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GASES Chemistry I Honors – Chapter 13

GASES Chemistry I Honors – Chapter 13. Importance of Gases (don’t copy). Airbags fill with N 2 gas in an accident. Gas is generated by the decomposition of sodium azide , NaN 3 . 2 NaN 3 --->2 Na + 3 N 2. General Properties of Gases. 1. There is a lot of “free” space in a gas.

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GASES Chemistry I Honors – Chapter 13

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  1. GASESChemistry I Honors – Chapter 13

  2. Importance of Gases (don’t copy) • Airbags fill with N2 gas in an accident. • Gas is generated by the decomposition of sodium azide, NaN3. • 2 NaN3--->2 Na + 3 N2

  3. General Properties of Gases 1. There is a lot of “free” space in a gas. 2. Gases can be expanded infinitely. (they will fill whatever “container” they are in.) 3. Gases fill containers uniformly and completely. 4. Gases diffuse and mix rapidly.

  4. Pressure Pressure is the amount of force per unit area. • Gas Pressure is the pressure caused by particles of gas striking an object. • Atmospheric pressure is the pressure of the column of atmosphere above you, pressing down on you.

  5. (COPY) • Measured with a BAROMETER (developed by Torricelli in 1643) • Hg rises in tube until force of Hg (UP) balances the force of atmosphere (DOWN). (Just like a straw in a soft drink) • (DON’T COPY) • Q: Why is Hg so good for use in barometer? • A: If you tried to use water, it would rise about 34 feet high! Atmospheric Pressure

  6. Atmospheric. Pressure COPY The column height (of the barometer) measures atmospheric pressure • 1 standard atmosphere (atm) * • = 101.3 kPa(SI unit is PASCAL) = 760 mm Hg = 14.7 pounds/in2 (psi) * These are the 2 units we will use most of the time From the height of mercury in the barometer

  7. Pressure Conversions A. What is 2.71 atm expressed in kPa? B. The pressure in a tire reads 262 kPa. What is this pressure in atm?

  8. Properties of Gases Gas properties can be modeled using math. The factors that affect gases are: • V = volume of the gas (L) • T = temperature (K) • n = amount (moles) • P = pressure (atm, kPa, mm Hg, or psi) All temperatures must be in Kelvins! No exceptions!

  9. Converting Temperatures to Kelvin • Kelvin = ˚Celsius + 273 OR • K = ˚C + 273 Copy onto P. Table Example: convert 25 ˚C to Kelvins. K = 25 ˚C + 273 = 298 K

  10. Properties of Gases, cont. • We can study the relationship between 2 variables if we keep the others the same. VARIABLES: Moles, temperature, pressure, and volume

  11. The relationship between:n (moles) & P (pressure) • Pressure is caused by the particles striking the walls of the container. • If the gas is in a rigid container, the volume is constant (not changing) • If you increase the number of gas particles (n), you increase the pressure. Animation

  12. SAMPLE PROBLEMn (moles) & P (pressure) Question: A container containing 2 moles of helium gas exerts a pressure of 0.75 atm on the container walls. Another 2 moles of helium are pumped into the container. What is the new pressure inside the container? Analysis: The number of (moles of) gas particles doubles, so the pressure doubles.

  13. Boyle’s Law: The relationship between: P(pressure) & V (volume) • All other variables kept the same (T, n, etc.) • This relationship is seen in a flexible/adjustable container • Ex: a rising balloon or a cylinder with a piston • Animation

  14. Boyle’s Law • P α 1/V • This means Pressure and Volume are INVERSELY PROPORTIONAL • P goes up as V goes down. • P1V1 = P2 V2 Robert Boyle (1627-1691). Son of Earl of Cork, Ireland. Copy onto P. Table

  15. Boyle’s Law A bicycle pump is a good example of Boyle’s law. As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire.

  16. Sample Problem 14.1 (p 419) • A balloon contains 30.0 L of helium gas at 103kPa. What is the volume of the helium when the balloon rises to an altitude where the pressure is only 25.0 kPa? SOLUTION STEP 1: Identify & label all your variables (given & unknown) SOLUTION STEP 2: Find the formula that has all the variables you are using. SOLUTION STEP 3: Rearrange the formula to solve for the unknown. Unknown: V2 = ? Givens: P1= 103 kPa P2 = 25.0kPa V1 = 30.0L P1V1= P2 V2 P2 P2 P1V1 = P2 V2

  17. SOLUTION, cont. • SOLUTION STEP 4: substitute values SOLUTION STEP 5: Calculate an answer SOLUTION STEP 6: Evaluate your answer. Does it make sense? P1V1= P2 V2 P2 P2 (103kPa)(30.0L) = V2 25.0kPa V2 = 123.6 V2 = 124 L

  18. Charles’s LawThe relationship between: V(volume)& T (temperature) • All other variables kept the same (P, n, etc.) • Example: balloon

  19. Charles’s Law • V α T • V and T are directly proportional. V1 V2 = T1 T2 • If one temperature goes up, the volume goes up! Jacques Charles (1746-1823). Isolated boron and studied gases. Balloonist.

