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Essential Physics II 英語で物理学の エッセンス II Lecture 4: 19-10-15 News Last week’s homework: 2015 / 10 / 26 (10月26日) This week’s homework: Also due: 2015 / 10 / 26 (10月26日) on http://masteringphysics.com Last lectures... Temperature difference, T internal energy increases, U Eheat moving energy: heat Eheat Ideal gases: Ideal gas law: pV = nRT # mols T measures K for random motion of molecules: R = 8.314 J/K · mol 3 1 ¯2 kT = mv 2 2 Last lectures... Quiz An ideal gas has volume,V and pressure, p. The thermal speed of the gas molecules is v. If both volume and pressure are doubled to 2V and 2p, what is the thermal speed of the gas molecules? (a) v (b) 2v (c) 4v (d) v/2 (e) v/4 Last lectures... Quiz An ideal gas has volume,V and pressure, p. The thermal speed of the gas molecules is v. If both volume and pressure are doubled to 2V and 2p, what is the thermal speed of the gas molecules? p1 V 1 T1 = Ideal gas law: pV = nRT (a) v nR p2 V 2 4p1 V1 = 4T1 T2 = = (b) 2v nR nR r (c) 4v 3 1 3kT Since: kT = mv¯2 v̄ = m 2 2 (d) v/2 r r 3k(4T1 ) 3kT2 = 2 v ¯ = v ¯ = 1 2 (e) v/4 m m Heat and Work Joule’s apparatus Quiz You stir water (vigorously) with a spoon. The water’s temperature, T ... (A) increases (B) decreases (C) unchanged No temperature difference between spoon and water: No heat is transferred. Spoon does work on the water. Mechanical work increases water’s internal energy T =0 Joule’s apparatus Joule’s apparatus Quiz What is the temperature increase, T ? (Assume all energy is converted to heat) coil = 1800 J/kg · K 2 kg (A) 10 C 1m (B) 1 C (C) 0.1 C (D) 0.01 C 1 litre ~ 1 kg (E) 0 C Joule’s apparatus Quiz What is the temperature increase, T ? (Assume all energy is converted to heat) coil = 1800 J/kg · K 2 kg Potential energy: U = mg h = (2 kg)(9.81 m/s2 )(1 m) 1m = 19.62 J Heat energy: Q = moil c T 1 litre ~ 1 kg 19.62 J T = ' 0.01 K (1 kg)(1800 J/kg · K) = 0.01 C (max) Joule’s apparatus Temperature can be increased by... Heating: Temperature difference between flame and water T Heat energy Q = mc T internal energy higher T Doing work: Water is stirred Mechanical energy W internal energy higher T Same final state: T is higher First Law of Thermodynamics Change in internal energy U =Q+W heat transferred work done on the system 1st law of thermodynamics: The change in internal energy of a system depends only on the net heat transferred to the system and the net work done on the system, independent of the particular processes involved. How the energy moves is not important First Law of Thermodynamics U =Q+W Rate of energy flow dU dQ dW = + dt dt dt Rate of change of internal energy Rate of heat transfer Rate at which work is done on the system First Law of Thermodynamics A power plant supplies energy at a rate of 3.0 GW. steam U 3 GW Q Ex. 1 GW W drives turbine for electricity Used steam is cooled in the water If the power plant produces electrical energy at a rate of 1.0 GW, what is the rate of heat transfer to the water? dU dQ dW = + dt dt dt dQ dU = dt dt = dW = dt 2.0 GW 3.0 GW ( 1.0 GW) extracting energy doing work First Law of Thermodynamics Quiz A heat source supplies heat to a gas at a rate of 187.0 W The gas does work at a rate of 130.9 W What rate does the internal energy (dU/dt) of the gas change? (a) 56.1 W dU dQ dW = + dt dt dt (b) 318 W = 187 W (c) -56.1 W (d) 187 W = 56.1W 130.9W First Law of Thermodynamics Quiz A gas expands at constant T to twice (x2) its original volume. During the expansion, the gas absorbs 200 kJ of heat. What is the change in internal energy of the gas during the expansion? What