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Section 3.7—Gas Behavior How does the behavior of gases affect airbags? What is PRESSURE? Force of gas particles running into a surface.

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Presentation on theme: "Section 3.7—Gas Behavior How does the behavior of gases affect airbags? What is PRESSURE? Force of gas particles running into a surface."— Presentation transcript:

1 Section 3.7—Gas Behavior How does the behavior of gases affect airbags? What is PRESSURE? Force of gas particles running into a surface.

2 Pressure is measured by a Barometer

3 If pressure is molecular collisions with the container… As # of moles increase, pressure increases Think about blowing up a balloon! Pressure and Moles (# of Molecules) As number of molecules increases, there will be more molecules to collide with the wall Collisions between molecules and the wall increase  Pressure increases

4 # of Gas Particles vs. Pressure

5 Pressure & Volume If pressure is molecular collisions with the container… As volume increases, pressure decreases. Think about how your lungs work! http://www.youtube.com/watch?v=q6-oyxnkZC0 As volume increases, molecules can travel farther before hitting the wall Collisions between molecules & the wall decrease  Pressure decreases

6 What is “Temperature”? Temperature – measure of the average kinetic energy of the molecules Energy due to motion (Related to how fast the molecules are moving) As temperature increases, Average Kinetic Energy Increases and Molecular motion increases

7 Pressure and Temperature If temperature is related to molecular motion… and pressure is molecular collisions with the container… As temperature increases, pressure increases As temperature increases, molecular motion increases Collisions between molecules & the wall increase  Pressure increases

8 Volume and Temperature If temperature is related to molecular motion… and volume is the amount of space the gas occupies… As temperature increases, volume increases Think of liquid nitrogen and the balloon. http://www.youtube.com/watch?v=QEpxrGWep4E As temperature increases, molecular motion increases molecules will move farther away from each other  Volume increases

9 Pressure In Versus Out Example: A bag of chips is bagged at sea level. What happens if the bag is then brought up to the top of a mountain. A container will expand or contract until the pressure inside equals atmospheric pressure outside The internal pressure is higher than the external pressure. The bag will expand in order to reduce the internal pressure. The internal pressure of the bag at low altitude is high At high altitude there is lower pressure Higher pressure Lower pressure Lower pressure

10 Can Explodes! When Expansion Isn’t Possible Example: An aerosol can is left in a car trunk in the summer. What happens? Rigid containers cannot expand The internal pressure is higher than the external pressure. The can is rigid—it cannot expand, it explodes! The temperature inside the can begins to rise. As temperature increases, pressure increases. Higher pressure Lower pressure

11 Air Pressure Crushing Cans http://www.csun.edu/scied/4-discrpeant-event/the_can_crush/index.htm

12 http://www.youtube.com/watch?v=Zz95_VvTxZM Another cool video http://www.youtube.com/watch?v=JsoE4F2Pb20 Air Pressure Crushing “Cans”

13 Kinetic Molecular Theory(KMT): explains gas behavior based upon the motion of molecules based on an ideal gas  IDEAL gases are IMAGINARY gases that follow the assumptions of the KMT

14 1 Assumptions of the KMT All gases are made of atoms or molecules that are in constant, rapid, random motion Gas particles are not attracted nor repelled from one another *** All gas particle collisions are perfectly elastic (no kinetic energy is lost to other forms) The volume of gas particles is so small compared to the space between the particles, that the volume of the particle itself is insignificant*** 2 3 4 5 The temperature of a gas is proportional to the average kinetic energy of the particles

15 So what is a “REAL” gas? Real gases, (like nitrogen), will eventually condense into a liquid when the temperature gets too low or the pressure gets too high BECAUSE: Assumption #3 Assumption #5 Gas particles do have attractions and repulsions towards one another Gas particles do take up space

16 Real Gases Deviate from Ideal Gas Behavior when at high pressure  The gas molecules are compressed making the volume they take up more significant than if they were spread out

17

18 Real Gases Deviate from Ideal Gas Behavior when at low temperature.  The lower kinetic energy causes the molecules to move slower and ATTRACTIVE FORCES that really exist start to take effect ---------------------------  Polar gases (HCl) deviate more than nonpolar gases (He or H 2

19 At Lower Temperature

20 Gas Movement: Effusion vs Diffusion Effusion –gas escapes from a tiny hole in the container Effusion is why balloons deflate over time!

