The Gas Laws

Apparatus for Studying the Relationship Between Pressure and Volume of a Gas As P (h) increases V decreases

Boyle’s Law P ∝ 1/V Constant temperature P x V = constant Constant amount of gas P x V = constant P1 x V1 = P2 x V2

Example 1 A helium balloon was compressed from 4.0L to 2.5L at a constant temperature. If the pressure of the gas in the 4.0L balloon is 210 kPA, what will the pressure be at 2.5L? Given: V1 = 4.0L V2 = 2.5L P1= 210 kPa P2 = ? P1V1 = P2V2 (210 kPa) × (4.0L) = P2 × (2.5L) P2 = 336 kPa ≈ 340 kPa

Example 2 A sample of neon gas occupies 0.200L at 0.860 atm. What will be its volume at 29.2 kPa pressure? P1V1 = P2V2 Given: V1 = 0.200L V2 = ? P1= 0.860 atm P2 = 29.2 kPa (87.1 kPa) × (0.200L) = (29.2 kPa) × V2 V2 = 0.597L **Units must match for each variable (doesn’t matter which one is converted) 0.860 atm 101.3 kPa 1 atm = 87.1 kPa

Day 2 Comparing volume and temperature Keep pressure constant

Variation in Gas Volume with Temperature at Constant Pressure As T increases V increases

Charles’s Law 𝑉 1 𝑇 1 = 𝑉 2 𝑇 2 𝑉 𝑇 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 **Temperature must be in Kelvin K = 0C + 273 𝑉 1 𝑇 1 = 𝑉 2 𝑇 2 𝑉 𝑇 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Constant pressure Constant amount of gas 𝑉∝𝑇

Example 1 A gas sample at 40.0°C occupies a volume of 2.32L. If the temperature is raised to 75.0°C, what will the volume be, assuming the pressure remains constant? Given: T1 = 40.0°C = 313K T2 = 75.0°C = 348K V1= 2.32L V2 = ? V1 V2 T1 T2 = 2.32L V2 313 348 = V2 = 2.58L

Example 2 A gas sample at 55.0°C occupies a volume of 3.50L. At what new temperature in Celcius will the volume increase to 8.00L? Given: T1 = 55.0°C = 328.0K T2 = ? V1= 3.50L V2 = 8.00L V1 V2 T1 T2 = 3.50L 8.00 328.0 T2 = T2 = 750K-273=477°C

Day 3

Gay-Lussac’s Law 𝑃 1 𝑇 1 = 𝑃 2 𝑇 2 𝑃 𝑇 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑃∝𝑇 Constant volume 𝑃 1 𝑇 1 = 𝑃 2 𝑇 2 𝑃 𝑇 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Constant volume Constant amount of gas **Temperature must be in Kelvin K = 0C + 273

Example 1 The pressure of a gas in a tank is 3.20 atm at 22.0°C. If the temperature rises to 60.0°C, what will be the gas pressure in the tank? Given: P1 = 3.20 atm P2 = ? T1= 22.0°C = 295.0K T2 = 60.0°C =333.0K P1 P2 T1 T2 = 3.20 atm P2 295.0K 333.0K = P2 = 3.61 atm

Example 2 A rigid container has a gas at constant volume at 665 torr pressure when the temperature is 22.0C. What will the pressure be if the temperature is raised to 44.6C? Given: P1 = 665 torr P2 = ? T1= 22.0°C = 295K T2 = 44.6°C =317.6K P1 P2 T1 T2 = 665 torr P2 295K 317.6K = P2 = 716 torr

STOP

Avogadro’s Law and Combine Gas Law

What is the relationship between number of moles and volume?

What is the same about these gases? What is different about these gases?

Avogadro’s Law 𝑉 𝑛 = constant 𝑉 1 𝑛 1 = 𝑉 2 𝑛 2 V ∝ n (number of moles) Constant temperature Constant pressure 𝑉 𝑛 = constant 𝑉 1 𝑛 1 = 𝑉 2 𝑛 2 1 mol of gas at STP = 22.4L

Gas Stoichiometry When gases are involved, the coefficients in a balanced chemical equation represent not only molar amounts (mole ratios) but also relative volumes (volume ratios).

Example 1 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what will be the new volume? Assume constant temperature and pressure. Given: V1 = 5.00 L V2 = ? n1= 0.965 mol n2 = 1.80 mol V1 V2 n1 n2 = 5.00 L V2 0.965 mol 1.80 mol = V2 = 9.33 L

Calculate the volume that 0.881 mol of gas at STP will occupy. Example 2 Calculate the volume that 0.881 mol of gas at STP will occupy. 0.881 mol 22.4 L 1 mol = 19.7 L

How many grams of carbon dioxide gas are in a 0.75 L balloon at STP? Example 3 How many grams of carbon dioxide gas are in a 0.75 L balloon at STP? 0.75 L CO2 1 mol CO2 44.01 g CO2 22.4 L CO2 1 mol CO2 = 1.5 g CO2

Example 4 What volume of oxygen gas is needed for the complete combustion of 4.00 L of propane gas (C3H8)? Assume constant temperature and pressure. 𝐶 3 𝐻 8 + 5 𝑂 2 →3 𝐶 𝑂 2 + 4 𝐻 2 𝑂 4.00 L C3H8 5 L O2 1 L C3H8 = 20.0 L

Gas Law Summary

Charles’s Law Gay-Lussac’s Law

Combined Gas Law Combine all 4 to make one master equation: 𝐵𝑜𝑦𝑙 𝑒 ′ 𝑠 𝑙𝑎𝑤:𝑃×𝑉=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑛 𝑎𝑛𝑑 𝑇 𝐶ℎ𝑎𝑟𝑙 𝑒 ′ 𝑠 𝑙𝑎𝑤: 𝑉 𝑇 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑛 𝑎𝑛𝑑 𝑃 𝐺𝑎𝑦−𝐿𝑢𝑠𝑠𝑎 𝑐 ′ 𝑠 𝑙𝑎𝑤: 𝑃 𝑇 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑛 𝑎𝑛𝑑 𝑉 𝐴𝑣𝑜𝑔𝑎𝑑𝑟 𝑜 ′ 𝑠 𝑙𝑎𝑤: 𝑉 𝑛 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑃 𝑎𝑛𝑑 𝑇) Combine all 4 to make one master equation: Putting all of our gas laws together we get…Because we will typically not change the number of moles in our reactions, we can say that n1=n2 so then we can now take out the number of moles and get a combined gas law 𝑃 1 𝑉 1 𝑛 1 𝑇 1 = 𝑃 2 𝑉 2 𝑛 2 𝑇 2 𝑃𝑉 𝑛𝑇 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑃 1 𝑉 1 𝑇 1 = 𝑃 2 𝑉 2 𝑇 2

Example 1 A gas at 110 kPa and 30.0˚C fills a flexible container with an initial volume of 2.00L. If the temperature is raised to 80.0˚C and the pressure increased to 440 kPa, what is the new volume? Given: P1 = 110 kPa P2 = 440 kPa T1 = 30.0˚C =303K T2 = 80.0˚C = 353K V1 = 2.00L V2 = ? 110 (2.00) 440 (V2) 303 353 = V2 = 0.583L ≈ 0.58L 𝑃 1 𝑉 1 𝑇 1 = 𝑃 2 𝑉 2 𝑇 2

Example 2 An unopened bottle of soda contains 46.0 mL of gas confined at a pressure of 1.30 atm and temperature of 5.00˚C. If the bottle is dropped into a lake and sinks to a depth at which the pressure and temperature changes to 1.52 atm and 2.90˚C, what will be the volume of gas in the bottle? 1.30 (46.0) 1.52 (V2) 278 275.9 𝑃 1 𝑉 1 𝑇 1 = 𝑃 2 𝑉 2 𝑇 2 = Given: V1 = 46.0 mL V2 = ? P1 = 1.30 atm P2 = 1.52 atm T1 = 5.00˚C = 278K T2 = 2.90˚C = 275.9K V2 = 39.0mL

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Warm Up A sample of gas of unknown pressure occupies 3.2 L at a temperature of 302 K. The sample of gas is then tested under known condition and has a pressure of 22.6 kPa and occupies 1.2 L at 310 K. What was the original pressure of the gas?