  20. Sample Problem 14.2 Using Charles Law (p 421) • A balloon inflated in a room at 24˚C has a volume of 4.00L. The balloon is then heated to a temperature of 48˚C. What is the new volume if the pressure remains constant? Givens: T1 = 24˚C + 273 = 297K T2 = 48 + 273 =321K V1 = 4.00 L Unknown: V2

  21. Solution • V1 = V2  V2 = V1T2 • T1 T2 T1 • V2 = 4.00L*321K • 297K • V2 = CW/HWPractice Problems, pp 419-423 # 7-10 p 439 #47-49

  22. Charles’s Law What will happen if the syringe is put in a hot water bath? A cold water bath?

  23. Gay-Lussac’s Law If n and V are constant, then P α T P and T are directly proportional. P1 P2 = T1 T2 • If one temperature goes up, the pressure goes up! Joseph Louis Gay-Lussac (1778-1850)

  24. Practice Problem (G-L) • A 20 L cylinder containing 6 atm of gas at 27 °C. What would the pressure of the gas be if the gas was heated to 77 °C? GIVENS UNKNOWN 20 L = V ? = P 6 atm = P 27 °C + 273 = 300.K = T 77 °C + 273 = 350.K= T 2 1 1 2

  25. Section 14.1 Assessmentp 417 1. Why is a gas easy to compress? 2. List 3 factors that can affect gas pressure. 3. Why does a collision with an air bag cause less damage than a collision with a steering wheel? 4. How does a decrease in temp affect the pressure of a contained gas? 5. If the temp is constant, what change in volume would cause the pressure of an enclosed gas to be reduced to ¼ of its original value? 6. Assuming the gas in a container remains at a constant temp, how could you increase the gas pressure in a container a hundredfold?

  26. Combined Gas Law • Since all 3 gas laws are related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION! P1 V1 = P2 V2 T1 T2 No, it’s not related to R2D2

  27. Combined Gas Law If you only need one of the other gas laws, you can cover up the item that is constant and you will get that gas law! = P2 Boyle’s Law Charles’ Law Gay-Lussac’s Law P1 V1 V2 T1 T2

  28. Combined Gas Law Problem A sample of helium gas has a volume of 180 mL, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°K) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm? Set up Data Table P1 = 0.800 atm V1 = 180 mL T1 = 302 K P2 = 3.20 atm V2= 90 mL T2 = ??

  29. Solution P1 = 0.800 atm V1 = 180 mL T1 = 302K P2 = 3.20 atm V2= 90 mL T2 = ?? P1 V1 P2 V2 = P1 V1 T2 = P2 V2 T1 T1 T2 T2 = P2 V2 T1 P1 V1 T2 = 3.20 atm x 90.0 mL x 302 K 0.800 atm x 180.0 mL T2 = 604 K - 273 = 331 °C = 604 K

  30. Learning Check A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

  31. One More Practice Problem A balloon has a volume of 785 mL on a fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?

  32. 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

  33. Try This One A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?

  34. twice as many molecules Avogadro’s Hypothesis Equal volumes of gases at the same T and P have the same number of molecules. V and n are directly related.

  35. IDEAL GAS LAW P V = n R T Brings together gas properties. BE SURE YOU KNOW THIS EQUATION!

  36. Using PV = nRT P = Pressure n = number of moles V = Volume T = Temperature R is a constant, called theIdeal Gas Constant = 8.31 L*kPa K*mol NOTE: We must convert the units to match R.

  37. Using PV = nRT • Ex: p 439, Q#55 • 1.24 moles of gas at 35 C and 96.2 kPa pressure. What is the volume the gas occupies? • V=? • n= 1.24 mol • T = 35 + 273 = 308K • P = 96.2kPa • R=8.31L*kPa/K*mol • (96.2kPa)V = (1.24 mol ) (8.31L*kPa/K*mol) (308K) • V=33.0L

  38. Learning Check Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mm Hg) in the tank in the dentist office?

  39. Learning Check A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?

  40. Gases in the Air (don’t copy) The % of gases in air Partial pressure (STP) 78.08% N2 593.4 mm Hg 20.95% O2 159.2 mm Hg 0.94% Ar 7.1 mm Hg 0.03% CO2 0.2 mm Hg PAIR = PN2 + PO2 + PAr + PCO2 = 760 mm Hg (Total Pressure = 760mm Hg)

  41. Dalton’s Law of Partial Pressures • 2 H2O2 (l) ---> 2 H2O (g) + O2 (g) • 0.32 atm 0.16 atm What is the total pressure in the flask? Ptotal in gas mixture = PA + PB + ... Therefore, Ptotal = PH2O + PO2 = 0.48 atm Dalton’s Law: total P is sum of PARTIAL pressures.

  42. Health Note When a scuba diver is several hundred feet under water, the high pressures cause N2 from the tank air to dissolve in the blood. If the diver rises too fast, the dissolved N2 will form bubbles in the blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O2 in scuba tanks used for deep descents.

  43. HONORS only GAS DIFFUSION AND EFFUSION • diffusionis the gradual mixing of molecules of different gases. • effusionis the movement of molecules through a small hole into an empty container.

  44. HONORS only GAS DIFFUSION AND EFFUSION (Don’t Copy) Graham’s law governs effusion and diffusion of gas molecules. Rate of effusion is inversely proportional to its molar mass. Thomas Graham, 1805-1869. Professor in Glasgow and London.

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