Chemistry 163B Winter Lectures 2-3. Heat and Work

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1 Chemistry 163B Winter 2014 Lectures 2-3 Heat and Work

2 Chemistry 163B reserve books 2014 (S&E Library) handout #7

3 heat capacity (E&R section 2.5) -d q C heat capacity 1 J K dt the amount of heat requires to raise substance 1K -d q nc molar heat capacity 1 1 J mol K dt the amount of heat requires to raise 1 mol substance 1K C generally depends on T and conditions for example ideal monatomic gas (independent of T) but 3 add heat at constant volume C R 2 5 add heat at constant pressure CP R 2 3

4 transfers of energy: HEAT and work (sec 2.3) - change of energy by heat transfer d q CdT ncdt ( C is extensive, q d q ncdt C path - path is intensive) C will generally depend on T and path q > 0 energy (heat) gained by system (endothermic) q < 0 energy (heat) lost by system (exothermic) 4

5 heat only transfer (also zeroth law; E&R p7) initial T 1, n 1,C 1 T 2, n 2,C 2 vacuum T f - - dq n C dt dq n C dt assume C ' s independent of T f f f q n C dt n C T T q n C dt n C T T T T 1 2 n C T T q 1q 20 T f T1 nc nc T final T f, n 1,C 1 T f, n 2,C 2 5

6 transfers of energy: heat and WORK (sec. 2.2) change of energy by work done ON system - - dw dwp dw dwp Pext d wp d w Pext d path path other w > 0 energy gained by system (work done ON system) w< 0 energy lost by system (work done ON surroundings) 6

7 processes: definitions of constraints isolated q=0; w=0 isothermal T=0 adiabatic q=0 against constant pressure P ext = const reversible process P int = P ext a (ideal) process that proceeds so slowly that an infinitesimal change of conditions causes the process to proceed in the opposite (reverse) direction irreversible process all other (real) processes proceeding at finite rate 7

8 derivation of w = - PD (work of expansion or P work) w* surr = work done ON SURROUNDINGS pressure=force/area P=F/A ; F= P A D= A h w* surr = Force Distance w* surr = F h w* surr = P A h w* surr = P D w = work ON SYTEM w= - PD to be consistent with work done ON system 8

9 ideal gas and energy, heat, work for IDEAL GAS U(E) depends ONLY on T isothermal, ΔT=0 ΔU=0=q+w q=-w adiabatic q=0, ΔU=w monatomic ideal gas U = (3/2) n RT C = (3/2) n R C P = (5/2) n R 9

10 Pressure-olume work (gene s lame lifting weights analogy to work done against system is gas inside piston; weights are surrounding atmospheric pressure) P=10 atm isothermal DT=0 P ext =1 atm expansion Δ heat expansion against constant pressure (irreversible) system is piston and gas inside; weights represent external pressure of surroundings P ext = (during volume change) ; Δ sys > 0; w sys = - P Δ sys < 0 work (<0) is done BY SYSTEM ON SURROUNDINGS (1 atm weight lifted) later (E for ideal gas depends only on T) isothermal ideal gas DT=0 ïdu=0 w< 0; - w= q > 0 ; heat absorbed by system 10

11 Pressure-olume work (gene s lame lifting weights analogy to work done against system is gas inside piston; weights are surrounding atmospheric pressure) P=10 atm adiabatic q=0 P ext =1 atm adiabatic expansion Δ brr! adiabatic expansion against constant pressure system is piston and gas inside; weights represent external pressure of surroundings P ext = (during volume change); Δ sys > 0; w sys = - P ext Δ sys < 0 work (<0) is done BY SYSTEM ON SURROUNDINGS (1 atm weight lifted) later (conservation of energy U) w< 0; q=0 (adiabatic); U sys < 0; energy (potential) of surroundings increases ï energy of system decreases ï gas cools 11

12 Pressure-olume work reversible expansion P=10 atm isothermal ΔT=0 reversible P ext = 10 atm Ø expansion initial heat initial final w on surr rev > w on surr irrev w= - P ext d P ext = P int = nrt/ w= - nrt/ d isothermal T=const w = - nrt ln ( final / initial ) 12

