Ideal Gas Law Calculations

Ideal Gas Law Calculations Missing Variable and Before & After Problems (and Stoichiometry)

Algorithm for Gas Law Problems Create a PVnT list One list means you have a Missing Variable problem Two lists means you have a Before & After problem Decide which form of the IGL to use For Missing Variable problem: PV = nrt [or (PV)/(nT) = R] For B & A problem: (P1V1)/(n1T1) = (P2V2)/(n2T2) Any variables that don t change will cancel, leaving the Combined Gas Law or a minor Gas Law (forget them) Attend to units For Missing Variable, all units must match R (especially P units) R =.0821 atm L/mol K OR 62.4 mmhg L/mol K For B & A, units must simply match each other (there is no R) T is always in Kelvins Solve for the missing variable (before or after substitution)

Gas Law - sample problem #1 A sample of gas has a volume of 22,400 ml at a temperature of 273 K and a pressure of 760. mmhg. How many moles of gas is in the sample? V = 22,400 ml T = 273 K P = 760. mmhg n =?

Gas Law - sample problem #1 P = 760. mmhg V = 22,400 ml n =? T = 273 K One list: use PV = nrt, solved for n = PV/RT Match units to R = 62.4 mmhg L/mol K Change 22,400 ml to 22.4 L n = (760.)(22.4)/(62.4)(273) = 1.00 = 1.00 mol

Gas Law - sample problem #2 What pressure is the exerted by 100. grams of H2 gas in a rigid 1.00 liter container at 100. C? V = 1.00 L T = 100. C + 273 = 373 K P =? n =? (two unknowns - yikes!) n = 100.g 2.02 g/mol = 49.5 mol One list ---> Use PV = nrt solved for P = nrt/v P = nrt/v = (49.5)(.0821)(373)/(1.00) = 1520 = 1520 atm

Gas Law - Sample Problem #3 Bacteria in sewage produce 37.6 ml of CH4 gas at 31 C and 753 mmhg. What is the volume of this gas at STP? V = 37.6 ml T = 31 C + 273 = 304 K P = 753 mmhg n =? mol But wait, there s more T2 = 0 C + 273 = 273 K P2 = 1 atm V2 =? L n2 =? mol

Gas Law - Sample Problem #3 V = 37.6 ml T = 31 C + 273 = 304 K P = 753 mmhg n =? mol Two lists ---> use PV/nT = PV/nT Since n isn t given, it isn t changing, Canceling n leaves P 1 V 1 /T 1 = P 2 V 2 /T 2 V2 =? T2 = 0 C + 273 = 273 K P2 = 1 atm n2 =? mol Match units: change P 2 to 760. mmhg, V 2 will be in ml Plug n Chug: (753)(37.6)/(304) = (760)V 2 /(273) V 2 = (753 x 37.6 x 273)/(304 x 760) = 33.5 V 2 = 33.5 ml

Gas Law - Sample Problem #3 V = 37.6 ml T = 31 C + 273 = 304 K P = 753 mmhg n =? mol Alternate Solution V2 =? ml T2 = 0 C + 273 = 273 K P2 = 1 atm = 760. mmhg n2 =? mol Multiply original Volume (37.6 ml) by P and T factors. Since P is increasing, that will cause V to decrease. The P factor must be less than 1. Since T is decreasing, that will cause V to decrease. The T factor must be less than 1. V 2 = (37.6 ml) (753/760) (273/304) = 33.5 ml

Stoichiometry with Gases You ve memorized the molar volume of a gas at STP = 22.4 L From Sample Problem #1, you notice that the molar volume factoid is simply a particular solution to the IGL. Forget it now. We are no longer restricted to STP. We can calculate the volume of a gas at any T & P by using PV=nRT Old: V=22.4 L New: V = nrt/p for n=1 mol, T=273K and P=1atm with no restrictions at all.