measures internal energy (U) ? (a) 0 kJ (b) 100 kJ (c) 200 kJ (d) 400 kJ Temperature (T). Does T change? No. Therefore does U change? No. First Law of Thermodynamics Quiz A gas expands at constant T to twice (x2) its original volume. During the expansion, the gas absorbs 200 kJ of heat. How much work does the gas do during the expansion? U =Q+W (a) 0 kJ 0 = 200 kJ + W (b) 100 kJ (c) 200 kJ work done on the gas: W = (d) 400 kJ work done by the gas: 200 kJ 200 kJ Reversible & Irreversible U =Q+W First law applies to any system but... Ideal gas: pV = nRT Only need 2 variables [ ( p , V ) or ( V , T ) or ( T , p ) ] can use pV diagram to show state: p p2 , V2 , T2 p1 , V1 , T1 V Reversible Process Gas T T System: Gas ball immersed (covered by) water in equilibrium Water If water T increases slowly: Gas and water change T together p Always in equilibrium p2 , V2 , T2 p1 , V1 , T1 V Change follows curve quasi-static process If water T decreases slowly: Change follows same curve reversible process p p2 , V2 , T2 p1 , V1 , T1 V Irreversible Process Gas T T System: Gas ball immersed (covered by) water in equilibrium Water If water T increases quickly: e.g. pour boiling water over cold gas ball Gas and water NOT in equilibrium Different p and T in different regions p p2 , V2 , T2 not well-defined values irreversible process p1 , V1 , T1 V Work Ideal gas: pV = nRT A Insulated cylinder (no heat loss, Eout = 0 ) Volume can change Pressure, p Cross-section area, A A Reversible process How much work is done on the gas? p Work A Force from gas: Fgas = pA A Gas does work: x p Wgas = Fgas x = pA x p =p V Newton’s 3rd Law Work done on the gas: W = Wgas = But... p maybe changes as V changes: dW = Z Z V2 Total work: W = dW = pdV = - (area under pv curve ) p V pdV small volume change: p ~ constant p V1 (+ if gas is compressed V2 < V1 , - if expands V1 < V2 ) V Work Quiz 2 gas cylinders start and end in the same state. (pi , Vi ) ! (pf , Vf ) p They move between the 2 states by different processes. What is the same for both cylinders? (a) V work done on or by the gas (b) heat added or removed (c) change in internal energy quantities that do not depend on the HOW you reach a state (path taken) = state variables depends only on current p,V Work W = Z V2 pdV V1 What happens when 1 of is constant (no change) ? work done on gas during volume change p (pressure) V (volume) T (temperature) Q (heat) Isothermal ( T = constant ) T = constant p 1 p/ V pV = nRT T1 T2 T3 System changes along isotherm: curve of constant T: T3 > T2 > T1 V p Work done on gas = -(area) W = = Z V2 pdV = V1 nRT Z V2 V1 Z Z dV V V2 V1 nRT dV V 1 dx = ln x x = nRT ln V ln x2 |VV21 ln x1 = ln = ✓ x2 x1 ◆ nRT ln V1 ✓ V2 V1 ◆ V2 V Isothermal ( T = constant ) Internal energy, U from from molecules’ kinetic energy (motion) Gas T Therefore, constant T = constant U 1st law of thermodynamics: since: W = nRT ln Q= ✓ V2 V1 U =0=Q+W ◆ W = nRT ln ✓ V2 V1 ◆ Q= isothermal process W Isothermal How much work does the bubble do as it rises to the surface? T = 300 K 8 mm p = 350 kPa Ex. 25 m Assume T = constant = 300 K patm = 101 kPa Isothermal process: W = nRT ln ?? ✓ V2 V1 ?? ◆ Isothermal Ex. How much work does the bubble do as it rises to the surface? T = 300 K 8 mm p = 350 kPa 25 m Assume T = constant = 300 K patm = 101 kPa Isothermal process: W = nRT ln 4 3 ideal gas: pV = nRT = ⇡r p 3 4 3 ⇡r 3 V2 V1 ◆ p1 V1 = constant = p2 V2 constant V2 p1 = 3.5 = V1 p2 p1 V1 = constant = p2 V2 4 3 W = ⇡r p ln 3.5 = 3 ✓ 0.94 J p1 V 1 = Isothermal Quiz An ideal gas expands isothermally at 300 K. Its volume increased from 0.020 m3 to 0.040 m3 . The final pressure is 120 kPa. (A) 3.3 kJ (B) 1.7 kJ (C) 3.3 kJ (D) 1.7 kJ (E) 0.0 kJ The heat transfer to the gas....? R = 8.314 J/mol · K Isothermal Quiz An ideal gas expands isothermally at 300 K. Its volume increased from 0.020 m3 to 0.040 m3 . The final pressure is 120 kPa. The heat transfer to the gas....? pV = nRT (A) constant 3.3 kJ (B) 1.7 kJ (C) 3.3 kJ (120 ⇥ 103 Pa)(0.040 m3 ) = nRT ✓ ◆ V2 W = nRT ln V1 ✓ 3 (D) (E) 1.7 kJ = (120 ⇥ 10 Pa)(0.040 m ) ln = 3.3 kJ 0.0 kJ 3 Q= W = 3.3 kJ 0.040 0.02 ◆ Constant volume,V constant volume: isometric isochoric isovolume volume does not change 1st law of thermodynamics: W =0 U =Q Introduce molar specific heat at constant volume, CV : Q = nCV # moles (constant-volume) T C per unit mass (last lecture) CV per mol U = nCV T (any process) U only depends on T: U = constant T Isobaric (p = constant) p System changes along isobar: curve of constant P Work done on gas = -(area) W = p(V2 V1 ) = p V V1 1st law of thermodynamics: Q = Since: U = nCV T U W = Q = nCV U +p V T +p V Introduce molar specific heat at constant pressure, CP : Q = nCP T nCp T = nCV T +p V isobaric process V2 V Isobaric (p = constant) How are CV and Cp related? Ideal gas law: pV = nRT p V = nR T Therefore: nCp T = nCV T +p V So: Cp = CV + R nCp T = nCV molar specific heat For solids and liquids, expansion is small Therefore, work is small: Cp ⇠ CV T + nR T Adiabatic (Q = 0) Q No heat flow : Q = 0 Q Q Q Quickly occurring processes ~ adiabatic (finished before heat transfer occurs) Q e.g. combustion engine 1st law of thermodynamics: U =W adiabatic process Adiabatic (Q = 0) Work / internal energy W done by gas lowers gas U If V increases, gas does W U decreases T decreases Since Q = 0 Since pV = nRT p decreases Isothermal changes only V and p p Adiabat : steeper than isotherms V1 V2 V pV = constant = Cp /CV Adiabatic (Q = 0) pV = constant Rewrite this equation in terms of T : (a) V =T (b) T V = constant (c) T V (d) T 1 1 = constant V = constant Quiz Adiabatic (Q = 0) pV = constant Rewrite this equation in terms of T : Ideal gas: pV = nRT constant nRT V V TV = constant 1 = constant nRT p= V Quiz Adiabatic (Q = 0) Ex. Diesel Engine Compressed gas raises T ! 500 C and ignites (starts burning) fuel Vb What is compression ratio, ? Va Initial T = 20 C and 1 TV Vb 1 Va 1 = constant Ta = Tb 1 = 1.4 1 Tb Vb = Ta Va before after ✓ ◆1/ 1 ✓ ◆1/0.4 Vb Ta 500 + 273 K = = 11 = Va Tb 20 + 273 K Adiabatic (Q = 0) Since W = Z V2 pdV and pV = constant = C V1 What is the work done by an adiabatic gas? Z V2 Z V2 C C W = CV dV dV = p= V V1 V V1 +1 V2 CV pV = C = + 1 V1 W = p2 V 2 p1 V 1 1 Ideal gas processes Cyclic Processes Systems that return periodically to the same state (p, V, T ) e.g. refrigerator liquid looses heat outside liquid removes heat from inside p A System moves between ‘A’ and ‘B’ B V Cyclic Processes Cyclic processes can involve: Isothermal constant-volume Isobaric Adiabatic processes p A B Net (total) work done on the gas V Ex. Cyclic Processes An ideal gas has = 1.4 , V = 4.0 L, T = 300 K and p = 100 kPa. 1. Compressed adiabatically to 0.25 V A!B 2. Cooled at constant-volume back to 300K 3. Expands isothermally to V B!C C!A How much work is done on the gas? Adiabat A ! B p B pV =C pB V B pB = pA (VA /VB ) C A (696.4⇥103 )(1⇥10 W = = 741 J V W = pB V B ??? = pA V A pA V A 1 = 696.4 kPa 3 ) (100⇥103 )(4⇥10 1.4 1 3 ) Ex. Cyclic Processes An ideal gas has = 1.4 , V = 4.0 L, T = 300 K and p = 100 kPa. 1. Compressed adiabatically to 0.25 V A!B 2. Cooled at constant-volume back to 300K 3. Expands isothermally to V B!C C!A How much work is done on the