13 w other (E & R p. 20) change of energy by work done ON system - - dw dwp dw dw Pext d dw - - w Pext d d w - other - other other Done lifting a weight against gravity (weight is surroundings) Mass (m), position (h) h 2 wm g dh h 1 kg m 2 s -2 = J 13

14 Isothermal expansion: P ext = const ideal gas (irreversible) P=10 atm isothermal DT=0 P ext =1 atm expansion Δ 1mole 300K 10 atm 1 1mole 300K 1 atm 2 14

15 Isothermal expansion: P ext = const; ideal gas; Graphical Interpretation 10 w system P d w Area under P curve system final - initial ext w Area under P curve surr ext ext P ext (atm) all Δ against P ext = 1 atm (L) 15

16 Isothermal expansion: P ext = const ideal gas (irreversible) P=10 atm isothermal DT=0 P ext =1 atm expansion Δ 1mole 300K 10 atm 1 w P d P=nRT final initial ext w -1 atm mole 300K 1 atm 300K R (1mol) 300K R (1mol) atm 10 atm 1 1 w= -300 K-mol1 atm R 2 1 atm 10 atm w= J= kj (- sign implies net work done ON surrounding) 16

17 Pressure-olume work reversible isothermal expansion; P ext =P int P ext = 10 atm Ø P=10 atm isothermal expansion final 1mole 300K 10 atm 1 1mole 300K 1 atm 2 17

18 isothermal irreversible vs isothermal reversible: which does more work on surroundings?? P ext =1 atm P=10 atm irrev 9 atm 1at m Δ irrev w sys = kj w surr =+2244 kj 1at m 9at m rev 9 atm P ext = 10 atm Ø 1at m rev w surr? > > < irrev w surr final 18

19 Isothermal expansion: P ext = P int ideal gas; Graphical Interpretation 10 P ext nrt P ext (atm) 1 w irrev at const P w system final - initial P d ext w Area under P curve system ext w Area under P curve surr ext 1 2 (L) 19

20 Pressure-olume work reversible isothermal expansion; P ext =P int P=10 atm isothermal ΔT=0 P ext = 10 atm Ø expansion 1mole 300K 10 atm 1 final nrt w - Pext d P ext = P int = initial final 2 nrt 1 w - d = - nrt d= - nrt ln initial 1 final 300K R (1mol) 300K R (1mol) atm 10 atm 10 atm w= K-mol R ln 1 atm w= J= kj (more work done ON surroundings by reversible than irreversible; w = kj) irrev 2 1 1mole 300K 1 atm 2 20

4K n=7 n=7 n=4 n=4 heating of He, constant w = 0; q > 0 1.

21 molecular picture of heat and work: constant volume heating (E&R p 23-24) He gas in 5 nm box ; 0.2 K He gas in 5 nm box ; 0.4K n=7 n=7 n=4 n=4 heating of He, constant w = 0; q > 0 1. energy levels same spacing: D=0, w=0 (no change in size of box) 2. greater number of atoms in higher energy levels: q> 0 raises U; 3. DU>0 21

energy levels further apart for smaller box : D <0, w>0, raises U (note E scale 10-24 J vs

22 molecular picture of heat and work: isothermal compression (E&R p 23-24) He gas in 10 nm box, 0.2K He gas in 5 nm box, 0.2K n=8 n=8 n=4 n=4 He, 0.2 K isothermal compression w > 0; q <0 1. energy levels further apart for smaller box : D <0, w>0, raises U (note E scale J vs J) 2. greater number of atoms in relatively lower energy levels: q<0, lowers U; 3. DU=0 22

23 lectures next Wednesday-Friday [3X] (Monday 20 th Jan HOLIDAY; exam Friday, 31 st Jan) better make it a triple 23

Problem: Calculate the entropy change that results from mixing 54.0 g of water at 280 K with 27.0 g of water at 360 K in a vessel whose walls are

Problem: Calculate the entropy change that results from mixing 54.0 g of water at 280 K with 27.0 g of water at 360 K in a vessel whose walls are perfectly insulated from the surroundings. Is this a spontaneous