CHAPTER 14 GAS LAWS Boyles Law Pressure k

CHAPTER 14 GAS LAWS

Boyles Law: Pressure (k. Pa) Volume (L) -for a given mass of gas at constant temperature, the volume of the gas varies inversely with pressure. – Pressure increases, Volume decreases – Inversely proportional – Smaller volume- more pressure b/c particles collide more often EX: syringe P 1 V 1=P 2 V 2

n where P 1 V 1=P 2 V 2 P 1 and V 1 are the pressure & volume before the gas expands n and P 2 and V 2 are the pressure & volume after the gas expands.

Syringe— Piston can slide in (towards water) or out (towards hand) No cap—Vol. Inc while Pr. Dec Piston slides out Capped—Vol. Dec while Pr. Inc Piston slides in

Boyles Law: Pressure (k. Pa) Volume (L) Practice: If you had a gas that exerted 202 k. Pa of pressure and took up a space of 3. 00 liters, & you decided to expand the space to 7. 00 liters, what would be the new pressure? (temperature remains constant) Answer: P 1 V 1 = P 2 V 2 So, P 1 = 202 k. Pa, V 1 = 3. 00 L, V 2 = 7. 00 L, and you need to solve for P 2, the new pressure. Plug the numbers into the equation, & you have: (P 1 V 1)/V 2 = P 2 (202 k. Pa) x (3. 00 L) = (P 2) x (7. 00 L). P 2 = 86. 6 k. Pa.

Charles’ Law: Temp. (K) Volume (L) -the volume of a fixed mass of gas is directly proportional to its Kelvin temperature if the pressure is kept constant. – Temperature increases, Volume increases – Directly proportional – Higher temperature, gas particles speed up and move farther away from one another EX: heating a sealed container V 1 = V 2 T 1 T 2

V 1 = V 2 T 1 T 2 n where V 1 and T 1 are the volume & temperature before the gas expands n and V 2 and T 2 are the volume & temperature after the gas expands.

Charles’ Law: Temp. (K) Volume (L) Practice: If you took a balloon outside that was at 20 o. C at 2 L in volume, & it heated up to 29 o. C, what would its volume be? Assume constant pressure. Answer: V 1 / T 1 = V 2 / T 2 – V 1 = 2. 0 L, T 1 = 20 o. C, T 2 = 29 o. C, you must solve for V 2. Wait!! You have to convert the temperatures to Kelvin So: T 1 = 20 o. C + 273 = 293 K T 2 = 29 o. C + 273 = 302 K – Now, you can plug in the numbers and solve for V 2. 0 L 293 K = V 2 302 K and V 2 = 2. 1 L

Gay-Lussac's Law : Temp. (K) Pressure (k. Pa) -the pressure of a fixed mass of gas is directly proportional to its temperature if the volume is kept constant. – Temperature increases, Pressure increases – Directly proportional P 1 = P 2 T 1 T 2

Combined Gas Law: -combined the three gas laws into a single expression. V P = V P 1 1 2 2 CHARLES’ LAW BOYLE’S LAW T 1 T 2 GAY-LUSSAC’S LAW The Vice President is higher than the Treasurer

COMBINED GAS LAW DOESN‘T: ACCOUNT FOR THE AMOUNT OF GAS IT JUST LOOKS AT PRESSURE, VOLUME, & TEMPERATURE

Ideal Gas Law: -allows you to solve for the number of moles of a gas(n). PV = n. RT P=pressure in k. Pa or atm V=volume in L T=temperature in K n=#of moles R=Ideal gas constant (R) is 8. 31 (L x k. Pa) (K x mol) 0. 0821 (L x atm) (K x mol)

Ideal Gas Law: PV = n. RT Practice: A sample of O 2 gas has a volume of 4. 52 L at a temp. of 10 o. C and a pressure of 110. 5 k. Pa. Calculate the number of moles of O 2 gas present in this sample. Answer: Rearrange to solve for n (number of moles): n = PV/RT. P = 110. 5 k. Pa, V = 4. 52 L, T = 10 o. C + 273 = 283 K, and R = 8. 31 L x k. Pa/K x mol. n= (110. 5 k. Pa)(4. 52 L)/(8. 31 L k. Pa/K mol)(283 K) =. 212 moles.

Do Now If 4. 50 g of methane gas (CH 4) is in a 2. 00 -L container at 35°C, what is the pressure inside the container? n Solution: PV=n. RT - Ideal Gas Law Equation. P=? V = 2. 00 L n n = 4. 50 g CH 4 x 1 mol / 16. 0 g CH 4 R = 8. 31 L * k. Pa mol * K T = 308 K = 0. 281 mol

Do Now If 4. 50 g of methane gas (CH 4) is in a 2. 00 -L container at 35°C, what is the pressure inside the container? n Solution: PV=n. RT - Ideal Gas Law Equation. P = 3. 60 x 102 k. Pa. V = 2. 00 L n n = 4. 50 g CH 4 x 1 mol / 16. 0 g CH 4 R = 8. 31 L * k. Pa mol * K T = 308 K = 0. 281 mol

Mathematical Relationship Each slice weighs 0. 5 lbs. Eight slices make up a pie. How much does this pie weigh? What can I ignore in my calculation?

