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+ Lab

by: Antonetta Hermann
Antonetta Hermann
GPA 3.69


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Class Notes
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This 80 page Class Notes was uploaded by Antonetta Hermann on Tuesday September 8, 2015. The Class Notes belongs to CH 417 at University of Oregon taught by Staff in Fall. Since its upload, it has received 64 views. For similar materials see /class/187295/ch-417-university-of-oregon in Chemistry at University of Oregon.


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Date Created: 09/08/15
Ohm39s Law EIR Where 0E is the electromotive force voltage applied to the circuit 01 is the current through the circuit and 0R is the resistance of the circuit Series resistors If resistors are placed in series in a circuit the overall resistance is the sum of the individual values R EQm 21 total Parallel resistors If resistors are placed in parallel in a circuit the overall resistance is obtained by taking the sum of the reciprocals of individual values then taking the reciprocal Kirchhoff s rst law At any point in a circuit the current owing in equals the current owing out 21in 210m If the current is represented as a signed quantity then 21120 139 DC Circuit Analysis The analysis of a DC circuit involves determining either the Voltage current or resistance between any two points As an example for the circuit Given he total Voltage drop between A and c he resistances R and R2 Find he expected Voltage drop between B and c DC C1rcu1t Analys1s 39 39 kn 39 A to C is the Voltage divided by the total resistance We further know that this is the me as the current owing from B to C Finally we know that the Voltage between B and C is the current times the resistance nu I VAC AC R1R2 IBCIAC VBCIBCR2 VAC R2 R1R2 DC Circuit Analysis For example suppose you are trying to measure the output Voltage of a source What is the voltage drop from A to B DC Circuit Analysis For example suppose you are trying to measure the output Voltage of a source What is the voltage drop from A to B R V V meter AB source R R source meter Notice that the Voltage you measure depends on the resistance of the meter R V V meter AB source R R source meter There are two important limiting cases of this equation One occurs if the resistance of the meter is large compared With the internal resistance of the source Rmeter gtgt R Z VAB z V SOLWC SOLWC The other limiting case occurs if the resistance of the meter is much less than the resistance of the source R 7 7 f r Rmeter ltlt R 2 AB me 6 source SDIWC R SDIWC It should be clear from the above example that the measured voltage depends on how the measurement was made DC voltage measurements We summarize the previous example as follows DC voltage measurements To emphasize the point consider the circuit shown below If no meter were attached to the circuit the voltage drop from point A to point B would be exactly half of the total applied voltage or 05V 1 1m I39M L 1 T 39lliLWi39r ql Whit i MD DC voltage measurements If a meter is attached however the resistance of the meter must be considered The total resistance between point A and point B is given by the expression La l L Emm ER 1 1 1 x 1 ma Us Walt R Z 1069 R total meter 4 q 391 ME DC voltage measurements Suppose the resistance of the meter is 22 MD a typical input resistance for a digital voltmeter Then the total resistance from A to B will be 0957MQ and the Voltage drop is A E Wm R H V 1V 0957MQ V 1th i m A 1MQ0957MQ 1m 04s9v i DC voltage measurements Now suppose you make the same measurement using an oscilloscope as a meter The input resistance is about 1M9 so the total AiB resistance is 05 M9 The Voltage drop from A to B will be VA 05MQ B 1MQ 05MQ g3 ll g DC voltage measurements Finally make the same measurement using a VOM The input resistance of this device is about 30KQ so the total AiB resistance is only 29KQ 0029MQ The Voltage drop from A to B will be VAB1V 0029M 1m Rm 1MQ0029MQ Viva gt 0028v M2 1 C The VOM is doing exactly as you asked but the answer is not What you wanted AC Circuit Analysis Many electronic signals encountered in physical measurements vary rapidly With time The circuits used to record these signals are similar to those we have just encountered but their analyses are somewhat more complicated they are analyzed as AC alternating current circuits rather than DC direct current circuits In general three quantities are necessary to specify an AC signal Amplitudez how big the signal is units Volts Frequencyz how quickly the signal changes units Hz Phasez a quotstarting pointquot for the signal units radians or degrees AC Circuit Analysis If the only circuit elements are resistors there is no formal difference between the analyses of AC and DC circuits The amplitude of the AC signal behaves in exactly the same way as a DC voltage and the frequency