Gas Laws Edward Wen, PhD

Properties of Gases expand to completely fill their container take the shape of their container low density much less than solid or liquid state compressible when pressure is changed. mixtures of gases are always homogeneous (common air) fluid

Kinetic Molecular Theory

Properties of Gas: Indefinite Shape & Volume Gas molecules have enough kinetic energy (~500 m/s) and little attractions among each other  keep moving around and spreading out fill the container of whatever shape Hit the surface of container

Pressure: Gases Pushing Surface as Gas molecules strike a surface, they push on that surface push = force Pressure of gas = Hitting force per Hitted area Pressure depends on: number of gas particles in a given volume volume of the container average speed of the gas particles

Measuring Air Pressure Barometer: Height of column of mercury supported by air pressure Force of the air on the surface of the mercury = Gravity on the column of mercury Standard pressure: exactly 1 atmosphere = 760 mmHg  14.7 psi (pound per square inch) (“mmHg” read as “millimeter mercury”) gravity

The Effect of Gas Pressure Wind: whenever there is a Pressure difference, a gas will flow from area of High pressure  area of Low pressure the bigger the difference in pressure, the stronger the flow of the gas Vacuuming & Drinking through straw: if there is something in the gas’ path, the gas will try to push it along as the gas flows

Atmospheric Pressure & Altitude Altitude↑ Atmospheric pressure↓ At the surface, P = 14.7 psi, At 10,000 ft altitude (Big Bear Lakes, Ca) P = 10.0 psi Rapid changes in atmospheric pressure may cause your ears to “pop”  an imbalance in pressure on either side of your ear drum (driving or flying) Demo: Can you make a piece of paper uphold a bottle of water?

Common Units of Pressure Average Air Pressure at Sea Level pascal (Pa) 101,325 kilopascal (kPa) 101.325 atmosphere (atm) 1 (exactly) millimeters of mercury (mmHg) 760 (exactly) torr (torr) pounds per square inch (psi, lbs./in2) 14.7

Practice: Convert Pressure between units 735.0 mmHg = ? atm 35. psi = ? torr Ans: 0.9671 atm Ans: 1.8 × 103 torr

Boyle’s Law For the gas contained Pressure of a gas is inversely proportional to its volume: P 1/V constant T and amount of gas P x V = constant P1 x V1 = P2 x V2

When you double the pressure on a gas, the volume reduces to one half, (as long as the temperature and amount of gas do not change)

Boyle’s Law Demonstrations Puffing Mushmellow Ballon (long video) Why unfinished drinking water bottle collapsed after coming home from Big Bear Lakes?

Example of Boyle’s law: A cylinder equipped with a moveable piston has an applied pressure of 4.0 atm and a volume of 6.0 L. What is the volume if the applied pressure is decreased to 1.0 atm? Information Given: P1 = 4.0 atm V1 = 6.0 L P2 = 1.0 atm Find: V2 = ? L Answer: 24 L

We’re losing altitude. Quick Professor, give your lecture on Charles’ Law!

Charles’ Law For the gas contained and at constant Pressure: Volume is directly proportional to temperature (Kelvin): V  T constant P and amount of gas graph of V vs T is straight line as T increases, V also increases Kelvin K = °C + 273 V = constant x T if T measured in Kelvin

Example of Charles’ Law: A gas has a volume of 2 Example of Charles’ Law: A gas has a volume of 2.80 L at an unknown temperature. When the sample is at 0°C, its volume decreases to 2.57 L. What was the initial temperature in kelvin and in celsius? Information Given: V1 = 2.80 L V2 = 2.57 L t2 = 0°C Find: temp1 in K and °C Answer: T = 297 K = 24°C

Avogadro’s Law Volume directly proportional to the number of gas molecules V = constant x n constant P and T more gas molecules = larger volume count number of gas molecules by moles Equal Volumes of gases contain Equal numbers of molecules the gas doesn’t matter

Avogadro’s Law: At constant P and T, more gas  more volume

Combined Gas Law Boyle’s Law : Pressure and Volume at constant temperature Charles’ Law : Volume and absolute Temperature at constant pressure  Volume of a sample of gas when both the Pressure and Temperature change

Apply Gas laws to solve real world problems Which Gas Law to use? Boyle’s law: study P and V at constant T Charles’ law: study V and T at constant P Combined Gas Law for problems where two of P/V/T changes

Convert temperature to Kelvin: Example of Combined Gas Law: A sample of gas has a volume of 158 mL at a pressure of 755 mmHg and a temperature of 34°C. The gas is compressed to a volume of 108 mL and heated to 85°C, what is the final pressure in mmHg? Information Given: V1 = 158 mL, P1 = 755 mmHg, t1 = 34°C V2 = 108 mL, t2 = 85°C Find: P2, mmHg Convert temperature to Kelvin: T1 = 307 K, T2 = 358 K,

