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Gas Laws

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Gas Pressure Just means that gas is “pushing” on something.

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**Gas Pressure Tire What’s going on inside? Air: Nitrogen 78% Oxygen 21%**

Argon ~1% Carbon Dioxide <1% Each of these particles are constantly flying around. Like a lotto ball! They slam against the container and keep the tire “full”. The particles press against the walls. Tire

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**Measuring Gas Pressure**

Air: Nitrogen 78% Oxygen 21% Argon ~1% Carbon Dioxide <1% Think of a giant ball pit miles and miles up. At the bottom of the ball pit, is like us walking around. That’s the atmospheric pressure.

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**Measuring Gas Pressure**

Vacuum Vacuum U-Tube It pushes down on this side, and it moves up on the other side. So how do we measure it? Can’t use it to measure atmospheric pressure, because atmospheric pressure presses on everything equally.

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**Measuring Gas Pressure**

The fluid that is contained in this U tube, is mercury. If we measure this at sea level, we get. 760mmHg between the bottom and the top. Vacuum 760 mmHg We can measure that! Take a ruler and measure low to high in milimeters!

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**Measuring Gas Pressure**

What if we go up a mountain or down into a mine? Think about that ball pit again. If you’re at the bottom of the ball pit will it weigh more or less than at the top? Sea Level More Pressure 760mmHg Less Pressure

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**Measuring Gas Pressure of Containers**

760 mmHg 800 mmHg 40 mmHg What if I snap off the vacuum bulb? Because atmospheric pressure is pushing down!

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**Measuring Gas Pressure**

Barometer Manometer

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**Gas Pressure Conversions**

How do we measure things? Lots of ways! Same goes with gas pressure. Gas Pressure Units mmHg atmosphere kilopascal Torr atm kPa Conversions 760 mmHg = 1 atm = 101.3kpa

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**Gas Pressure Conversions**

The pressure inside a car tire is 225 kPa. Express this value in both atm and mmHg. 760 mmHg = 1 atm = kPa 225 kPa x 1 atm 101.3 kPa =2.22 atm 225 kPa x 760 mmHg 101.3 kPa =1688 mmHg

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Boyle’s Law If we keep the temperature the same, we can predict what pressure and volume will do.

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**Boyle’s Law Pressure and Volume What about volume?**

Gas particles have a bunch of room. Gas particles are squeezed into smaller space. P= Low V=High P=High V=Low As pressure goes up, volume goes down. That means inverse relationship.

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**Boyle’s Teeter Totter When volume is high, pressure is low**

When the volume is low, pressure is high An Inverse relationship.

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**Boyle’s Law Boyle’s law is explained by the equation P1V1=P2V2**

Let’s get right to it! At 1.70 atm, a sample of gas takes up 4.35 L. If the pressure on the gas is increased to 2.40 atm, what will the new volume be? P1V1 = P2V2 (before) (after) What do you know? Last p 2, 3, 5 P1 (before pressure) = 1.70 atm (1.70 atm)(4.35L)=(2.40 atm)V2 V1 (before volume)= 4.35 L 7.40atm/L = (2.40atm)V2 P2 (after pressure) = 2.4 atm V2 =3.01L V2 = ??

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**Boyle’s Law Does that answer make sense?**

At 1.70 atm, a sample of gas takes up 4.35 L. If the pressure on the gas is increased to 2.40 atm, what will the new volume be? We increased the pressure, so we pushed down that piston. We squeezed the molecules into a smaller space. So the volume should go down!

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Boyle’s Law If I have 5.6 liters of gas in a piston at a pressure of 1.5 atm and compress the gas until its volume is 4.8 L, what will the new pressure inside the piston be? P1V1 = P2V2 (before) (after) P1 (before pressure) = 1.5 atm V1 (before volume)= 5.6 L (1.5atm)(5.6L) = (P2)(4.8L) P2 (after pressure) = ? 8.4 atm/L = (4.8L)P2 V2 = P 7 4.8L 1.8 atm = P2

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Charles’ Law Charles’ law relates volume and temperature, while keeping pressure the same V1 = V2 T T2

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Charles’ Law How could we test the theory that temperature and volume are related? Think about kinetic theory and molecules.

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**Charles’ Law COLD HOT What’s going on with the temp? T= High T = Low**

V= High V = Low Period 8 Charles’ law says that as the temp increases, so does volume. A direct relationship.

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**Charles’ Law So now we can relate volume and temperature. V1 = V2**

T T2 MUST ALWAYS USE KELVIN TEMPERATURE in gas laws A balloon takes up 625 L at 0°C. If it is heated to 80°C, what will its new volume be? Must convert to Kelvin. 0 °C = 273K 80 °C = 353K V1 = 625 L T1 = 0 °C T2 = 80 °C V2 = ??