gas? p Constant-volume B ! C B volume does not change C A V W =0 Ex. Cyclic Processes An ideal gas has = 1.4 , V = 4.0 L, T = 300 K and p = 100 kPa. 1. Compressed adiabatically to 0.25 V A!B 2. Cooled at constant-volume back to 300K 3. Expands isothermally to V B!C C!A How much work is done on the gas? p B C Isotherm: C ! A W = nRT ln constant ⇣ VA VC Ideal gas: pV = nRT 3 = pA VA = (100 ⇥ 10 )(4 ⇥ 10 = 400 J A W = (400 J)(ln 4) = 555 J V ⌘ 3 ) Ex. Cyclic Processes An ideal gas has = 1.4 , V = 4.0 L, T = 300 K and p = 100 kPa. 1. Compressed adiabatically to 0.25 V A!B 2. Cooled at constant-volume back to 300K 3. Expands isothermally to V B!C C!A How much work is done on the gas? p WABCA = WAB + WBC + WCA B = 741 J + 0 J C = 186 J A V 555 J Quiz Cyclic Processes p What is the work done? 5.0 atm isothermal 1 L = 0.001 m3 1 atm = 101, 325 Pa 1.0 atm 10 L 50 L V (a) 8120 J (c) (b) 0J (d) 4050 J 4101 J Quiz Cyclic Processes p What is the work done on the ideal gas? 5.0 atm isothermal constant V 1.0 atm 1 atm = 101, 325 Pa isobaric 10 L isobaric: W = = 50 L p(V2 V V1 ) (101, 325 Pa)( 40 ⇥ 10 = 4053 J constant V: 1 L = 0.001 m3 W =0 3 m3 ) Quiz Cyclic Processes p What is the work done on the ideal gas? 5.0 atm constant V 1.0 atm isothermal isobaric 1 L = 0.001 m3 1 atm = 101, 325 Pa V 10 L 50 L ✓ ◆ V2 isothermal: W = nRT ln V1 constant pV = (5.0 ⇥ 101, 325 Pa)(10 ⇥ 10 3 )✓ = nRT ◆ 50 3 W = (5.0 ⇥ 101, 325 Pa)(10 ⇥ 10 ) ln 10 = 8153.8 J Quiz Cyclic Processes p What is the work done on the ideal gas? 5.0 atm constant V 1.0 atm isothermal isobaric 10 L Total work: 50 L = 4053 + 0 = 4100.8 J 1 L = 0.001 m3 1 atm = 101, 325 Pa V 8153.8 J Specific heats: Last lecture: & CV Cp 1 ¯2 3 mv = kT 2 2 average K of molecule (for 1 molecule) gas temperature Total internal energy of n mols: U = nNA ✓ 1 ¯2 mv 2 3 = nNA kT 2 3 = nRT 2 From U = nCV T 1 CV = n 3 U = R 2 T ◆ R Specific heats: and: Cp = CV Therefore: = CV 5 3R 3 2R 5 = = 1.67 3 H2 Ne N2 Ar O2 Great! But.... Cp CV + R = CV He 5 = 3 & SO2 N O2 7 = 5 ' 1.3 Specific heats: 1 ¯2 3 mv = kT 2 2 CV & Cp to find this... Assumed: Gas molecules have no internal structure nothing inside He OK for monatomic molecules: Ne 1 molecule = 1 atom Ar y Can only move in 3 directions: = 3 degrees of freedom z x Specific heats: CV & Cp Diatomic molecules can move in 5 directions x y translational motion 5 degrees of freedom 1 molecule = 2 atoms z ✓ rotational motion Equipartition Theorem In thermodynamic equilibrium, 1 Average energy / molecule = kT for each degree of freedom 2 Equipartition theorem e.g. monatomic molecule, 3 degrees of freedom: ✓ ◆ 1 3 U =3 kT = kT 2 2 since all energy is kinetic: 1 ¯2 3 U = K = mv = kT 2 2 Equipartition Theorem e.g. diatomic molecule, 5 degrees of freedom: ✓ ◆ 1 5 kT = kT Av. energy / molecule: U = 5 2 2 Total internal energy of n mols: 1 Therefore: CV = n (as before) and: 5 U = nNA kT 2 5 = nRT 2 U 5 = R T 2 Cp CV + R 7 = = = = 1.4 CV CV 5 Equipartition Theorem e.g. polyatomic molecule, 6 degrees of freedom: 3 translational (x, y, z) 3 rotational U = 3nRT CV = 3R 4 = ' 1.33 3 Equipartition Theorem At very high T, diatomic molecules can also vibrate: Adds kinetic energy (K) ... and potential energy + 2 degrees of freedom Total: 7 degrees of freedom (only at high T) Equipartition Theorem Ex. A gas mixture has 2.0 mol of oxygen ( O2 ) and 1.0 mol of argon (Ar) Find the volume specific heat, CV O2 : Ar : 5 degrees of freedom 5 5 U = nRT = 2.0RT = 5RT 2 2 3 degrees of freedom 3 3 3 U = nRT = 1.0RT = RT 2 2 2 3 13 Total internal energy: U = 5RT + RT = RT 2 2 U 13 = R T 2 1 13 1 U = R = 2.2R CV = 2.0 + 1.0 2 n T