First Law of Thermodynamics Introduction First Law of Thermodynamics Calculation of Work PVT diagrams Thermodynamic processes Simple Examples The “everything” problem More examples

Perspective Governs theoretical energy & statistical limits, not engineering details (we’re not “Click and Clack” www.cartalk.com) If fuel releases 100 kJ energy cannot get more than 100 kJ mechanical work (First Law) cannot get even 100 kJ mechanical work (Second Law) (some heat must flow from hot to cold) If heat pump does 100 kW mechanical work Can obtain at least 100 KW heat (First Law) Can obtain more than 100 kW heat (Second Law) (can combine work with heat “pumped” from cold to hot) 2 Fundamental Rules Energy must be conserved - First Law ∆𝑈=𝑄−𝑊 Must be statistically probable - Second Law ∆𝑆= 𝑄 𝑇 ≥0

First Law of Thermodynamics First Law Thermodynamics ∆𝑈=𝑄−𝑊 U internal energy, ΔU change in internal energy. Q is non-mechanical heat, in or out. W is mechanical work, in or out. Don’t let screwy sign convention spook you! Q is positive in, negative out W is negative in, positive out. Negative sign in first law “flips” that Negative work in increases U Positive work out decreases U Convention used so expansion does positive work Example 15-1 Heat added to system 2500 J, work done on system -1800 J ∆𝑈=𝑄−𝑊=2500 𝐽 − −1800 𝐽 =4300 𝐽 +Q -W +W -Q U

Work by expanding gas Work done on piston for constant pressure 𝑊𝑜𝑟𝑘=𝐹 ∙𝑑𝑥 (Physics 103) 𝑊𝑜𝑟𝑘= 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ∙𝐴𝑟𝑒𝑎 ∙𝑑𝑥 𝑊𝑜𝑟𝑘=𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ∙ 𝐴𝑟𝑒𝑎∙𝑑𝑥 𝑊𝑜𝑟𝑘=𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ∙𝑑𝑉 In general 𝑊𝑜𝑟𝑘= 𝑃 𝑑𝑉

PVT diagrams – Viewing P,V,T Plot pressure vs. volume curves at constant temperature. Different PV curves for different temperatures. T is a “parameter”. Constant temperature curves called “isotherms” Isotherms also lines of constant internal energy. 𝑈= 3 2 𝑛𝑅𝑇. If you return to same isotherm, you return to same temperature and internal energy. ∆𝑈= 3 2 𝑛𝑅∆𝑇=0

PVT diagrams – Viewing Work For constant pressure 𝑊𝑜𝑟𝑘=𝑃∆𝑉 Positive moving to right (volume increasing) Negative moving to left (volume decreasing) Also area under curve For non-constant pressure Work= 𝑃𝑑𝑉 Area under curve Zero (volume remaining constant)

PVT diagrams – Filling in the Heat Constant pressure expansion ∆𝑈=𝑄−𝑊 Work is positive Temperature, Internal Energy change positive. Heat must be doubly positive Constant temperature expansion Temperature, Internal Energy change zero Heat in equals work out Q is fudge factor that makes up difference.

Thermodynamic Processes I 1. Isovolumetric pressure increase Pressure increase at constant volume 𝑊𝑜𝑟𝑘=𝑃∆𝑉= 𝑧𝑒𝑟𝑜 Higher isotherm, Higher temperature and internal energy Heat in ∆𝑈=𝑄−𝑊 2. Isobaric expansion Volume increase as constant pressure. 𝑊𝑜𝑟𝑘=𝑃∆𝑉= 𝑜𝑢𝑡 Heat doubly in ∆𝑈=𝑸−𝑊

Thermodynamic Processes II 3. Isothermal expansion Volume increase as constant temperature. 𝑊𝑜𝑟𝑘=∫ P dV= out Same isotherm, constant temperature and internal energy Heat in equals work out ∆𝑈=𝑄−𝑊 4. Adiabatic expansion Volume increase with no heat exchange. Drop to lower temperature and internal energy to match work out. ∆𝑈=𝑄−𝑊

4 thermodynamic paths A Isovolumetric D Isobaric C Isothermal Volume constant. Pressure drop. Lower temperature and U. No work. D Isobaric Volume increase. No Pressure change. Higher temperature and U. Work out. C Isothermal Pressure decrease. No temperature U and change. B Adiabatic Temperature and U decrease.