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Charles’ Law A balloon takes up 625 L at 0°C. If it is heated to 80°C, what will its new volume be? V1 = V2 T T2 V1 = 625 L T1 = 273K T2 = 353K V2 = ??L 625L = V2 273K K 2.29L/K= V2 353K 808L = V2

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Charles’ Law At °C a gas has a volume of 6.00 L. What will the volume be at °C? V1 = V2 T T2 What’s the equation? V1= 6.00 L Must convert to Kelvin. 27 °C = 300K 150°C = 423K T1= 27 °C V2= ?? T2= 150.0 °C

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Charles’ Law At °C a gas has a volume of 6.00 L. What will the volume be at °C? V1 = V2 T T2 6.00L = V2 300K K V1= 6.00 L T1= 300K 0.02L/K = V2 423K V2= ?? T2= 423K 8.46L = V2

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**Avogadro’s Law Relationship between: Amount of gas (n) and the Volume.**

What happens to one, when I change the other? I start with the first balloon, and then blow more air into it…will the volume increase? Yes, a direct relationship

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**Avogadro’s Law As the amount (in moles) goes up, so does the volume.**

If we double the amount, it doubles the volume.

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**Avogadro’s Law We only changed TWO things.**

The volume and the amount of particles. We didn’t mess with the pressure or the temperature, they were held constant. V1 = V2 n n2

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**Avogadro’s Law V1 = V2 n1 n2 Let’s try!**

In a sample of gas, 50.0 g of oxygen gas (O2) take up 48L of volume. Keeping the pressure constant, the amount of gas is changed until the volume is 79 L. How many mols of gas are now in the container? n1= n2 = V1 = V2 = 50g mol? 79L 40L When doing Avogadro's law, “n” MUST be in moles!

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**Avogadro’s Law Before After n1=50g n2 = g? V1 = 48L V2 = 79L V1 = V2**

n n2 1.6mol When doing Avogadro's law, “n” MUST be in moles! 50g O2 x 1 mol O2 32g O2 = 1.6 mol O2 1.6 mol O2 48L = n2 79L 0.03 = n2 79L 2.6 mol = n2

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Gay-Lussac’s Law The pressure and Kelvin temperature of a gas are directly proportional, when the volume remains constant.

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Gay Lussac’s Law This law only applies to gases held at a constant volume. Only the pressure and temperature will change. Pi =initial pressure Pf = final pressure Ti = initial temperature (kelvin) Tf = final temperature (kelvin) P1 = P2 T1 T2 The pressure in a sealed can of gas is 235 kPa when it sits at room temperature (20C). If the can is warmed to 48C, what will the new pressure inside the can be?

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Gay Lussac’s Law The pressure in a sealed can of gas is 235 kPa when it sits at room temperature (20°C). If the can is warmed to 48°C, what will the new pressure inside the can be? P1 = P2 T1 T2 P1 = 235 kPa Must convert to Kelvin 20°C = 293K 48°C = 321K P2 = ? T1 = 20°C T2 = 48°C

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The pressure in a sealed can of gas is 235 kPa when it sits at room temperature (20°C). If the can is warmed to 48°C, what will the new pressure inside the can be? P1 = P2 T1 T2 P1 = 235 kPa 235 293 = Pf 321 P2 = ? 0.80 = Pf 321 T1 = 293K T2 = 321K 257.5 kPa = Pf

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**How to use these formulas**

Charle’s Law V1 = V2 T T2 They are all pretty much the same equation, just different variables! Avogadro’s Law V1 = V2 n n2 Gay Lussac’s Law P1 = P2 T T2

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**Combined Gas Law Charle’s Law V1 = V2**

T T2 What if I had a balloon. I wanted to increase the pressure and cool it down. What is the volume? Do we have an equation for that? P, T, V. Boyle’s Law (P1)(V1) = (P2)(V2) We can combine the laws! Gay Lussac’s Law P1 = P2 T T2 Combined Gas Law (P1)(V1) = (P2)(V2) T T2

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Combined Gas Law A 40.0L balloon is filled with air at sea level (1.00 atm, 25.0 °C). It's tied to a rock and thrown in a a cold body of water, and it sinks to the point where the temperature is 4.0 ° C and the pressure is atm. What will its new volume be? Convert to Kelvin 25°C = 298K 4°C = 277K (P1)(V1) = (P2)(V2) T T2 P1 = 1 atm P1= 1 atm P2= 11 atm V1= 40 L V2= ?? T1= 298K T2= 277K (1)(40) = (11)(V2) 298K 277K P2 = 11 atm V1 = 40 L 0.13 = (11)(V2) 277K V2 = ?? T1 = 25°C 36.01 = (11)(V2) T2 = 4°C 3.27 L = V2

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Ideal Gas Law How can we describe what’s going on in this container? What variables can we think of? Temperature (T) 313K Pressure (P) 3.18 atm Volume (V) 95.2 L Amount of Gas (n) 7.5 mol Did you know that if we know 3 of the 4 variables, we can find the last one?