Ideal Gas Law Combined Gas Law + Avgadro’s Law  Ideal Gas Law R is called the Gas Constant the value of R depends on the units of P and V R = 0.0821 atm · L/K · mol convert P to atm and V to L Application of Ideal Gas law: when T, P, V of a gas all changes

First, convert P or V or T to standard units: atm, Liter, Kelvin Example of Ideal Gas Law : Calculate the number of moles of gas in a basketball inflated to a total pressure of 24.2 psi with a volume of 3.2 L at 25°C Information Given: V = 3.2 L, P = 24.2 psi, t = 25°C Find: n, mol Eq’n: PV = nRT SM: P,V,T,R → n First, convert P or V or T to standard units: atm, Liter, Kelvin T = 298 K P = 24.2 psi x ________ = 1.646 atm n = 0.22 mol

Applying Ideal Gas Law: Molar Mass of a Gas Determination of the molar mass of an unknown substance: Heat a weighed sample into gaseous state Measure temperature T, pressure P and volume V of the gas Use the Ideal Gas Law: #mole n = PV/RT

Example: Find Molar Mass A sample of a gas has a mass of 0. 311 g Example: Find Molar Mass A sample of a gas has a mass of 0.311 g. Its volume is 0.225 L at a temperature of 55°C and a pressure of 886 mmHg. Find its molar mass. Information Given: V = 0.225 L, P =, t = 328 K, m = 0.311 g Find: molar mass, (g/mol) Eq’n: PV = nRT; MM = mass/moles SM: P,V,T,R → n & mass → mol. mass Molar Mass = mass/moles Mass: given To find mole of gas? PV = nRT P = 886 mmHg x _____________ Mole of gas n = ____________

Air: Mixtures of Gases Air is a mixture (N2 , O2) Each gas in the mixture behaves independently of the other gases though all gases in the mixture have the same volume and temperature all gases completely occupy the container, so all gases in the mixture have the volume of the container Gas % in Air, by volume nitrogen, N2 78 argon, Ar 0.9 oxygen, O2 21 carbon dioxide, CO2 0.03

Collecting gas over water Zn metal reacts with HCl(aq) to produce H2(g). The gas flows through the tube and bubbles into the jar, where it displaces the water in the jar. Pgas = PH2O + PH2 Because water evaporates, some water vapor gets mixed in with the H2.

Reactions Involving Gases Reaction stoichiometry (Chapter 8) can be combined with the Gas Laws for reactions involving gases in reactions of gases, the amount of a gas is often given as a Volume instead of moles as we’ve seen, must state pressure and temperature Ideal Gas Law: from Volume of the gas + P & T  Moles; then we can use the coefficients in the equation as a mole ratio

Example: Gases in Chemical Reactions

gram of KClO3  mol KClO3  mol O2 (gas)  volume of O2 Example: How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 308 K Information Given: 294 g KClO3 PO2 = 755 mmHg, TO2 = 308 K Find: VO2, L Eq’n: PV=nRT Gram A  mole A  mole A  gram B Plan: gram of KClO3  mol KClO3  mol O2 (gas)  volume of O2 Molar mass KClO3 = 122.5 g/mol n = 3.60 mol O2 P = 0.993 atm V = 90.7 L

Standard Conditions (STP) Common reference points for comparing Standard Temperature & Pressure Standard Pressure = 1.00 atm Standard Temperature = 0°C = 273 K

Molar Volume of a Gas at STP Definition: The volume of 1 (exact) mole gas at STP Use the Ideal Gas Law: PV = nRT 1 mole of any gas at STP will occupy 22.4 L ==> Molar volume can be used as a conversion factor as long as you work at STP 1 mol gas  _______ L

Molar Volume So much empty space between molecules in the gas state,  the volume of the gas is not effected by the size of the molecules, (under ideal conditions).

Density of Gas at STP Since every exactly one mole of any gas has a volume of 22.4 L, whereas the mass of such gas would be as the molar mass in grams Density of Gas = Molar mass / Molar Volume Example: Density of Oxygen gas at STP = 32.00 g/mol / 22.4 L/mol = 1.43 g/L

Density of Common Gases At STP, the density of common gases (in g/L) as: H2 0.0900 He 0.179 CH4 0.716 N2 1.25 Air 1.29 O2 1.43 CO2 1.96 Cl2 3.17 Which one, hydrogen gas or helium gas, is better in blimps in providing lift? Why carbon dioxide is used in putting out fire? What if its density is less than the air?

Volume H2  mol H2  mol H2O  gram H2O Example of Using Molar Volume: How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2? Information Given: 1.24 L H2 Find: g H2O CF: 1 mol H2O = 18.02 g 1 mol H2  22.4 L STP? 1 mol gas = _____ L Volume H2  mol H2  mol H2O  gram H2O 0.988 g H2O

Real Gases Ideal gas laws assume No Attractions between gas molecules No Volume: gas molecules do not take up space based on the Kinetic-Molecular Theory Real gases: often do not behave like Ideal gases at High pressure (“Squeezed”) or Low temperature (“Frozen”)

Ideal vs. Real

Puffing Mushmellow & Ballon Soda bottle Submarine Pressure surge when bottle being squeezed Volume of the air inside the dropper decreases Water filling up the dropper Increased density of dropper (Massdropper/Volume) Dropper sinking!