Example 15-5 (1) 𝑊=0 (no volume change) ∆𝑈=0 (no temperature change) Work done in path BD 𝑊=(2.02 ∙ 10 5 𝑁 𝑚 2 )(−8∙ 10 −3 𝑚 3 ) =−1600 𝐽 (in) Work done in path DA 𝑊=0 (no volume change) Internal Energy change BDA ∆𝑈=0 (no temperature change) Heat Flow BDA 𝑄=𝑊=−1600 𝐽 (out)

Example 15-5 (2) ∆𝑈=0 (no temperature change) Work done in path AB 𝑊= 2𝐿 10𝐿 𝑃 𝑑𝑉 = 2𝐿 10𝐿 𝑛𝑟𝑇 𝑑𝑉 𝑉 =𝑛𝑅𝑇 2𝐿 10𝐿 𝑑𝑉 𝑉 =𝑛𝑅𝑇 ln⁡ 10 𝐿 2 𝐿 = 𝑃 𝐵 𝑉 𝐵 ln⁡ 5 =3250 𝐽 (work out) Internal Energy change BA ∆𝑈=0 (no temperature change) Heat Flow BA 𝑄=𝑊=3250 𝐽 (heat in) More work out during AB than in during BDA Heat in during AB, Heat out during BDA

Example 15-6 Internal Energy change First Law for Q= 0 ∆𝑈=𝑄−𝑊=−𝑊 ∆𝑈= 𝑈 𝑓 − 𝑈 𝑖 = 3 2 𝑛𝑅( 𝑇 𝑓 − 𝑇 𝑖 ) = 3 2 0.25 𝑚𝑜𝑙 8.314 𝐽 𝑚𝑜𝑙−𝐾 400 𝐾−1150𝐾 =−2300 𝐽 First Law for Q= 0 ∆𝑈=𝑄−𝑊=−𝑊 𝑊=+2300 𝐽

The “everything” problem One mole of an ideal gas undergoes the series of processes shown in the figure. (a) Calculate the temperature at points A, B, C and D. (b) For each process A  B, B  C, C  D, and D  A, calculate the work done by the process. (c) Calculate the heat exchanged with the gas during each of the four processes. (d) What is the net work done during the entire process? (e) What is the net heat added during the entire process? (f) What is the thermodynamic efficiency?

“Everything” problem – calculate temperatures (K) & Internal Energies 𝑇 𝑎 = 𝑃𝑉 𝑛𝑅 = (2× 10 5 𝑃𝑎) (0.01 𝑚 3 ) (1 𝑚𝑜𝑙)(8.314 𝐽 𝑚𝑜𝑙 𝐾) =240. 55𝐾 𝑈 𝑎 = 3 2 𝑛𝑅 𝑇 𝑎 = 3 2 1 𝑚𝑜𝑙 8.314 𝐽 𝑚𝑜𝑙 𝐾 240.6 𝐾 =3000.0 𝐽 𝑇 𝑏 = 𝑃𝑉 𝑛𝑅 = (2× 10 5 𝑃𝑎) (0.03 𝑚 3 ) (1 𝑚𝑜𝑙)(8.314 𝐽 𝑚𝑜𝑙 𝐾) =721.67 𝐾 𝑈 𝑏 = 3 2 𝑛𝑅 𝑇 𝑏 = 3 2 1 𝑚𝑜𝑙 8.314 𝐽 𝑚𝑜𝑙 𝐾 721.7 𝐾 =9000. 0𝐽 𝑇 𝑐 = 𝑃𝑉 𝑛𝑅 = (1× 10 5 𝑃𝑎) (0.03 𝑚 3 ) (1 𝑚𝑜𝑙)(8.314 𝐽 𝑚𝑜𝑙 𝐾) =360.83 𝐾 𝑈 𝑐 = 3 2 𝑛𝑅 𝑇 𝑐 = 3 2 1 𝑚𝑜𝑙 8.314 𝐽 𝑚𝑜𝑙 𝐾 360.8 𝐾 =4500.0 𝐽 𝑇 𝑑 = 𝑃𝑉 𝑛𝑅 = (1× 10 5 𝑃𝑎) (0.01 𝑚 3 ) (1 𝑚𝑜𝑙)(8.314 𝐽 𝑚𝑜𝑙 𝐾) =120.27 𝐾 𝑈 𝑑 = 3 2 𝑛𝑅 𝑇 𝑑 = 3 2 1 𝑚𝑜𝑙 8.314 𝐽 𝑚𝑜𝑙 𝐾 120.3 𝐾 =1500.0 𝐽

“Everything” problem – calculate works For paths ab and cd pressure is constant 𝑊 𝑎𝑏 =𝑃∆𝑉= (2 𝑎𝑡𝑚)( 10 5 𝑃𝑎 𝑎𝑡𝑚) 20𝐿 (.001 𝑚 3 𝐿) =+4000 𝐽 𝑊 𝑐𝑑 =𝑃∆𝑉= (1 𝑎𝑡𝑚)( 10 5 𝑃𝑎 𝑎𝑡𝑚) −20𝐿 (.001 𝑚 3 𝐿) =−2000 𝐽 For paths bc and da volume is constant 𝑊 𝑏𝑐 =𝑃∆𝑉=0 𝑊 𝑑𝑎 =𝑃∆𝑉=0