Gas Stoich. - Sample Problem (#5.69 on Page 226) LiOH is used in spacecraft to absorb the CO2 exhaled by astronauts. The reaction produces lithium carbonate and water. What volume of CO2 gas at 21 C and 781 mmhg could be absorbed by 327 g of LiOH? Eq n: 2 LiOH + 1 CO2 ---> 1 Li2CO3+ 1 H2O Change given to moles: 327 g LiOH 23.9 g/mol =13.7 mol LiOH Change moles LiOH to moles CO2: 13.7 mol LiOH x 1CO2/2 LiOH = 6.84 mol CO2 Change moles CO2 to liters CO2 (not at STP) V = nrt/p = (6.84 x 62.4 x 294)/781 = 161 L

Dalton s Law of Partial Pressures Deals with mixtures of gases Assigns a partial pressure to each component of the mixture based on the mole fraction The simple concept: If one third of the moles in a mixture is H2 gas (i.e., mole fraction H2=1/3), then one third of the total pressure is H2 s partial pressure The sum of all partial P s = total P The Key: In dealing with a mixture of gas, know whether you re using Ptotal or Ppartial npartial = PpartialV/(RT) ntotal = PtotalV/(RT) 2nd Key: There is no Vpartial. Every gas occupies all of the Volume

An Illustration of Dalton s Law of Partial Pressures Copyright Houghton Mifflin Company. All rights reserved 5-16AB

Partial Pressure - Sample Problem (#5.79 on Page 226) The gas from a certain volcano has the following composition in mole percent: 65.0% CO2, 25.0% H2, 5.4% HCl, 2.8% HF, 1.7% SO2, and 0.1% H2S. What is the partial pressure of each gas if the atmospheric pressure is 760. mmhg? PCO2 =.650 x 760 mmhg = 494 mmhg PH2 =.250 x 760 mmhg = 190 mmhg PHCl =.054 x 760 mmhg = 41 mmhg PHF =.028 x 760 mmhg = 21 mmhg PSO2 =.017 x 760 mmhg = 12 mmhg PH2S =.001 x 760 mmhg =.8 mmhg Ptot = (494+190+41.0+21.3+12.9+.8) = 760.mmHg

Partial Pressure - Sample Problem (#5.75 on Page 226) A mixture of gas contains 0.0200 mol He and 0.0100 mol H2 in a 5.00 L container at 10.0 C. Calculate PHe, PH2, and Ptotal. PHe = nhert/v = (.0200 x 62.4 x 283)/5.00 = 70.7 mmhg PH2 = nh2rt/v = (.0100 x 62.4 x 283)/5.00 = 35.3 mmhg Ptotal = ntotalrt/v = (.0300 x 62.4 x 283)/5.00 = 106.0 mmhg OR Ptotal = PHe + PH2 = (70.7 + 35.3) = 106.0 mmhg Also: Mol Fraction He x Ptotal = (.0200/.0300)106.0 mmhg = 70.7 mmhg Mol fraction H2 x Ptotal = (.0100/.0300)106.0 mmhg = 35.3 mmhg

Collection of a Gas over Water Copyright Houghton Mifflin Company. All rights reserved 5-17

Partial Pressure - Sample Problem Collecting Gas over Water (#5.81 on Page 226) Formic acid, HCOOH, is a convenient source of carbon monoxide. When warmed with sulfuric acid, formic acid decomposes to give H2O and CO. If 3.85 L of CO was collected over water at 25 C and 689 mmhg, how many grams of formic acid was consumed. Eq n: 1 HCOOH ----> 1 H2O + 1 CO Change given to moles: An IGL calc ---> Make list, etc Remember to subtract VP of H2O since gas is collected over water n = PV/(RT) = (689-24)mmHg(3.85 L)/(62.4 x 298 K) n =.138 mol CO Change mol CO to mol HCOOH:.138 mol CO x 1 HCOOH/1 CO =.138 mol CO Change mol HCOOH to g HCOOH:.138 mol HCOOH x 46.0 g/mol = 6.348 g HCOOH