CHEM 170 1st Edition Lecture 2Outline of Last Lecture I. Kinetic Theory of GasesII. Root Mean Squared Speed (Urms)III. Graham’s Law of EffusionOutline of Current Lecture IV. Last Lecture: Graham’s Law of Effusiona. exampleV. Pressurea. exampleVI. Boyle’s LawCurrent LectureChapter 11 – GasesLast Lecture: Graham’s Law of EffusionTo compare the speed of two gases we can use the equation for Urms. urms=√u2=√3 RTMIf the temperature was held constant and there was gas A and gas B, thenurms(A)urms(B)=√3 RTMA√3 RTMB=√MB√MAThese 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.ex. Argon gas trapped in a box escapes through a porous barrier in 40 seconds. An unknown gas of the same volume escapes from the same box in 120 seconds.u=tdspeed is equal to time divided by distanceurms(A)urms(B)=tAdtBdsince distance is constant¿tBtAthen set up a proportion so thattBtA=√MB√MA120 s40 s=√MB√39.95gmolMB=(120 s ×√39.95gmol40 s)2=359.55gmolPressure (P)Evangelista Torricelli (1608-1647) built the barometer. A barometer measures pressure.Force=(mass) (acceleration)F=maVolume=Areaheight → Area=VolumeheightA=Vhdensity=ρ=massVolumeρ=mVP=ForceArea=(ma)( A)=(mg)( A)=(mgh)(V )= ρgh ; where g =gravity =9.8ms2ex. If mercury (density = 1.3359 × 104 kg/m3) was at sea-level (0.76 m) what would be the pressure?P=ρgh=(1.3359 ×10 4kgm3)(9.8ms2)(0.76 m)¿1.01325× 105kgm× s2¿1.01325× 105pascalsNote: 1.01325× 105pascals=1 atm=760 mmHg=760 torrBoyle’s Law (1662) from “The Spring of the Air and its Effect”Boyle placed mercury in a J-shaped tube and used it to measure pressure.Pressure and volume are inversely related (V ∝1P) and at constant temperature and number of moles the relationship can be linearized so thatV =cP , where c is a
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