“Everything” problem – fill in 1st law table ΔU Work Q a -> b +6000 J +4000 J b -> c -4500 J 0 J c -> d -3000 J -2000 J d -> a +1500 J

“Everything” problem – fill in Q’s ΔU Work Q a -> b +6000 J +4000 J 10000 J b -> c -4500 J 0 J c -> d -3000 J -2000 J -5000 J d -> a +1500 J 1500 J entire cycle 2000 J ∆𝑈=𝑄−𝑊 𝑄=∆𝑈+𝑊 Net work during entire cycle 2000J Net heat absorbed/expelled during cycle 2000 J Change in internal energy during cycle 0 J Net Heat in equals net work out. Change in internal energy zero

“Everything” problem – getting efficiency ΔU Work Q a -> b +6000 J +4000 J 10000 J b -> c -4500 J 0 J c -> d -3000 J -2000 J -5000 J d -> a +1500 J 1500 J entire cycle 2000 J Net work during entire cycle 2000J Net heat absorbed/expelled during cycle 2000 J Change in internal energy during cycle 0 J Efficiency (common sense) 𝑒= 𝑏𝑒𝑛𝑒𝑓𝑖𝑡 𝑐𝑜𝑠𝑡 = 𝑚𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡 𝑓𝑢𝑒𝑙 𝑏𝑖𝑙𝑙 𝑖𝑛 = 𝑛𝑒𝑡 𝑤𝑜𝑟𝑘 ℎ𝑒𝑎𝑡 𝑖𝑛 = 2000 𝐽 11500 𝐽 =17.3%

“Everything” problem – Shortcut for ΔUs Work Q a -> b +6000 J +4000 J b -> c -4500 J 0 J c -> d -3000 J -2000 J d -> a +1500 J ∆𝑈 𝑎𝑏 = 3 2 𝑛𝑅∆ 𝑇 𝑎𝑏 = 3 2 𝑃∆ 𝑉 𝑎𝑏 = 3 2 𝑊𝑜𝑟𝑘 𝑎𝑏 =6000 𝐽 ∆𝑈 𝑏𝑐 = 3 2 𝑛𝑅∆ 𝑇 𝑏𝑐 = 3 2 𝑉∆ 𝑃 𝑏𝑐 = 3 2 .03 −10 5 =−4500 𝐽 ∆𝑈 𝑐𝑑 = 3 2 𝑛𝑅∆ 𝑇 𝑐𝑑 = 3 2 𝑃∆ 𝑉 𝑐𝑑 = 3 2 𝑊𝑜𝑟𝑘 𝑐𝑑 =−3000𝐽 ∆𝑈 𝑑𝑎 = 3 2 𝑛𝑅∆ 𝑇 𝑑𝑎 = 3 2 𝑉∆ 𝑃 𝑑𝑎 = 3 2 .01 +10 5 = +1500 𝐽

Animations Add heat, increase temperature/pressure at constant volume. Add heat, increase temperature/volume at constant pressure. Add and remove heat, expand and compress gas. (Otto cycle)

Problem 10 Do A and C agree on temperature? What is work done abc? No. Point C should be 10.7 L What is work done abc? What is change of internal energy abc? What is heat flow abc?

Problem 10 (cont) What is work done ac? = 6.8 𝐿 10.7 𝐿 𝑃 𝑑𝑉 = 6.8𝐿 10.7𝐿 𝑛𝑟𝑇 𝑑𝑉 𝑉 =𝑛𝑅𝑇 6.8𝐿 10.7𝐿 𝑑𝑉 𝑉 =𝑛𝑅𝑇 ln⁡ 10.7 𝐿 6.8 𝐿 = 𝑃 𝐵 𝑉 𝐵 ln⁡ 1.57 What is change of internal energy ac? What is heat flow ac?

Problem 11 Find temperature at 1 and 2 Find 1-step work from 1 to 2 Find 1-step internal energy change 1 to 2 Find 2-step internal energy change 1 to 2

Problem 12 Find ΔU along curvy path ∆𝑈=𝑄−𝑊=−63 𝐽 − −35 𝐽 =−28 𝐽 Work along abc -48 J, find heat along path abc ∆𝑈=𝑄−𝑊 𝑄=∆𝑈+𝑊=−28 𝐽+ −48 𝐽 =−76 𝐽 Pc = ½ Pb, find heat along path cda 𝑊=+24 𝐽 𝑄=∆𝑈+𝑊=+28 𝐽+24 𝐽=52 𝐽 Since 𝑈 𝑑 − 𝑈 𝑐 =5 𝐽 → 𝑈 𝑎 − 𝑈 𝑑 =23 𝐽 𝑄 𝑑𝑎 =∆ 𝑈 𝑑𝑎 +𝑊=23 𝐽

Questions?