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**Ideal Gas Law P =nRT V Ideal gas law: PV = nRT Temperature (T) 313K**

Pressure (P) 3.18 atm ?? Volume (V) 95.2 L Amount of Gas (n) 7.5 mol ?? What if we needed the amount of gas (n)? How would we rearrange the problem to find P? P =nRT V PV = n RT

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**Ideal Gas Law PV = nRT So what is R? R is a constant! For most cases,**

R = L▪atm/mol ▪K Those units look familiar. V = L P = atm T = K n = mol The units on “R” MUST match the units in the problem!

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**Ideal Gas Law “R” will come in many forms. R = 62.4 L▪mmHg /K ▪mol**

R = 8.31 L▪kPa /K ▪mol NOT A BIG DEAL! The “R” constant will always be given, just use the right constant.

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Ideal Gas Law 2.3 moles of Helium gas are at a pressure of 1.70 atm, and the temperature is 41°C. What is volume of the gas? PV = nRT Convert to Kelvin 41°C = 314K P = 1.70 atm V = ?? n = 2.3 mol R = L▪atm/K ▪mol T = 41°C

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Ideal Gas Law 2.3 moles of Helium gas are at a pressure of 1.70 atm, and the temperature is 41°C. What is volume of the gas? PV = nRT Rearrange the equation. P = 1.70 atm V = nRT P V = ?? V = (2.3 mol)(314K) x L ▪atm 1.70 atm K ▪ mol n = 2.3 mol R = L▪atm/K ▪mol V = 59.3 1.7 T = 314K V = 34.9 L

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Ideal Gas Law At a certain temperature, 3.24 moles of CO2 gas at 2.15 atm takes up a volume of L. What is this temperature (in Celsius)? Do the units given match the R? P = 2.15 atm V = 35.28 L T = ?? n = 3.24 mol R = L▪atm/K ▪mol

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Ideal Gas Law PV = nRT At a certain temperature, 3.24 moles of CO2 gas at 2.15 atm takes up a volume of L. What is this temperature (in Celsius)? Rearrange the equation. P = 2.15 atm T = PV nR V = 35.28 L T = ?? n = 3.24 mol T = (2.15 atm)(35.28L) X K ▪ mol (3.24 mol) L ▪ atm R = L▪atm/K ▪mol

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**Ideal Gas Law Charle’s Law V1 = V2 T1 T2**

Who wants to memorize all of these?!?! Avogadro’s Law V1 = V2 n n2 Gay Lussac’s Law P1 = P2 T T2 Ideal Gas Law PV = nRT You don’t have to! Combined Law (P1)(V1) = (P2)(V2) T T2

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**Gas Law Just memorize one! Can use it for any of the gas law problems!**

Ideal Gas Law PV = nRT Warning: If this blows your mind and you get totally confused, just memorize the equations.

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**Gas Law Rearrange the ideal equation so that the variables given are**

on the same “side” Before After P1 = 3 atm P2 = 7atm T1 = ?? T2 = 150k You’ve found the equation you need to use. You don’t need “n, R, or V”. PV =nRT PV = nRT V V P = nRT T VT P1 = P2 T T2 P= nR T V

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**Gas Law PV = nRT P1 = 1,217 mmHg P2 = 732 mmHg V1 = ?? V2 = 42L**

Rearrange the equation so the variables you’re looking for are on the same side of the equation. Easy! PV is already on the same side. Now just double it. P1V1 = P2V2

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**Gas Law PV = nRT V1 = 7.5L V2 = 1.2L n1= 32 mol n2 = ?**

Rearrange the equation so V and n are on the same side. PV = nRT P P V = nRT P V = nRT n Pn V1 = V2 n1 n2

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**Gas Law Before After V1 = ? V2 = 54L P1 = 96 kPa P2 = 112 kPa**

T1 = 12K T2 = 42K PV = nRT Rearrange so V, T, P are on same side. PV = nRT T T P1V1 = P2V2 T T2

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Three More Laws. A. Ideal Gas Law The 4 th variable that considers the amount of gas in the system is P 1 V 1 T 1 n = P 2 V 2 T 2 n Equal volumes of gases.

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