and phase remain unchanged If capacitors or inductors are present in the circuit these will cause the circuit to respond differently to an AC signal than to a DC signal Capacitorsz will quotpassquot high frequency AC but not low frequency or DC signals Inductors or coils behave in the opposite way an inductor passes DC or low frequency AC but resists high frequency AC Circuit Analysis There are two ways of analyzing an AC circuit Which give equivalent results 1 Solve the differential equations Which govern the charging of capacitors and the passage of current through coils 2 Trick Ohm39s laW into solving the differential equations for you The second method involves treating the time dependent part of an AC voltage as an imaginary exponential and the troublesome AC circuit elements capacitors and inductors as resistors With an imaginary resistance Which depends on frequency AC Circuit Analysis Method 1 Consider the following circuit Fla We would like a way of expressing the voltage l xl between any two elements of the circuit as a J function of time The total voltage applied to the I circuit Et is related to the resistance R1 the capacitance C2 the current I through the resistor and the charge Q on the capacitor the equation i r w E l E II 1 2 Et IR1 C 2 d t Or letting Et Eoem we obtain Eoem R1 t 4quot which is solved by Qoelthw Q A N V 393 N AC Circuit Analysis Method 1 Consider the following circuit Fla WW 2 We would like a way of expressing the voltage l xl between any two elements of the circuit as a H 1 m function of time The total voltage applied to the circuit Et is related to the resistance R1 the capacitance C2 the current I through the resistor and the charge Q on the capacitor the equation Ill 1 EtIR1g C2 or letting Em 2 R1 dim 39 t 2 h39 h39 1 db m wquot w 1c 1s so ve y Qt Q06 l EEC geimlw AC Circuit Analysis Method 1 1I39 Ib R1 Wl T Summary Knowing R1 D t 11 and C2 allows us to solve for J Q0 and 1 We can then nd E AB and EBC as functions of time I n AC Circuit Analysis Method 2 e Given R1 C2 and Et nd EL R1 EBCt An Ohm39s Law like iiiL solution would have to look like i this c E t AC 2 Z1 Z2 BC IAC EBC I IBCZ2 E I E BC I AC Circuit Analysis Method 2 In order to make this method work 7 we must de ne the impedance Z C2 of a circuit element as follows E ZRiwLLRiwLaL iwC wC Th impedance is just an extension of the concept ofresistance whic accounts for some of the peculiarities of AC circuits Notice that 1 the impedance has units of Ohms just like resistance 2 impedance unlike resistance is a complex quantity d to the appearance of the m AC Circuit Analysis Method 2 l lagt l The three terms of the impedance g are called the resistance the inductive reactance and the capacitive reactance 7 ill ill y QH x a capacitive reactance AC Circuit Analysis Method 2 The AC Voltage applied to the 7 circuit is represented as if E t E e C EEl cosaz isinat T E with u2rrf By convention a measurement of the Voltage or current will correspond to taking the real part ofEt or t Having made these de nitions we can write EU for any AC or DC circuit IrZ AC Circuit Analysis Method 2 Now see how this method works m Let R11KQ if C2 1 11F C2 f lKHz 7 First calculate the impedance then calculate the current 2 R 1K0 zZ 37 1471592i0 21r10 Hz10 F Note that Hz lF391Q AC Circuit Analysis Method 2 The magnitude of the total impedance is Ztotal URZ X2 NOW since Z lzle p R iX Zcos isin R 1X 1 NEHI H gr lL tomi quot We are able to solve for 0 5 00801 cot X sin tan 1 tan l 1 1 R 21thR 580 AC Circuit Analysis Method 2 Now solve for the current through the circuit D1 I T ww r I Z i 139 0 l C Z6 IZI T Finally use the current to calculate the voltages across Z1 and 22 EAB 1R1 EEC 24122 E0eiwti Eoem q i Z R1 z 275fC2 W tram it 1Ve 10009 z W e 15929 2 18809 1880Q Eggmd 053v cosat E 1ng 085V cosat AC Circuit Analysis Method 2 EAB 1R1 EEC 122 Eoemi R E0eiwi i Z 1 Z 275sz we 1 W 41 e 2 1880Q 1880Q measured EAB 053V 008W Eggmd 085V cosat TEL 6381191 1913 i39 AC Circuit Analysis Method 2 the phase angle which represents the phase shift between EBC and E AC can also be measured E AC 2 E t Eggamd 085V cosat 1 Eoel 0t Eg asured COSUI 580 Er asured COSDt COSUI 1480 Why does it work Multiplying E t E 06 2 E0 cosat isinat by i0 has the same effect as differentiating with respect to t Dividing by i0 has the same effect as integrating with respect to t Thus we can make complex arithmetic mimic calculus in manipulating the time dependent voltage How is this method better or worse 0 The concept of impedance is a straightforward extension of the concept of resistance You can use much of your intuition about resistance to handle AC signals 0 On the other hand the physics of the circuits can be obscured by the algebra Electrons don t really pass through a capacitor and the reactive elements store energy rather than dissipating it as resistors do Compressibility of a real gas The compressibility