Apply the Solution Map: Example: A gas has a volume of 2.80 L at an unknown temperature. When the sample is at 0°C, its volume decreases to 2.57 L. What was the initial temperature in kelvin and in celsius? Information Given: V1 = 2.80 L V2 = 2.57 L t2 = 0°C Find: temp1 in K and °C Eq’n: SM: V1, V2 T2 → T1 Apply the Solution Map:

Apply the Solution Map: Example: A gas has a volume of 2.80 L at an unknown temperature. When the sample is at 0°C, its volume decreases to 2.57 L. What was the initial temperature in kelvin and in celsius? Information Given: V1 = 2.80 L T1 = 297 K V2 = 2.57 L t2 = 0°C, T2 = 273 K Find: temp1 in K and °C Eq’n: SM: V1, V2 T2 → T1 Apply the Solution Map: convert to celsius

Example: A sample of gas has a volume of 158 mL at a pressure of 755 mmHg and a temperature of 34°C. The gas is compressed to a volume of 108 mL and heated to 85°C, what is the final pressure in mmHg? Information Given: V1 = 158 mL, P1 = 755 mmHg, t1 = 34°C V2 = 108 mL, t2 = 85°C Find: P2, mmHg Eq’n: SM: P1, V1, V2, T1, T2 → P2

Apply the Solution Map: Example of Ideal Gas Law : Calculate the number of moles of gas in a basketball inflated to a total pressure of 24.2 psi with a volume of 3.2 L at 25°C Information Given: V = 3.2 L, P = 24.2 psi, t = 25°C Find: n, mol Eq’n: PV = nRT SM: P,V,T,R → n Apply the Solution Map: convert the units

Apply the Solution Map: Example: Calculate the number of moles of gas in a basketball inflated to a total pressure of 24.2 psi with a volume of 3.2 L at 25°C Information Given: V = 3.2 L, P = 1.6462 atm, T = 298 K Find: n, mol Eq’n: PV = nRT SM: P,V,T,R → n Apply the Solution Map:

Example: How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 308 K Information Given: 294 g KClO3 PO2 = 755 mmHg, TO2 = 308 K Find: VO2, L Eq’n: PV=nRT CF: 1 mole KClO3 = 122.5 g 2 mole KClO3  3 moles O2 SM: g → mol KClO3 → mol O2 → L O2

Apply the Solution Map: Example: How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 308 K Information Given: 294 g KClO3 PO2 = 755 mmHg, TO2 = 308 K, nO2 = 3.60 moles Find: VO2, L Eq’n: PV=nRT CF: 1 mole KClO3 = 122.5 g 2 mole KClO3  3 moles O2 SM: g → mol KClO3 → mol O2 → L O2 Apply the Solution Map: convert the units

Apply the Solution Map: Example: How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 308 K Information Given: 294 g KClO3 PO2 = 0.99342 mmHg, TO2 = 308 K, nO2 = 3.60 moles Find: VO2, L Eq’n: PV=nRT CF: 1 mole KClO3 = 122.5 g 2 mole KClO3  3 moles O2 SM: g → mol KClO3 → mol O2 → L O2 Apply the Solution Map:

Example: How many grams of water will form when 1 Example: How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2? Information Given: 1.24 L H2 Find: g H2O CF: 1 mol H2O = 18.02 g 2 mol H2O  2 mol H2 1 mol H2  22.4 L Write a Solution Map: L H2 mol H2 mol H2O g H2O

Apply the Solution Map: Example: How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2? Information Given: 1.24 L H2 Find: g H2O CF: 1 mol H2O = 18.02 g 2 mol H2O  2 mol H2 1 mol H2  22.4 L SM: L → mol H2 → mol H2O → g H2O Apply the Solution Map:

Apply the Solution Map: Example: Find Molar Mass A sample of a gas has a mass of 0.311 g. Its volume is 0.225 L at a temperature of 55°C and a pressure of 886 mmHg. Find its molar mass. Information Given: V = 0.225 L, P = 1.1658 atm, t = 328 K, m = 0.311 g Find: molar mass, (g/mol) Eq’n: PV = nRT; MM = mass/moles SM: P,V,T,R → n & mass → mol. mass Apply the Solution Map:

Example of Using Molar Volume: How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2? Information Given: 1.24 L H2 Find: g H2O CF: 1 mol H2O = 18.02 g 2 mol H2O  2 mol H2 1 mol H2  22.4 L Write a Solution Map: L H2 mol H2 mol H2O g H2O

Apply the Solution Map: Example: How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2? Information Given: 1.24 L H2 Find: g H2O CF: 1 mol H2O = 18.02 g 2 mol H2O  2 mol H2 1 mol H2  22.4 L SM: L → mol H2 → mol H2O → g H2O Apply the Solution Map: