The Ideal Gas Law & Co.

A Reminder… assume ideal We that we live in an world where: Gas particles have no mass Gas particles have no volume Gas particles have elastic collisions These assumptions are used when trying to calculate the AMOUNT of a gas we have!

Why are these assumptions important? PV = nRT Image source: thefreedictionary.com

PV = nRT P V n R T The Ideal Gas Law RESSURE OLUME MOLES OF GAS GAS CONSTANT EMPERATURE Image source: popartuk.com

The MysteRious R 62.4 mmHg · L mol · K 8.31 kPa · L mol · K R is a constant (doesn’t change). Number value of R depends on other units. Units of R are a combination of many units. 62.4 mmHg · L mol · K 8.31 kPa · L mol · K 0.0821 atm · L mol · K Image source: toysrusemail.com

Ummm… What? PV = nRT P V R = n T (kPa) (mm Hg) (L) (atm) R = (mol) (K) Solve for R: P V R = n T Plug in units: (kPa) (mm Hg) (L) (atm) R = (mol) (K)

Gas Laws, Gas Laws Everywhere! T1 = V2 T2 Charles' Law Boyle's Law P1 x V1 = P2 x V2 P1 V1 P2 V2 = T1 T2 Combined Gas Law Ideal Gas Law Used with CHANGING CONDITIONS P V = n R T Used with only ONE SET OF CONDITIONS

When to Use PV = nRT Calculating amount of gas in moles Calculating P, V, or T if moles of gas are known. IMPORTANT! We must have 3 out of 4 pieces of information: P V n T

Practice with the Ideal Gas Law A gas sample occupies 2.62 L at 285ºC and 3.42 atm. How many moles are present in this sample? PV = nRT P V n = R T (3.42 atm) (2.62 L) n = = 0.196 mol 0.0821 L · atm mol · K (558 K)

But Let’s Be Practical… We don’t usually measure in moles! We usually measure quantities in GRAMS! PV = nRT PVM = gRT

PVM = gRT P V M g R T RESSURE OLUME OLAR MASS OF GAS (g/mol) RAMS OF GAS GAS CONSTANT EMPERATURE Image source: popartuk.com

Practice with the Ideal Gas Law A balloon is filled with 0.2494 g of helium to a pressure of 1.26 atm. If the desired volume of the balloon is 1.250 L, what must the temperature be in ºC? P V M PVM = gRT T = g R 4.00 g mol (1.26 atm) (1.250 L) T = 308 K = - 273 0.0821 L · atm mol · K (0.2494 g) 35 ºC

PV=nRT vs. PVM=gRT Use PV=nRT when: Use PVM=gRT when: You are given moles in the problem. You are searching for moles as an answer. Use PVM=gRT when: You are given grams in the problem. You are searching for grams as an answer.

What Else Happens Under Unchanging Conditions? At constant V and T, pressure is easy to calculate! “The sum of the individual pressures is equal to the total pressure.” Total Pressure = Pressure of gas 1 + Pressure of gas 2 + Pressure of gas 3 + Pressure of gas 4 … Ptotal = P1 + P2 + P3 + … Dalton's Law of Partial Pressures

Partial Pressures Practice A sample of hydrogen gas is collected over water at 25ºC. The vapor pressure of water at 25ºC is 23.8 mmHg. If the total pressure is 523.8 mmHg, what is the partial pressure of the hydrogen? Ptotal = PH2 + PH2O = 523.8 mm Hg PH2 + 23.8 mm Hg = PH2 500.0 mm Hg Source: 2003 EOC Chemistry Exam

What do Changing Conditions Affect? We have learned that we can change 3 variables: Temperature, Volume, and Pressure. If MASS remains constant… …But VOLUME changes… Then DENSITY CHANGES! D = M V

Two Types of Density Problems: At STP: Not at STP: molar volume of any gas at STP = Determine new volume (V2 ) using Combined Gas Law 22.4 Liters P1 V1 P2 V2 = T1 T2 STP Values non-STP Density at STP = Density at non-STP = molar mass molar mass molar volume 22.4 Liters V2

Practice with Density Problems: Determine the density of ethane (C2H6) at STP: Determine the density of C2H6 at 3.0 atm and 41ºC. P1 V1 P2 V2 = T1 T2 V2 = 8.6 L molar mass D (at STP) = molar volume D = molar mass V2 V2 = P1 V1 T2 molar mass = 30.08 g T1 P2 P1 = 1.0 atm V1 = 22.4 L T1 = 273 K P2 = 3.0 atm V2 = ? T2 = 314 K molar volume = 22.4 L V2 = (1.0 atm) (22.4 L) (314 K) D = 30.08 g 8.6 L = 3.5 g/L (273 K) (3.0 atm) 30.08 g D = = 1.34 g/L 22.4 L V2 = 8.6 L

What do Changing Conditions Affect? Density Stoichiometry problems So Far: Now… STP Non-STP

Mass-Volume at Non-STP Two parts to solving these problems: Use Stoichiometry Use Gas Law