factor of a gas and by extension any uid is defined as PVm RT For an ideal gas the compressibility factor is always 1 For a real gas the deviation of the compressibility factor from 1 is a convenient measure of nonideality This eXperiment allows you to measure the compressibility factor of carbon dioxide at temperatures just above and just below the critical point and to compare your observations with a simple model The model The van der Waals equation of state is a commonly used empirical equation that is more exible than the ideal gas equation P 12 Vm b Vm The constant a is interpreted as a measure of the attractive force between molecules While the constant 9 is a measure of the excluded volume resulting from the volume of the condensed gas The model The van der Waals equation can be rewritten in terms of the compressibility factor Z PVm l a RT 1 b RTVm V m Notice that as Vm becomes large Z approaches 1 The gas Carbon dioxide has been an important industrial chemical for many years It now is receiving attention as a solvent The fact that its critical point is only slightly above room temperature means that its solvation properties can be tailored by changing its density in a way that cannot be accomplished With ordinary liquids The equipment The instrument is the Phase Monitor from Supercritical Fluid Technologies It is designed to provide an adjustable volume 3 to 33 mL and temperature room temperature to 100 at pressures up to 10 000 PSI The equipment The instrument is the Phase Monitor from Supercn39tical Fluid Technologies It is designed to provide an mm adjustable volume 3 to 33 mL and temperature room temperature to 100 at pressures up to 10 000 PSI Part 1 an isotherm The first part of the experiment will involve recording pressure as a function of volume at a constant temperature or as near to one as we can manage The procedure will be to maintain a constant thermostatted temperature while reducing the volume of the chamber Choosing a temperature below the critical point we eXpect condensation to occur at the vapor pressure of CO2 at that temperature During condensation the pressure will not change as volume is changed except as a result of a change in the temperature Part 1 an isotherm The Van der Waals equation makes a somewhat s prediction about 5 mm the shape of the 5 mm isotherm Plat pv 39vl 273 5 27 coz hcoz nnm nnnzu n n nun 5 mm 5 mm Graphlcs Part 1 an isotherm Plat pv 39vl 273 5 27 coz hcoz 1 n nnm nnz Clearly the oscillatory Segment should be replaced by a t 39 h horizontal line connecting the outer and inner curves Where 75 should the line be drawn mm Where to draw the line The solution was given by Maxwell who reasoned as follows The Helmholtz free energy is given by dA Pd V SdT So a finite Change AA is given by AA AZ A1 f12PdV and a round trip AA is given by AA 121161de L 2PWsz which must be zero since A is a state function Where to draw the line The solution we seek is then the pressure PW for which 2Pdw P dV0 vap V1 V Where Volumes V1 and V2 are the inner and outer solutions of the Van der Waals equation at the temperature of interest Where to draw the hue This is equivalent to constructing PV so that the areas included between the curve and the horizontal line are equal above and below the line This allows us to calculate the Vapor pressure of CO2 for as many temper tures as we like which brings us to the next part of the experiwent Part 2 the vapor pressure curve You can easily collect pressures as a function of temperature for a range of 5 C or so or more if you want to take the time Below the critical point these pressures are the vapor pressure Whenever the liquid and vapor coeXist Above the critical point no such interpretation is possible The treatment in terms of the van der Waals equation is straightforward enough but it is a bit tedious computationally Why not use the more convenient vapor pressure equation the Clausius Clapeyron equation dep AH dT RT2 vap WP RT lanap The Clapeyron equation The C lapeyron equation is an exact thermodynamic relationship d P AS 1 dT AV This holds for any phase transition The C lausiusC lapeyron equation is derived from it by assuming The volume of the liquid is negligible compared With that of the gas The gas behaves ideally so Vgas can be replaced by RTP Neither of these assumptions is good The Clausius Clapeyron equation You Will have demonstrated in part 1 that the gas is not behaving ideally so the second assumption is testably unsound What about the assumption that the volume of the gas is much larger than that of the liquid At the critical point the molar volumes of the vapor and liquid phases are equal So are the densities the refractive indices the enthalpy of formation and all other physical properties The distinction between the gas and the liquid vanishes Part 3 An isotherm above the critical point Slightly above the critical point an isotherm looks very much like an isotherm just below the critical point There is one subtle difference There is no longer a at condensation region Instead there is a broad region With a sl1ght slope MK m 1 39F Egtlt 10 1 KR 7 4x 10 cm 39TI392gtlt1 x 000012 000014 000016 000018 00002 The critical point Just at the critical point there is an in ection point in the PV isotherm for the van der Waals gas r j 113 R V a 13 10 RR 1 ax H s K an 753111 quot N aha 5 E xxx Hack 7410 39 E a 5 1 ha 5x xquot H E xx 7 239 1 DE WERE any 3 xJ o 39 5 x xxx 0013012 0013014 0001316 BEEN95H inane an E 2 1 3935 an Active and passive circuit elements The circuit elements we have encountered up to now have been for the most part passive circuit elements These include for example resistors and capacitors They alter the ow of electrons through a circuit bit they do not add electrons to the circuit In contrast quotactivequot circuit elements amplifiers transistors vacuum triodes inject electrons into a circuit in response to some signal This has the effect of altering the voltage the current or both Active and passive circuit elements Passive circuit elements are relatively uniform stable and cheap They are easy to design and manufacture to relatively strict performance standards and their performance does not change much with use Short of destruction a resistor will not usually change its resistance by more than 1 for years Active circuit elements are more expensive to produce with tight tolerances and they are inherently less stable than resistors and capacitors Their lack of uniformity can cause amplifiers from the same assembly line to differ by a factor of two or more Furthermore their performance tends to drift with temperature load and age The operational ampli er The operational amplifier is a differential ampli er whose output is stabilized using passive circuit elements In this way a relatively cheap ampli er can be made to perform with a high degree of accuracy and reliability The strategy is to direct part of the amplifier39s output into its own inverting input so that the amplifier quotknows when to quitquot The simplified schematic symbol for an opamp is inverting input output noninverting input The operational ampli er The output of an opamp is proportional to the difference in the voltage between the two inputs v w vol gain 1 Mg in Usually the proportionality constant the gain is chosen to be very large a million or so so that the output swings to its maximum or minimum in response to a just a few microvolts difference between the inverting and noninverting inputs The voltage follower Suppose we now connect the output to the inverting input I Vol gam Mg V 1 B in in The output will rise or fall until it is only a few microvolts different from the noninverting input then it will stop The few microvolts difference between the two inputs will be sufficient to maintain the output voltage The microvolt or so difference is usually ignored for practical purposes the output voltage is the same as This device is a voltage follower The voltage follower If the output voltage is the same as the input voltage what good is a voltage follower akv out 2 The answer lies in the amount of current the opamp draws and provides Usually opamps are designed to have a very high input impedance so that almost no current is drawn In addition the opamp can provide a great deal of current the more expensive the ampli er the higher the current it has a low output impedance The voltage follower The output of the voltage follower may be predicted as follows m in V1 I V0ut gainV V v 11 1 vow gaitwy 1106 0 v5 vol v v 106 0vg Mg 2 1000000 w z w m 1000001 m m An ideal opamp From this relatively simple example we can extract the most important features of an operational ampli er The gain is very high It draws no input current It delivers unlimited output current These are characteristics of an ideal operational ampli er If thee characteristics are met and if the feedback is properly connected to the inverting input then The opamp will do whatever is necessary to equalize the voltage at the two inputs These four rules are all that is necessary to analyze most op amp circuits A simple ampli er circuit As a further example consider the following circuit What is the output voltage Bout as a function of the input voltage Em How does this voltage depend on the values of the quotinputquot and quotfeedbackquot resistor A simple ampli er circuit First note that the output tries to equalize the two inputs so that EE Since EO ground EO A simple ampli er circuit Next calculate the current through the input resistor R1 Em E IinRi Em IinRi 1D I in 1 A simple ampli er circuit and then the current through the feedback resistor E E IfRf out ECU I R t f f Rf Eout If R Rf quot 1WW D v out A simple ampli er circuit The current into the opamp itself is easy to calculate It is zero That39s how the opamp was designed Thus the feedback current