ideal Gas Constant = 8.31 L*kPa = 8.31 L*kPa K*mol K*mol NOTE: We must convert the units to match R. 36

Using PV = nRT Ex: p 439, Q#55 Ex: p 439, Q# moles of gas at 35 C and 96.2 kPa pressure. What is the volume the gas occupies? 1.24 moles of gas at 35 C and 96.2 kPa pressure. What is the volume the gas occupies? V=? V=? n= 1.24 mol n= 1.24 mol T = = 308K T = = 308K P = 96.2kPa P = 96.2kPa R=8.31L*kPa/K*mol R=8.31L*kPa/K*mol (96.2kPa)V = (1.24 mol ) (8.31L*kPa/K*mol) (308K) (96.2kPa)V = (1.24 mol ) (8.31L*kPa/K*mol) (308K) V=33.0L V=33.0L 37

Learning Check Dinitrogen monoxide (N 2 O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mm Hg) in the tank in the dentist office? 38

Learning Check A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder? 39

Gases in the Air (don’t copy) The % of gases in air Partial pressure (STP) 78.08% N mm Hg 20.95% O mm Hg 0.94% Ar 7.1 mm Hg 0.03% CO mm Hg P AIR = P N 2 + P O 2 + P Ar + P CO 2 = 760 mm Hg (Total Pressure = 760mm Hg) 40

Dalton’s Law of Partial Pressures What is the total pressure in the flask? P total in gas mixture = P A + P B +... Therefore, P total = P H 2 O + P O 2 = 0.48 atm Dalton’s Law: total P is sum of PARTIAL pressures H 2 O 2 (l) ---> 2 H 2 O (g) + O 2 (g) 0.32 atm 0.16 atm 0.32 atm 0.16 atm

Dalton’s Law 42 John Dalton

Health Note When a scuba diver is several hundred feet under water, the high pressures cause N 2 from the tank air to dissolve in the blood. If the diver rises too fast, the dissolved N 2 will form bubbles in the blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O 2 in scuba tanks used for deep descents. 43

Gases and Stoichiometry 2 H 2 O 2 (l) ---> 2 H 2 O (g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the volume of O 2 at STP? 44 Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

Gases and Stoichiometry 2 H 2 O 2 (l) ---> 2 H 2 O (g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the volume of O 2 at STP? Solution 1.1 g H 2 O 2 1 mol H 2 O 2 1 mol O L O 2 34 g H 2 O 2 2 mol H 2 O 2 1 mol O 2 34 g H 2 O 2 2 mol H 2 O 2 1 mol O 2 45 = 0.36 L O 2 at STP

Gas Stoichiometry: Practice! A. What is the volume at STP of 4.00 g of CH 4 ? B. How many grams of He are present in 8.0 L of gas at STP? 46

What if it’s NOT at STP? 1. Do the problem like it was at STP. (V 1 ) 2. Convert from STP (V 1, P 1, T 1 ) to the stated conditions (P 2, T 2 ) 47

Try this one! How many L of O 2 are needed to react 28.0 g NH 3 at 24°C and atm? 4 NH 3 (g) + 5 O 2 (g) 4 NO(g) + 6 H 2 O(g) 48

GAS DIFFUSION AND EFFUSION diffusion is the gradual mixing of molecules of different gases. diffusion is the gradual mixing of molecules of different gases. effusion is the movement of molecules through a small hole into an empty container. effusion is the movement of molecules through a small hole into an empty container. 49 HONORS only

GAS DIFFUSION AND EFFUSION (Don’t Copy) Graham’s law governs effusion and diffusion of gas molecules. 50 Thomas Graham, Professor in Glasgow and London. Rate of effusion is inversely proportional to its molar mass. HONORS only