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

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Presentation on theme: "Gas Laws. Gas Pressure Just means that gas is “pushing” on something."— Presentation transcript:

1 Gas Laws

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

3 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.

4 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.

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

6 Measuring Gas Pressure Vacuum We can measure that! Take a ruler and measure low to high in milimeters! 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. 760 mmHg

7 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 760mmHgLess Pressure

8 Measuring Gas Pressure of Containers 800 mmHg 40 mmHg What if I snap off the vacuum bulb? Because atmospheric pressure is pushing down! 760 mmHg

9 Measuring Gas Pressure BarometerManometer

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

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

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

13 Boyle’s Law Pressure and Volume Gas particles have a bunch of room. Gas particles are squeezed into smaller space. What about volume? V=High V=Low As pressure goes up, volume goes down. That means inverse relationship. P= Low P=High

14 Boyle’s Teeter Totter When volume is high, pressure is low When the volume is low, pressure is high An Inverse relationship. Like when I buy clothes Pressure Volume

15 Boyle’s Law Boyle’s law is explained by the equation P 1 V 1 =P 2 V 2 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? P 1 V 1 = P 2 V 2 (before) (after) What do you know? P 1 (before pressure) = V 1 (before volume)= P 2 (after pressure) = V 2 = ?? (1.70 atm)(4.35L)=(2.40 atm)V 2 7.40atm/L = (2.40atm)V 2 V 2 =3.01L 1.70 atm 4.35 L 2.4 atm

16 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!

17 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? P 1 V 1 = P 2 V 2 (before) (after) P 1 (before pressure) = V 1 (before volume)= P 2 (after pressure) = V 2 = (1.5atm)(5.6L) = (P 2 )(4.8L) 8.4 atm/L = (4.8L)P 2 1.8 atm = P 2 1.5 atm 5.6 L ? 4.8L

18 Charles’ Law Charles’ law relates volume and temperature, while keeping pressure the same V 1 = V 2 T 1 T 2

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

20 Charles’ Law HOT COLD T= HighT = Low V= High V = Low Charles’ law says that as the temp increases, so does volume. A direct relationship. What’s going on with the temp?

21 Charles’ Law So now we can relate volume and temperature. V 1 = V 2 T 1 T 2 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 + 273 = 273K 80 °C + 273 = 353K 625 L 0 °C ?? V 1 = T 1 = T 2 = V 2 = 80 °C

22 Charles’ Law V 1 = V 2 T 1 T 2 A balloon takes up 625 L at 0°C. If it is heated to 80°C, what will its new volume be? V 1 = 625 L T 1 = 273K T 2 = 353K V 2 = ??L 625L = V 2 273K 353K 2.29L/K= V 2 353K 808L = V 2

23 Charles’ Law At 27.00 °C a gas has a volume of 6.00 L. What will the volume be at 150.0 °C? What’s the equation? V 1 = V 2 T 1 T 2 V1=V1= T1=T1= V2=V2= T2=T2= 6.00 L 27 °C ?? 150.0 °C Must convert to Kelvin. 27 °C + 273 = 300K 150°C + 273 = 423K

24 Charles’ Law At 27.00 °C a gas has a volume of 6.00 L. What will the volume be at 150.0 °C? V 1 = V 2 T 1 T 2 V1=V1= T1=T1= V2=V2= T2=T2= 6.00 L ?? 300K 423K 6.00L = V 2 300K 423K 0.02L/K = V 2 423K 8.46L = V 2

25 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

26 Avogadro’s Law As the amount (in moles) goes up, so does the volume. If we double the amount, it doubles the volume.

27 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. n 1 = n 2 V 1 V 2

28 Avogadro’s Law n 1 = n 2 V 1 V 2 Let’s try! In a sample of gas, 50.0 g of oxygen gas (O 2 ) 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? n 1 =n 2 = V 1 = V 2 = When doing Avogadro's law, “n” MUST be in moles! 50g 40L mol? 79L

29 Avogadro’s Law n 1 = n 2 V 1 V 2 BeforeAfter n 1 =50gn 2 = g? V 1 = 48LV 2 = 79L When doing Avogadro's law, “n” MUST be in moles! 50g O 2 x 1 mol O 2 32g O 2 = 1.6 mol O 2 1.6mol 1.6 mol O 2 48L = n 2 79L 0.03 = n 2 79L 2.6 mol = n 2

30 Gay Lussac’s Law This law only applies to gases held at a constant volume. Only the pressure and temperature will change.volumepressuretemperature P i = P f T i T f P i =initial pressure P f = final pressure T i = initial temperature (kelvin) T f = final temperature (kelvin) 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?

31 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? P i = P f T i T f P i = 235 kPa P f = ? T i = 20°C T f = 48°C Must convert to Kelvin 20°C + 273 = 293K 48°C + 273 = 321K P i = 235 kPa P f = ? T i = 293K T f = 321K 235 293 = P f 321 0.80 = P f 321 257.5 kPa = P f

32 How to use these formulas Charle’s Law V 1 = V 2 T 1 T 2 Avogadro’s Law V 1 = V 2 n 1 n 2 Gay Lussac’s Law P 1 = P 2 T 1 T 2 They are all pretty much the same equation, just different variables! Write the equations for each variable and for each law.

33 Combined Gas Law Charle’s Law V 1 = V 2 T 1 T 2 Boyle’s Law (P 1 )(V 1 ) = (P 2 )(V 2 ) Gay Lussac’s Law P 1 = P 2 T 1 T 2 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. I can combine the laws! Combined Gas Law (P 1 )(V 1 ) = (P 2 )(V 2 ) T 1 T 2

34 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 11.00 atm. What will its new volume be? (P 1 )(V 1 ) = (P 2 )(V 2 ) T 1 T 2 P 1 = 1 atm P 2 = 11 atm V 1 = 40 L V 2 = ?? T 1 = 25 °C T 2 = 4 °C Convert to Kelvin 25°C + 273 = 298K 4°C + 273 = 277K P 1 = 1 atm P 2 = 11 atm V 1 = 40 L V 2 = ?? T 1 = 298K T 2 = 277K (1)(40) = (11)(V 2 ) 298K277K 0.13 = (11)(V 2 ) 277K 36.01 = (11)(V 2 ) 3.27 L = V 2


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