Physics 2101 Section 3 April 26th: Chap. 18 : Chap Ann n ce n e t nnt : Exam #4, April Exam #4,

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1 Physics 2101 Section 3 April 26 th : Chap Announcements: n nt Exam #4, April 28 th (Ch ) 18.8) Final Exam: May 11 th (Tuesday), 7:30 AM Make up Final: May 15 th (Saturday) 7:30 AM Class Website: 3/

2 The First Law of Thermodynamics E int = sum total of the energy of particles (molecules/atoms) in system Internal Energy or Thermal Energy E int (increases) if work done to system or heat added to system E int (decreases) if work done by system or heat taken from system Although W and Q are path dependent, E int is not. ΔE int = E int, f E int,i = Q W by de int = dq dw dw by In thermodynamics, Work is defined as done by system: ΔE int = Q + W on

3 V f Path Dependence of Work and ΔE int W by = dw = pdv isobaric: W = pδv ΔE int = E int, f E int,i = Q W V i isochoric: W = 0 de int = dq dw by Volume increases, Pressure decreases: area > 0 > W > 0 (gas expands) Two step: Volume increases then Pressure decreases: area > 0 > W > 0 Work and Heat are NOT CONSERVATIVE: depends on path ΔE int does NOT depend on path!! Figure shows four paths on p V diagram along which a gas can be taken from state ito state f. Rank: 1) ΔE int? 2) Work done by gas? 3) Heat transferred? NET WORK, W net, done by system during a complete cycleis shaded area. It can be pos., neg, or zero depending on path. Around a closed cycle ΔE int is zero.

4 Special cases of First Law of Thermodynamics ΔE it = E itf E iti = Q W int int, f int,i 1) Adiabatic processes NO TRANSFER OF ENERGY AS HEAT Q = 0 a) rapid expansion of gasses in piston no time for heat to be transferred b) if work is done by system (W>0), then ΔE int decreases c) NOTE: temperature changes!! [ ΔE int = W] adiabatic 2) Constant volume processes (isochoric) NO WORK IS DONE W = 0 a) if heat is absorbed, the internal energy increases b) NOTE: temperature changes!! W by = V f =V i pdv = 0 V i ΔE int = Q 3) Cyclical process (closed cycle) ΔE int,closed cycle =0 a) net area in p V curve is Q ΔE int = 0 Q = W 4) Free Expansion : adiabatic process with no transfer of heat a) happens suddenly b) no work done against vacuum ΔE int = Q = W = 0 c) non thermal equilibrium process 5) Isothermal: Temperature does not change We ll talk about this later

5 Sample Problem 18 5 Let 1.0 kg of liquid at 100 C be converted to steam at 100 C by boiling at twice atmospheric pressure (2 atm) as shown. The volume of the water changes from an initial value of m 3 as a liquid to m 3 as a gas. Here, energy is transferred from the thermal reservoir as heat until the liquid water is changed completely to steam. Work is done by the expanding gas as it lifts the loaded piston against a constant atmospheric pressure. a) How much work is done by the system during the process? How do we calculate work? W by = V f ( ) pdv = p V f V i V i = ( 2 atm) ( 5 N/m 2 atm)1.671 m m 3 = 338 kj ( ) b) How much energy is transferred as heat during the process? What is the heat added? no temperature change only phase change Q = ml V = ( 1.0 kg)2256 ( kj/kg) 2260 kj c) What is the change in the system s internal energy during the process? ΔE int = Q W by Using1st Lawof Thermo: = 2260 kj 338 kj 1920 kj Positive! Energy mostly (85 %) goes into separating H 2 O molecules

6 More Example 18 43: Gas within a closed chamber undergoes the cycle shown in the p V diagram. Calculate the net energy added to the system as heat (Q) during on complete cycle. In one complete cycle, ΔE int,cycle = 0 so Q=W. To find Q, calculate W! W by = W A B + W B C + W C A V B V C = p A B dv + p B C dv + p C AdV = V A V B V C ( V )dV 1 + ( 30)dV + p C A dv 4 V A 1 1 p A B (V ) = 20 3 V m Pa 3 p B C (V ) = 30 Pa ΔV C A = 0 4 m ( 3 = ( 20 1 V 2 ) ( V) ) 1 m m 3 ( 30( V) ) + 0 = 30 J 4 m 3 W = Q = 30 J Or just add up area enclosed! CCW negative CW positive

7 Chapt. 19: Kinetic TheoryofGasesof Thermodynamics = macroscopic picture Gases = micro > macro picture IDEAL GAS LAW n = number of moles N = number of particles pv= nrt pv = NkT k = J/K R = kn A R = J/(mol K) = (L atm)/(mol K) = 1.99 calories/(mol K) Monoatomic ideal gas : He, Ar, Ne, Kr (no potential energies) E = N( 3 kt )= 3 nrt The internal energy int,monotonic 2 2 ΔE int,monotonic = 3 2 ( ) 2 of an ideal gas depends d only on the temperature nr( ΔT)

8 Avogadro s Number One mole is the number of atoms in 12 g sample of carbon 12 C(12) 6 protrons, 6 neutrons and 6 electrons 12 atomic units of mass assuming m P =m n N A =6.02 x mol 1 So the number of moles n is given by Another way to do this is to know the mass of one molecule: then N=M(sample)/mN A n=n/n A

9 19-1: Gold has a molar mass of 197 g/mol. (a) How many moles of gold are in 2.50g sample of pure gold? (b) How many atoms are in the sample? Au has 79 protons and ~118 neutrons. n is number of moles (19-3) n= M sample l M sample M (molar mass) m(one( atom) N A M sample 2.5g n= = = mol. M (molar mass) 197g / mol. ( b) Number of atoms Eqn N n= or N=n N A N A ( )( ) N = x10 = 7.64x

10 19.2: Find the mass in kilograms of 7.5 x atoms of arsenic, which has a molar mass of 74.9 g/mole. As has 33 protons and ~42 neutrons Each atom has a mass m= M N A Where M is the molar mass The molar mass of As is 74.9 g/mol. ( ) 3 Total M= 7.5x x10 3 kg / mol 6.02x10 24 mol 1 M=0.933 kg ( ) ( )

11 19-3: The best laboratory vacuum has a pressure of about 1.00 x Pa or atm. How many ygas molecules are there per cubic centimeter in such a vacuum at 293 K? Get units straight V=1.0x10 6 m 3 p = Pa T=293 K Ideal gas Law pv = nrt R gas constant R=8.31 J/mol. K ( 13 )( 10 Pa 10 6 m 3 ) pv = = = RT 8.31 J / mol. K 293K 23 n x10 mole ( )( ) ( 23 )( 23 1) N = nn = 4.1x10 mole 6.02x10 mole = 25 molecules A