CHEM 1061 1st EditionLecture 12Outline of Last Lecture I. TemperatureII. Avogadro’s LawIII. Gas Behavior and Standard Conditionsa. Standard Temperature and Pressureb. Standard Molar VolumeIV. Ideal Gas LawV. Applications of Ideal Gas LawOutline of Current Lecture I. Molar Mass of a GasII. Partial Pressure of a Gas III. Collecting Gases Over WaterCurrent LectureMolar Mass of a GasPV=nRTPV=(mass/molar mass)RTmolar mass = (mass)RT/PVEx: Unknown Liquid is in a flask. Find the MOLAR MASS.P(atm)=P(gas)=749mmHGV(flask)=495mLmass=2.56gT=325K1atm=760mmHG2.56g x 0.0821 L*atm/mol*K x 325K = 68.3072/0.4878 = 140.038 g/mol0.9855atm x 0.495 LPartial Pressure of GasesThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Gases mix completelyAssumes no interactions between gasesP is proportional to nP(total)V=n(total)RTP(total)=n(total)RT/Vn(total)=n1+n2+n3+etcMole Fraction: (X) = fraction of moles of that gasEx: X(a) = mole fraction A = 4/12 = 0.33 = 33%Collecting Gases Over WaterYou need to correct for the vapor pressure of water.P(gas)+P(H2O)=P(total)P(gas)=P(total)-P(H2O)P(H2O) found on Table 5.2 and is dependent on temperature.EX: How many moles of H2 are formed?P(tube) = P(total) = 745mmHGV = 50.0mL = 0.050LT = 26 degree C = 299KP(H2) =P(total) – P(H2O) = 745 mmHG – 25.2 mmHG (amount found on Table 5.2) =719.8mmHG0.947 atm x 0.050 L = 299 K x 0.0821 L*atm/mol*K x n(H2)n(H2) = 1.93 x 10^-3 mol
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