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GASES Importance of Gases 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.

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Presentation on theme: "GASES Importance of Gases 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."— Presentation transcript:

1

2 GASES

3 Importance of Gases 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

4 Atmosphere The atmosphere is composed of nitrogen, oxygen, argon, water vapor, and a number of trace gases. Chemical reactions maintain the ratios of major constituents of the atmosphere to each other. For example, oxygen is released into the atmosphere by photosynthesis and consumed by respiration.

5 Altitude vs. Temperature

6 Altitude vs. Pressure

7 General Properties of Gases There is a lot of “free” space in a gas. There is a lot of “free” space in a gas. Gases can be expanded infinitely. Gases can be expanded infinitely. Gases fill containers uniformly and completely. Gases fill containers uniformly and completely. Gases diffuse and mix rapidly. Gases diffuse and mix rapidly.

8 Pressure Pressure of air is measured with a BAROMETER (developed by Torricelli in 1643) Hg rises in tube until force of Hg (down balances the force of atmosphere(pushing up). (Just like a straw in a soft drink) P of Hg pushing down related to Hg density Hg density column height column height

9 PressureColumn height measures Pressure of atmosphere 1 standard atmosphere (atm) * = 760 mm Hg (or torr) * = inches Hg * = 14.7 pounds/in2 (psi) = kPa * Memorize these!

10 Pressure Conversions A. What is 475 mm Hg expressed in atm? 1 atm 760 mm Hg B. The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg? 760 mm Hg 14.7 psi = 1.52 x 10 3 mm Hg = atm 475 mm Hg x 29.4 psi x

11 Pressure Conversions A. What is 2 atm expressed in torr? B. The pressure of a tire is measured as 32.0 psi. What is this pressure in kPa?

12 Kinetic Molecular Theory To fully understand the world around us requires that we have a good understanding of the behavior of gases. The description of gases and their behavior can be approached from several perspectives. The Gas Laws are a mathematical interpretation of the behavior of gases. However, before understanding the mathematics of gases, a chemist must have an understanding of the conceptual description of gases. That is the purpose of the Kinetic Molecular Theory.

13 Kinetic Molecular Theory The Kinetic Molecular Theory is a single set of descriptive characteristics of a substance known as the Ideal Gas. All real gases require their own unique sets of descriptive characteristics. Considering the large number of known gases in the World, the task of trying to describe each one of them individually would be an awesome task. In order to simplify this task, the scientific community has decided to create an imaginary gas that approximates the behavior of all real gases. In other words, the Ideal Gas is a substance that does not exist. The Kinetic Molecular Theory describes that gas.

14 KINETIC MOLECULAR THEORY Gases are composed of particles- usually molecules or atoms: Small, hard spheres, relatively far apart from each other Particles in a gas move rapidly in constant random motion. Move in straight paths, changing direction only when colliding with one another or other objects Collisions are perfectly elastic- meaning kinetic energy is transferred without loss from one particle to another- the total kinetic energy remains constant

15 ALL TEMPERATURE HAS TO BE IN KELVIN

16 And now, we pause for this commercial message from STP 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

17 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 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

18 As the pressure increases Volume decreases Volume decreases

19 How does Pressure and Volume of gases relate graphically? Volume Pressure PV = k Temperature, # of particles remain constant Temperature, # of particles remain constant

20 Boyle’s Mathematical Law: since PV = k P 1 V 1 = P 2 V 2 Eg: A gas has a volume of 3.0 L at 2 atm. What is its volume at 4 atm? What if we had a change in conditions?

21 1)determine which variables you have: P and V = Boyle’s Law 2)determine which law is being represented:  P 1 = 2 atm  V 1 = 3.0 L  P 2 = 4 atm  V 2 = ?  P 1 = 2 atm  V 1 = 3.0 L  P 2 = 4 atm  V 2 = ?

22 3) Rearrange the equation for the variable you don’t know 4) Plug in the variables and chug it on a calculator: P 1 V 1 = V 2 P2P2 P2P2 (2.0 atm)(3.0L) = V 2 (4atm) V 2 = 1.5L

23 Charles’s 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. 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.

24 Volume of balloon at room temperature Volume of balloon at 5°C

25 Temp How does Temperature and Volume of gases relate graphically? Volume V/T = k Pressure, # of particles remain constant Pressure, # of particles remain constant

26 Charles’s Mathematical Law: since V/T = k Eg: A gas has a volume of 3.0 L at 127°C. What is its volume at 227 °C? V 1 V 2 T 1 T 2 = What if we had a change in conditions?

27 1)determine which variables you have: T and V = Charles’s Law 2)determine which law is being represented:  T 1 = 127°C = 400K  V 1 = 3.0 L  T 2 = 227°C = 5ooK  V 2 = ?  T 1 = 127°C = 400K  V 1 = 3.0 L  T 2 = 227°C = 5ooK  V 2 = ?

28 4) Plug in the variables: (500K)(3.0L) = V 2 (400K) V 2 = 3.8L 3.0L V 2 400K 500K = = 5) Cross multiply and chug

29 Gay Lussac’s Law Old man Lussac determined the relationship between temperature and pressure of a gas. He measured the temperature of air at different pressures, and observed a pattern of behavior which led to his mathematical law. During his experiments volume of the system and amount of gas were held constant. Old man Lussac determined the relationship between temperature and pressure of a gas. He measured the temperature of air at different pressures, and observed a pattern of behavior which led to his mathematical law. During his experiments volume of the system and amount of gas were held constant.

30 Pressure Gauge Pressure Gauge Car before a trip Think of a tire... Let’s get on the road Dude!

31 Car after a long trip Think of a tire... WHEW! Pressure Gauge Pressure Gauge

32 Temp Pressure How does Pressure and Temperature of gases relate graphically? P/T = k Volume, # of particles remain constant Volume, # of particles remain constant

33 Lussac’s Mathematical Law: What if we had a change in conditions? since P/T = k P 1 P 2 T 1 T 2 = Eg: A gas has a pressure of 3.0 atm at 127º C. What is its pressure at 227º C?

34 T and P = Gay-Lussac’s Law  T 1 = 127°C = 400K  P 1 = 3.0 atm  T 2 = 227°C = 500K  P 2 = ?  T 1 = 127°C = 400K  P 1 = 3.0 atm  T 2 = 227°C = 500K  P 2 = ? 1)determine which variables you have: 2)determine which law is being represented:

35 4) Plug in the variables: (500K)(3.0atm) = P 2 (400K) P 2 = 3.8atm 3.0atm P 2 400K 500K = = 5) Cross multiply and chug

36 LAW RELAT- IONSHIP LAW CON- STANT Boyle’s P VP VP VP V P 1 V 1 = P 2 V 2 T, n Charles’ V TV TV TV T V 1 /T 1 = V 2 /T 2 P, n Gay- Lussac’s P TP TP TP T P 1 /T 1 = P 2 /T 2 V, n


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