Heat is transferred in or out of system, but temperature may NOT change: Change of phase. sublimation

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3 Heat is transferred in or out of system, but temperature may NOT change: Change of phase sublimation Heat of Fusion Heat of Vaporization Q = ± L m ΔT = 0 L F Heat of fusion Solid to liquid (heat is adsorbed : atomic bonds are broken L V Heat of Vaporization Liquid to gas (heat is adsorbed) L S Heat of Sublimation Solid to gas (heat is adsorbed) H 2 O L F = 79.5 cal/g = 6.01 kj/mol = 333 kj/kg L V = 539 cal/g = 40.7 kj/mol = 2256 kj/kg Amount of heat transferred during phase change depends on L and mass (M)

4 H 2 O c ice = 2000 J/kg C L F = 33.5x10 4 J/kg c water = 4186 J/kg C L V = 22.6x10 5 J/kg Melt or Not? 1 kg Will all the ice melt? At a party, a 0.5 kg chunk of ice at - 10 C is placed in a 3.0 kg of tea at 20 C. At what temperature and in what phase will the final mixture be? 1) How much heat must be removed from tea(water) to reach 0 C? Q = c WATER m TEA ΔT = (4186 J )(3kg)( 20 C) = 2.5 kg C 105 J 2) How much heat must added to raise temperature of ice to 0 C? Q = c ICE m ICE ΔT = (2000 J )(0.5kg)(10 C) kg C =104 J 3) How much heat must added to melt ice? Q = L F m ICE = ( J )(0.5kg) =1.675 Less than kg 105 J 4) What now? Ice warms to 0 C and melts at 0 C taking Q =10 4 J J = J 5) What is final temperature? The melted ice warms and the tea cools down; the final temperature is Q cool tea =Q warm ice +Q melt ice +Q warm water c WATER m TEA (20 C T) = c WATER m ICE (T 0 C) J 60 3T = 0.5T 0 C (kg C) T = 5.0 C

5 V f W by = dw = pdv V i isobaric: W = pδv isochoric: W = 0 Volume increases, Pressure decreases: area > 0 -> W by > 0 (gas expands) Two step: Volume increases then Pressure decreases: area > 0 -> W by > 0 Here W by > 0, but less than before Work is NOT CONSERVATIVE: depends on path Volume decreases, Pressure increases: W by < 0 Work done by system is NEGATIVE (gas is compressed) NET WORK, W net, done by system (gas) during a complete cycle is shaded area. It can be pos., neg, or zero depending on path

6 18 10: The First Law of Thermodynamics ΔE int = total energy of particles (molecules/atoms) in system Internal Energy or Thermal Energy ΔE int (increases) if work done to system or heat added to system ΔE int (decreases) if work done by system or heat taken from system Although W and Q are path-dependent, ΔE int is not. ΔE int = E int, f E int,i = Q W by = Q + W on de int = dq dw by = dq + dw on ΔE int = E int, f E int,i = Q W by -- The 1 st Law of Thermodynamics

7 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 ( ) ( )( m m 3 ) = ( 2 atm) N/m 2 atm = 338 kj b) How much energy is transferred as heat during the process? What is the heat added? c) What is the change in the system s internal energy during the process? Using 1st Law of Thermo: no temperature change only phase change Q = ml V ( )( 2256 kj/kg) 2260 kj = 1.0 kg ΔE int = Q W by = 2260 kj 338 kj 1920 kj Positive! Energy mostly (85 %) goes into separating H 2 O molecules

8 Work and ΔE int V f W by = dw = pdv V i Work and Heat are NOT CONSERVATIVE-depends on path ΔE int does NOT depend on path!! ΔE int = E int, f E int,i = Q W by = Q + W on Figure shows four paths on p-v diagram along which a gas can be taken from state i to state f. Rank: 1) ΔE int? ccw loop 2) Work done by gas? 3) Heat transferred? NET WORK, W net, done by system (gas) during a complete cycle is shaded area. It can be pos., neg, or zero depending on path. Around a closed cycle ΔE int is zero.

9 Problem#18 45 A gas sample expands from V 0 to 4.0V 0 while its pressure decreases from p 0 to p 0 /4. If V 0 =1.0m 3 and p 0 =40.0 Pa, how much work is done by the gas if its pressure changes with volume via (a) path A, (b) path B, and (c) path C? Path A W=pΔV No work done constant V Path B W= pdv p = a + bv Path C: Constant pressure W=pΔV

10 Checkpoint For one complete cycle as shown in the p-v diagram, Which ones are positive? Negative? Or Zero? ΔE int? Net work done by gas W? Net energy transferred Q?

11 Special cases of First Law of Thermodynamics ΔE int = E int, f E int,i = Q W by = Q + W on 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 by >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 by 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

12 Or just add up area enclosed! Problem 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 by. 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 Q = W = - 30 J

13 Problem A thermodynamic system is taken from state A to state B to state C, and then back to A, as shown in the figure. (a) (g) Complete the table on the right by inserting a plus sign, a minus sign, or a 0 in each indicated cell. (h) What is the net work done by the system as it moved once through the cycle ABCA? A B : W>0 Since ΔE int > 0 ΔE int = Q W (a) Q A B > 0 C A : B C : Q>0 W B C = pδv = 0 ΔE int = Q W = Q ( B C) > 0 ΔE int W C A = pdv < 0 ΔE int (loop) = 0 = ΔE int (A B) + ΔE int (B C) + ΔE int (C A) So ΔE int (C A) < 0 C A : find Q ΔE int = Q W ΔE int (loop) = 0 = Q A B Q c A W A B W C A Q c A = W A B + W C A Q A B Q c A < 0

14 Chapter 19: The KineHc Theory of Gases Thermodynamics = macroscopic picture Gases micro -> macro picture Avogadro s Number N A =6.02 x mol -1 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 So the number of moles n is given by n=n/n A Another way to do this is to know the mass of one molecule: then N = M(sample) m N A

15 Problem#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 M (molar mass) M sample m(one atom) N A n= M sample M (molar mass) = 2.5g 197g / mol. = mol. (b) Number of atoms Eqn n= N N A or N=n N A n = ( ) ( 6.02x10 23 ) = 7.64x10 21

16 Problem#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. ( )( 74.9x10 3 kg / mol) 6.02x ( mol 1 ) Total M= 7.5x1024 M=0.933 kg