The Ideal Gas Law Ideal Gas Real Gas Follows some gas laws under some conditions of temperature and pressure. Does not conform to the Kinetic Molecular Theory. Real gases have a volume and attractive and repulsive forces. A real gas differs from an ideal gas the most at low temperature, high pressure, and high concentration. Low T gas is moving less High P gas is more attractive High Conc has more collisions Ideal Gas Follows all gas laws under all conditions of temperature and pressure. Follows all conditions of the Kinetic Molecular Theory (KMT) An ideal gas does not exist in real life

(R is the universal gas constant) Ideal Gas law 𝑃𝑉 𝑛𝑇 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (R is the universal gas constant) PV = nRT R = PV nT = (1 atm)(22.4L) (1 mol)(273K) 𝑅 = 0.0821𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 𝐾 = 8.314𝐿 𝑘𝑃𝑎 𝑚𝑜𝑙 𝐾 = 62.4𝐿 𝑚𝑚𝐻𝑔 𝑚𝑜𝑙 𝐾 = 62.4𝐿 𝑡𝑜𝑟𝑟 𝑚𝑜𝑙 𝐾

Example 1 Calculate the number of moles of gas contained in a 3.00L vessel at 298K with a pressure of 1.50 atm. PV=nRT V = 3.00L T = 298K P = 1.50atm n = ? R = 0.0821 L atm mol K (1.50)(3.00) = n (0.0821) (298) n= 0.184 mol

Example 2 What will the pressure (in kPa) be when there are 0.400 mol of gas in a 5.00L container at 17.0˚C? V = 5.00L T = 17.0˚C = 290K P = ? kPa n = 0.400 mol R = 8.314 L kPa mol K PV=nRT P(5.00) = (0.400)(8.314)(290) P= 193 kPa

STOP

Stoichiometry

(3.00 atm)(5.00 L) = (n)(0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 𝑘 )(298 K) Example 1 If 5.00L of nitrogen reacts completely with excess hydrogen at a constant pressure and temperature of 3.00 atm and 298K, how many grams of ammonia are produced. 𝑁 2 + 3𝐻 2 →2𝑁 𝐻 3 PV=nRT (3.00 atm)(5.00 L) = (n)(0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 𝑘 )(298 K) n= 0.613 mol N2 0.613 mol N2 2 mol NH3 17.04 g NH3 1 mol N2 1 mol NH3 = 20.9 g NH3

(2.00 atm)(6.75 L) = (n)(0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 𝑘 )(298 K) Example 2 How many grams of calcium carbonate will be needed to form 6.75L of carbon dioxide at a pressure of 2.00 atm and 298K? CaCO3(s)  CO2(g) + CaO(s) PV=nRT (2.00 atm)(6.75 L) = (n)(0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 𝑘 )(298 K) n= 0.552 mol CO2 0.552 mol CO2 1 mol CaCO3 100.09 g CaCO3 1 mol CO2 1 mol CaCO3 = 55.2 g CaCO3

(3.50 atm)(V) = (1.13 mol)(0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 𝑘 )(225 K) Example 3 How many liters of chlorine will be needed to make 95.0 grams of C2H2Cl4 at 3.50 atm and 225K? 2Cl2(g) + C2H2(g)  C2H2Cl4(l) 95.0 g C2H2Cl4 1 mol C2H2Cl4 2 mol Cl2 167.84 g C2H2Cl4 1 mol C2H2Cl4 = 1.13mol Cl2 PV=nRT (3.50 atm)(V) = (1.13 mol)(0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 𝑘 )(225 K) V= 5.97 L Cl2

Graphic Organizer for PV=nRT and Stoichiometry

d is the density of the gas in g/L Density (d) Calculations d = PM RT n V = P RT m M V = P RT so n = m M m is the mass of the gas in g d = m V M is the molar mass of the gas d is the density of the gas in g/L Molar Mass (M ) of a Gaseous Substance dRT P M =

Example 1 What is the molar mass of a pure gas that has a density of 1.40gL at STP? d = PM RT

Example 2 Calculate the density a gas will have at STP if its molar mass is 39.9g/mol.

Test Review Using the principles behind Graham’s Law, knowing that M H2 = 2.02 g/mol and M N2 = 28.02 g/mol which gas would diffuse faster in a closed container? Hydrogen 2. Why do real gases deviate from ideal gas behavior at high pressure, low temperature and high concentration ? High P-increased collisions, increase temp High C- particles collisions increase and so doespressure Low T-slow particles down, less collisions and pressure 3. When can you use 22.4 L = 1 mole and when can you use volume to volume ratios? At STP = 1 atm and 273 K

Test Review 4. Given the two problems below, which problem starts with stoichiometry and then uses PV=nRt, and which problem starts with PV=nRt and then uses stoich? N2(g) + 3H2 (g) 2NH3(g) A. What is the mass of NH3 gas is produced at 55.5oC and 670 torr, when 1L of H2 reacts with an excess of N2? PV=nRT, then Stoich B. What volume of NH3 gas is produced at 55.5oC and 670 torr when 0.6789 g of H2 reacts with N2? Stoich, then PV=nRT