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Gases Entry Task: Oct 25 th Block 2 Question: What is the relationship between pressure and temperature? You have 5 minutes!!

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Presentation on theme: "Gases Entry Task: Oct 25 th Block 2 Question: What is the relationship between pressure and temperature? You have 5 minutes!!"— Presentation transcript:

1 Gases Entry Task: Oct 25 th Block 2 Question: What is the relationship between pressure and temperature? You have 5 minutes!!

2 Gases Agenda: QUICKLY discuss Ch. 10 sec 1-3 Gas Inquiry Lab HW: B, C, G-L and Combo gas law ws (review from last year!)

3 Gases BREAK OUT AP EQUATION SHEET These formulas are rarely or not at all on the AP Exam Ideal gas law Van der Waals equation Daltons Partial pressure Moles= molar mass/molarity Kelvin/Celsius Combination gas law

4 Gases BREAK OUT AP EQUATION SHEET These formulas are rarely or not at all on the AP Exam

5 Gases Chapter 10 Gases

6 Gases I can… Describe the characteristics of gases Interconvert the various SI units for pressure Manipulate mathematically the 4 variables that pertain to gases.

7 Gases Characteristics of Gases Unlike liquids and solids, they  Expand to fill their containers  Are highly compressible.  Create homogeneous mixtures  Molecules of gases are very far apart thus having extremely low densities.

8 Gases Pressure is the amount of force applied to an area. Pressure Atmospheric pressure is the weight of air per unit of area. P = FAFA

9 Gases Units of Pressure Pascals  1 Pa = 1 N/m 2 Bar  1 bar = 10 5 Pa = 100 kPa mm Hg or torr  These units are literally the difference in the heights measured in mm ( h ) of two connected columns of mercury. Atmosphere  1.00 atm = 760 torr Derivation of Pressure / barometer height relationship

10 Gases Manometer Used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel. Manometer at P of 1 atm

11 Gases Standard Pressure Normal atmospheric pressure at sea level. It is equal to  1.00 atm  760 torr (760 mm Hg)  101.325 kPa

12 Gases The Gas Laws The 4 variables needed to define the physical conditions of gas are:  Temperature (T) in kelvin  Pressure (P)  Volume (V)  Number of moles (n)

13 Gases Boyle’s Law The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.

14 Gases Boyle’s Law at Work… Doubling the pressure reduces the volume by half. Conversely, when the volume doubles, the pressure decreases by half.

15 Gases As P and V are inversely proportional A plot of V versus P results in a curve. Since V = k (1/P) This means a plot of V versus 1/P will be a straight line. PV = k

16 Gases Application of Boyle’s Law A gas has a volume of 3.0 L at 2 atm. What will its volume be at 4 atm? P 1 = 2 atm V 1 = 3.0 L P 2 = 4 atm V 2 = X (2 atm) (3.0L) = (4 atm) (X)

17 Gases Finishing the algebra (2 atm) (3.0 L) = (4 atm) (X) (4 atm) (6 L) = (X) (4) X =1.5 L

18 Gases P 1 V 1 = P 2 V 2 In a thermonuclear device, the pressure of 0.050 liters of gas within the bomb casing reaches 4.0 x 10 6 atm. When the bomb casing is destroyed by the explosion, the gas is released into the atmosphere where it reaches a pressure of 1.00 atm. What is the volume of the gas after the explosion? P 1 = 4.0 x10 6 atm V 1 = 0.050 L P 2 = 1 atm V 2 = X (4.0 x10 6 atm) (0.050L) = (1 atm) (X)

19 Gases Finishing the algebra (4.0 x10 6 atm) (0.05 L) = (1 atm) (X) (1 atm) (200000 L) = (X) (1) X =200000 L or 2.0 x 10 5 L

20 Gases If some neon gas at 121 kPa were allowed to expand from 3.7 dm 3 to 6.0 dm 3 without changing the temperature, what pressure would the neon gas exert under these new conditions? (121 kPa) (3.7 dm 3 ) = (X) (6.0 dm 3 ) (121 kPa)(3.7 dm 3 )= (X) 6.0 dm 3 75 kPa

21 Gases Charles’s Law The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature. A plot of V versus T will be a straight line. i.e., VTVT = k

22 Gases Charles’ Law at Work… As the temperature increases, the volume increases. Conversely, when the temperature decreases, volume decreases.

23 Gases WHY must you convert Celsius to Kelvin? For the math to work AND to show a proportional relationship, the absolute temperature (Kelvin) is needed.

24 Gases Converting K to C˚ and C˚ to K Kelvin temp = C ˚ + 273 Its 32 C ˚, what is this temp in Kelvin? 32 + 273 = 305 K Celsius temp = K - 273 Its 584 K, what is the temp in C ˚? 584 - 273 = 311 C ˚

25 Gases Kelvin Practice 0°C = _______ K 100°C = _______ K 100 K = _______ °C –30°C= _______ K 300 K = _______ °C 403 K = _______ °C 25°C = _______ K 0 K = _______ °C 273 -173 27 298 373 243 130 -273

26 Gases V 1 / T 1 = V 2 / T 2 If a 1.0 L balloon is heated from 22°C to 100°C, what will its new volume be? Need to convert C ° to K V 1 = 1.0 L T 1 = 22 °C + 273 = 295 K V 2 = X T 2 = 100 °C + 273 = 373 K 1.0L / 295K = X / 373K

27 Gases Finishing the algebra (295K) (X) = (1.0 L) (373 K) (295K) X =1.26 L X = (1.0 L) 373 295

28 Gases The temperature inside my refrigerator is about 4 0 Celsius. If I place a balloon in my fridge that initially has a temperature of 22 0 C and a volume of 0.5 liters, what will be the volume of the balloon when it is fully cooled by my refrigerator? (0.5 L) (295 K) (X L) (0.5 L)(277 K)= (X) 295 K 0.469 L 277 K =

29 Gases A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 0 C, what will the volume of the balloon be after he heats it to a temperature of 250 0 C? (0.4 L) (293 K) (X L) (0.4 L)(523 K)= (X) 293 K 0.714 L 523 K =

30 Gases Charles’ Law: Summary Volume / Temperature = Constant V 1 / T 1 = V 2 / T 2 With constant pressure and amount of gas, you can use these relationships to predict changes in temperature and volume.

31 Gases Avogadro’s Law The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas. Mathematically, this means V = kn

32 Gases Gay-Lussac’s Law combines pressure and temperature. This is a review from last year. When volume remains constant, pressure and temperature have a direct relationship. P 1 /T 1 = P 2 /T 2

33 Gases Gay-Lussac’s Law The pressure of 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? P 1 = 3.20 atm T 1 = 22 + 273 = 295 K P 2 = X T 2 = 60 + 273 = 333 K 3.20 atm / 295 K = X / 333 K Need to convert C ° to K

34 Gases Now the ALGEBRA!!! Set it up- Algebraically 3.20 atm / 295 K = X / 333 K Cross multiply (295 K) X = 3.20 atm = X 333 K295 K (3.20 atm)(333 K)

35 Gases Finishing the algebra (295 K) (X) = (3.20 atm) (333 K) (295K) X =3.61 atm X = (3.20 atm) 333 295 (295 K) X =(3.20 atm)(333 K)

36 Gases Gay-Lussac’s Law A rigid container has an initial pressure of 1.50 atm at 21 o C. What will the pressure be if the temperature is increased to 121 o C? Need to convert C ° to K P 1 = 1.50 atm T 1 = 21 + 273 = 294 K P 2 = X T 2 = 121+ 273 = 394 K 1.50 atm / 294 K = X atm / 394 K

37 Gases Now the ALGEBRA!!! Set it up- Algebraically 1.50 atm / 294 K = X atm / 394 K Cross multiply (294 K) X atm = 1.50 atm = X atm 394 K294 K (1.50 atm)(394 K)

38 Gases Finishing the algebra (294K) (X) = (1.50 atm) (394 K) (294K) X =2.01 atm X = (1.50 atm) 394 294 (294 K) X atm = (1.50 atm)(394 K)

39 Gases 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- in Celsius? P 1 = 125 kPa T 1 = 30 + 273 = 303K P 2 = 201 kPa T 2 = X 60903 = (X) (125 kPa) 125 kPa 303 K 487 – 273 = 214 ˚C 201 kPa X = 60903 K = X 125

40 Gases The pressure in an automobile tire is 1.88 atm at 25 ˚C. What will be the pressure if the temperature warms up to 37.0 ˚C? P 1 = 1.88 atm T 1 = 25 + 273 = 298K P 2 = X T 2 = 37 + 273 = 310K 583 = (X) (298K) 1.88 atm 298 K 1.95 atm X 310 K = 583 atm = X 298

41 Gases Explain Gay-Lussac’s law of combining volumes. At a given pressure and temperature the volumes of gases that react with one another are in ratio of small whole numbers

42 Gases Explain the difference between Avogadro’s hypothesis and Avogadro’s Law Avogadro’s hypothesis states that there are at equal volumes with P and T constant, they have the same number of gas molecules Avogadro’s Law takes it one step further, the equal volumes means equal number of moles.

43 Gases Explain V= constant x n If you double the number of moles of gas (n) then the volume will double- proportionally.

44 Gases Gas Law Inquiry Lab HW: B, C, G-L and combination gas law ws.


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