must exactly cancel the input current at the junction Where the three currents meet This is a consequence of Kirchhoffs law conservation of charge IinIf EEout R1 R hamp E R A simple ampli er circuit Thus the gain of an inverting ampli er is just the ratio of the resistances in the feedback and input parts of the circuit A simple ampli er circuit If we replace the resistors with some combination of resistors capacitors and inductors we must replace the resistance with an impedance Other than this minor change the analysis of the circuit is identical Z Z 39 I 1 39MM E E E II out Z in in IVM D A summing circuit As another example a summing circuit is constructed as follows R R E1 1 f wa WW E2 O IWW E3 crwwR O C Eout A summing circuit As another example a summing circuit is constructed as follows R R E 1 f E 1 WW If 2 E2 O IWW E3 crwwR O C Eout A summing circuit As another example a summing circuit is constructed as follows E E R1 Rf Iin 2 1 1 E Am E R1 E2 O IWW f o C Eout A summing circuit As another example a summing circuit is constructed as follows R R E E 1 f Iin 1 1 E AM EZRi E2 O IWW If 2 E3 R Rf o W 0 If Ln 0 Eout A summing circuit As another example a summing circuit is constructed as follows E E R1 Rf Iin 2 1 1 Hg AM E R1 E2 O IWW If out E3 R Rf O W 0 If Ln 0 Boat Other operations Other more complicated examples of operational ampli ers exist These include integrators differentiators and bandpass lters to name only a few Historically these operational ampli ers were developed to perform numerical operations in analog computers so an operational ampli er circuit has probably been designed to perform almost any numerical operation on a signal you might want to perform As you have seen these circuits can appear rather complicated they may all however be analyzed by applying the four rules introduced ear 1er The gain of an opamp is very high The opamp draws no input current from the signal The opamp delivers unlimited output current The opamp will do whatever is necessary to equalize the voltage at the two inputs Given two periodic functions yAsincolt zsinwzt Determine Whether 1102 First multiply the functions together yz Asinooltsinoo2 t Acosw1t 032 t cosw1t 032 t Next average over the signals for a long time If 01 and 02gt0 cos031t032t averages to zero If 01 032 cos031t032t averages to zero If 01 032 cosolt02t 1 1A if 001 2002 2 Average 0 otherwise NOW take the next most complicated case suppose yAsinoltq zsinwzt Again multiply the two signals yz Asinoa1tlsinw2t Asin 031 tcos I coswltsinc sinwzt Acosq sinwltsin 032 t A sinq coswltsinwzt 31 sin 01 tsinwzt az coswltsinwzt alcosm1 032t cosoa1 032 t 2i azsinoa1 032 t sinw1 032 t l 211 if 001 002 Average 2 O 0therw1se The other part of the amplitude of yA sinx111 can be determined by comparing with zcosw2tz yz Asinw1tcosoo2t Asinw1tcosq cos wltsino coswzt Acosq sinwltcoswz t A sinq cosoa1 tcoswzt a1 sin 01 tcos 032t az coswltcosw 2t 2 alsinoa1 032 t sinoo1 032 t 2lazcosw1 oa2t cosoa1 w2t gag if 01 02 Average 0 otherwise It takes two functions to determine the amplitude of a periodic function with arbitrary phase The sine and cosine transforms The procedure above is formally de ned by the sine and cosine transforms of a function 111 Cf Th cos2nftdt S f h t sin2 nftdt The Fourier transform Using complex notation with em Cos27tfi i sin27tfg we can wrlte 2 ft Hf jhte 7 dt H is the Fourier transform of ht It turns out that this transform can be inverted simply ht j H fe2mdt H m and My are said to be a Fourier transform pair Fourier transforms are convenient for representing periodic functions as sines and cosines decomposing periodic functions into harmonic contributions A Fourier transform represents data in a reciprocal domain time H frequency lengthH wavenumber This becomes useful if an eXperiment is easy to perform in one domain but analysis is more convenient in the other Examples Ion resonance mass spectrometer Fourier transform infrared spectrometers Fourier transform NMR spectrometers Digital signal processing for communication devices Encoding and compression of audio signals The ion cyclotron spectrometer The resonant frequency of an ion moving in a circular orbit in a magnetic eld is IBI 03 27m q for a single kind of ion thensignal will be a periodic function with frequency f q 2am For several species the signal will be a sum of several such signals A Fourier transform will provide information about the number and identity of the several kinds of ions Forms of the FT To extract all the frequency and phase information from a timedependent signal use the full Fourier transform HOP J htesztdt If you only care about the amplitude and not the phase you don39t need the complex H 9 Instead specify the power density as I 2 Po IHltrgtF IHlt rgtl this also is called the onesided power spectral density the power function or the power spectrum


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