Mathematical Relationship Each lego block has dimensions 4 cm x 2 cm x 1 cm How tall is the stack of lego blocks?

History n n n John Dalton (1766 -1844) was an English chemist and physicist born in Cumberland, England. Early in life, influenced by meteorology. Researched color-blindness a. k. a. “Daltonism. ” Most notably known for compiling fundamental ideas into a universal atomic theory. His interest in gases and gas mixtures lead him to investigate humidity. This ultimately lead to Dalton’s Law.

Observe and Find a Pattern Partial Pressures of Compressed Air (Assuming air is 80% Nitrogen and 20% Oxygen) Depth (meters) Absolute Pressure P O 2 P N 2 0 14. 7 2. 94 11. 76 33 29. 4 5. 88 23. 52 66 44. 1 8. 82 35. 28 99 58. 8 11. 76 47. 04 132 73. 5 14. 70 58. 80 165 88. 2 17. 64 70. 56 198 102. 9 20. 58 82. 32 231 117. 6 23. 53 94. 08 264 132. 3 26. 46 105. 8 297 147. 0 29. 40 117. 6

Explain n The Ideal Gas Law holds for virtually any gas, whether pure or a mixture, at ordinary conditions for two reasons: – Gases mix homogeneously in any proportions. – Each gas in a mixture behaves as if it were the only gas present. n Explain why we can use a gas mixture, such as air, to study the general behavior of an ideal gas under ordinary conditions.

Predict and Test What is the relationship between the number of particles in containers A and C and the partial pressures of A and C. n Predict what the reading will be for container T. n

Derive n n Based on the outcome of the experiment relative to the prediction, make a judgment about whether the experiment disproved the hypothesis or not. Write a mathematical expression that relates absolute (total) pressure to the individual gas pressures. – Ptotal = P 1 + P 2 + P 3 + …this is known as Dalton’s Law of Partial Pressures. How can we qualitatively explain this mathematical relationship?

Mathematical Application n It was a dark and stormy night and your “friend” has locked you in the lab without a key. You remembered that the oldest window in the lab is always cracked open on a diagonal that won’t ever seem to budge completely. Someone calculated that you need a total pressure of 7. 10 x 106 k. Pa to exert a strong enough force to bust open the window. Coincidentally, you have 2. 00 moles of Ne, 4. 00 moles of Xe, and 6. 00 moles of Ar in a 5. 00 -L vessel at 27°C standing next to your lab station. Is the total pressure of the mixture enough for you to escape and make it to the club in time or will you be left alone in the lab synthesizing aspirin to relieve you of your torment?

Observe and Explain KCl. O 3 Gas produced is less dense than water, so it replaces the water in the test tube. n Gas collected is not pure because it contains vapor from the water. n

Dalton’s Law of Partial Pressure: -for a mixture of gases in a container, the total pressure exerted is the sum of the pressures that each gas would exert if it were alone. Ptotal=P 1 + P 2 + P 3 +…. .

Dalton’s Law of Partial Pressure: Practice: Air contains O 2, N 2, CO 2, and trace amounts of other gases. What is the partial pressure of O 2 (PO 2) at 101. 3 k. Pa of pressure if PN 2 =79. 10 k. Pa, PCO 2 =0. 040 k. Pa, and Pothers =0. 94 k. Pa? Answer: Ptotal = 101. 3 k. Pa -Ptotal = PO 2 + PN 2 + PCO 2 + Pothers -PO 2 = Ptotal - (PN 2 + PCO 2 + Pothers) -PO 2 = 101. 3 k. Pa - (79. 10 k. Pa + 0. 040 k. Pa + 0. 94 k. Pa) -PO 2 = 21. 22 k. Pa

Ideal vs. Real Gases An ideal gas is a hypothetical concept. No gas exactly follows the ideal gas law.

Ideal Gases n Always a gas n Not attracted to one another n Have no volume n Follow the gas laws for all conditions of pressure and temperature n Ideal gases DO NOT exist!!!!

Real Gases n Can be liquified and sometimes solidified by cooling and applying pressure § (ideal gases can not) n Have a finite volume n Are attracted to one another, especially at low temperatures

Graham’s Law of Effusion: the rate of effusion and diffusion of a gas is inversely proportional to the square root of its molar mass. – – – Lighter gas goes faster. Heavier gas goes slower. To figure out how much faster take the square root of the gfm of the heavier gas and divide by the square root of the gfm of the lighter gas.

Graham’s Law of Effusion: Practice: Does He effuse faster or slower than O 2? What is the relative rate of diffusion of He compared to O 2? Answer: = 2. 8, He is 2. 8 times much faster