21 Diffusion –gas moves across a space from high to low concentration Diffusion is the reason we can smell perfume across the room

22 Effusion, Diffusion & Particle Mass How are particle size (mass) and these concepts related? As mass of the particles increases, rate of effusion and diffusion is lowered. As particle size (mass) increases, the particles move slower  it takes them more time to find the hole or to go across the room

23 Rate of Diffusion & Particle Mass Watch as larger particles take longer to get to your nose H2H2 CO 2

24 Section 3.8-3.9 —Gas Laws How can we calculate Pressure, Volume and Temperature of our airbag?

25 Pressure Units Several units are used when describing pressure UnitSymbol atmospheresatm Pascals, kiloPascals millimeters of mercury pounds per square inch Pa, kPa mm Hg psi 1 atm = 101300 Pa = 101.3 kPa = 760 mm Hg = 14.7 psi

26 Conversions Between Different Pressure Units 1 atm = 760 mmHg = 101.3 kPa Examples 1.Convert 654 mm Hg to atm 1.Convert 879 mm Hg to kPa 1.Convert 15.6 atm to kPa 654 mmHg x 1atm = 760 mmHg.861 atm 879 mmHg x 101.3 Kpa = 760 mmHg 1.16 Kpa 15.6 atm x 101.3 Kpa = 1atm 1580 Kpa

27 Temperature Unit used in Gas Laws Kelvin (K)– temperature scale with an absolute zero Temperatures cannot fall below an absolute zero Examples 1.Convert 15.6 °C into K 2. Convert 234 K into °C 15.6 + 273 = K 288.6  289 K °C + 273 = 234-39 °C

28 Standard Temperature & Pressure (STP) the conditions of:  1 atm (or the equivalent in another unit)  0°C (273 K) Problems often use “STP” to indicate quantities…don’t forget this “hidden” information when making your list!

29 GAS LAWS: “Before” and “After” This section has 5 gas laws which have “before” and “after” conditions. For example: 1= initial amount 2= final amount P= Pressure V= Volume T=Temperature n= moles(molecules)

30 Boyle’s Law Pressure Increases as Volume Decreases

31 Boyles’ Law Volume & Presssure are INVERSELY proportional when temperature and moles are held constant P = pressure V = volume The two pressure units must match and the two volume units must match! Example: A gas sample is 1.05 atm when at 2.5 L. What volume is it if the pressure is changed to 0.980 atm?

32 Boyles’ Law ***The two pressure units must match & the two volume units must match! Example: A gas sample is 1.05 atm when 2.5 L. What volume is it if the pressure is changed to 0.980 atm? P 1 = 1.05 atm V 1 = 2.5 L P 2 = 0.980 atm V 2 = ? L V 2 = 2.7 L

33 Boyles Law: Graph

34 Charles’ Law Volume Increases as Temperature Increases

35 Charles’ Law Volume & Temperature are DIRECTLY proportional when pressure and moles are held constant. V = Volume T = Temperature The two volume units must match & temperature must be in Kelvin! Example: What is the final v olume if a 10.5 L sample of gas is changed from 25  C to 50  C? V 1 = 10.5 L T 1 = 25  C V 2 = ? L T 2 = 50  C Temperature needs to be in Kelvin! 25  C + 273 = 298 K 50  C + 273 = 323 K

36 Charles’ Law ***The two volume units must match & temperature must be in Kelvin! Example: What is the final volume if a 10.5 L sample of gas is changed from 25  C to 50  C? V 1 = 10.5 L T 1 = 25  C V 2 = ? L T 2 = 50  C V 2 = 11.4 L = 298 K = 323 K

37 Charles Law: Graph

38 Gay-Lussac’s Law Temperature decreases as Pressure decreases

39 Gay-Lussac’s Law Pressure & temperature are DIRECTLY proportional when moles and volume are held constant P = Pressure T = Temperature The two pressure units must match and temperature must be in Kelvin! Example: A sample of hydrogen gas at 47  C exerts a pressure of.329 atm. The gas is heated to 77  C at constant volume and moles. What will the new pressure be? P 1 =.329 atm T 1 = 47  C P 2 = ? atm T 2 = 77  C Temperature needs to be in Kelvin! 47  C + 273 = 320 K 77  C + 273 = 350 K

40 Gay-Lussac’ Law Example: A sample of hydrogen gas at 47  C exerts a pressure of.329 atm. The gas is heated to 77  C at constant volume and moles. What will the new pressure be? P 1 =.329 atm T 1 = 47  C P 2 = ? atm T 2 = 77  C P 2 =.360 atm = 320 K = 350 K

41 Gay Lussac Law: Graph

42 Avogadro’s Law Moles and Volume are directly proportional when temp. & pressure are held constant V = Volume n = # of moles of gas Example: A sample with 0.15 moles of gas has a volume of 2.5 L. What is the volume if the sample is increased to 0.55 moles? The two volume units must match!

43 Avogadro’s Law Example: A sample with 0.15 moles of gas has a volume of 2.5 L. What is the volume if the sample is increased to 0.55 moles? The two volume units must match! n 1 = 0.15 moles V 1 = 2.5 L n 2 = 0.55 moles V 2 = ? L V 2 = 9.2 L

44 Combined Gas Law P = Pressure V = Volume n = # of moles T = Temperature Each “pair” of units must match and temperature must be in Kelvin! Example: What is the final volume if a 15.5 L sample of gas at 755 mmHg and 298 K is changed to STP?

45 Combined Gas Law P = Pressure V = Volume T = Temperature Moles is not mentioned so remove it from equation! Example: What is the final volume if a 15.5 L sample of gas at 755 mmHg and 298K is changed to STP? P 1 = 755 mmHg V 1 = 15.5 L T 1 = 298 K P 2 = 760mmHg V 2 = ? L T 2 = 273 K V 2 = 14.1 L STP is standard temperature (273 K) and pressure (1 atm)

46 The combined gas law can be used for all “before” and “after” gas law problems! For example, if volume is held constant, then and the combined gas law becomes: Why you really only need 1 of these

47 Watch as variables are held constant and the combined gas law “becomes” the other 3 laws Hold pressure and temperature constant Avogadro’s Law Hold moles and temperature constant Boyles’ Law Hold pressure and moles constant Charles’ Law Transforming the Combined Law

48 Dalton’s Law

49 Each gas in a mixture exerts its own pressure called a partial pressure = P 1, P 2 …. Total Pressure = P T Example: A gas mixture is made up of oxygen(2.3 atm) and nitrogen(1.7 atm) gases. What is the total pressure? P T = 2.3 atm + 1.7 atm 4.0 atm

50 2KClO 3 (s) 2KCl (s) + 3O 2 (g) P T = P O + P H O 22 Modified Dalton’s Law When a gas is Collected over water, the total pressure of the mixture collected is a combination of water vapor and the gas you are collecting!

51 Modified Dalton’s Law Example : What is the pressure of the water vapor if the total pressure of the flask is 17.5 atm and the pressure of the oxygen gas is 16.1 atm? 17.5 = 16.1 atm + P H2O 1.4 atm

52 The Ideal Gas Law ( an “ AT NO W” equation ) The volume of a gas varies directly with the number of moles and its Kelvin temperature P = Pressure V = Volume n = moles R = Gas Law Constant T = Temperature There are three possibilities for “R”! Choose the one with units that match your pressure units! Volume must be in Liters when using “R” to allow the unit to cancel!

53 The Ideal Gas Law Example: A sample with 0.55 moles of gas is at 105.7 kPa and 27°C. What volume does it occupy?

54 The Ideal Gas Law Example: A sample with 0.55 moles of gas is at 105.7 kPa and 27°C. What volume does it occupy? n = 0.55 moles P = 105.7 kPa T = 27°C + 273 = 300 K V = ? R = 8.31 L kPa / mole K V 2 = 13 L The Ideal Gas Law does not compare situations—it describes a gas in one situation. Chosen to match the kPa in the “P” above

55 The Ideal Gas Law Example 2: What mass of hydrogen gas in grams is contained in a 10.0 L tank at 27°C and 3.50 atm of pressure? n = ? P = 3.50 atm T = 27°C + 273 = 300 K V = 10.0 L R =.0821 L atm /mole K Chosen to match the atm in the “P” above n = 1.42 mol  1.42 mol x 2.02 g = 2.87 g 1 mol


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