Chapter 11, The First Law of Thermodynamics Mechanical Work dW F.d s Work in a volume change. dF = P.dA d' W PdAdS or d' W PdV An inexact differential (path dependant) Note that the integral is -ve for work done on the system, +ve for work done by the system. W Thursday, March 20, 2014 Vb Va PdV Phys235 : Thermodynamics 3 1 P = const. (isobaric) WP Ideal gas at constant T (Isothermal) W Vb Va dV P(Vb Va ) Vb Vb nRT dV nRTn Va Va V Van der Waals gas a P 2 V b nRT V P nRT a 2 V b V Vb b 1 1 W nRTn a Va b Va Vb Thursday, March 20, 2014 Phys235 : Thermodynamics 3 2 Work depends on the path W abd' W The total work around the loop is Vb Va PdV W PdV Sometimes the configuration changes without doing work. e.g free expansion of gas Thursday, March 20, 2014 Phys235 : Thermodynamics 3 3 The First Law of Thermodynamics The total work is always the same in all adiabatic processes between any two equilibrium states having the same kinetic and potential energy. d Thursday, March 20, 2014 Phys235 : Thermodynamics 3 4 Internal Energy The difference between the internal energy in state a and b is equal to the work done by the system along any adiabatic path from a to b. dU = -d’Wab Ua-Ub=Wab Thursday, March 20, 2014 Phys235 : Thermodynamics 3 5 Example: An ideal gas, and a block of copper, have equal volumes of 0.5 m3 at 300 K and atmospheric pressure. The pressure on both is increased reversibly and isothermally to 5 atm. (a) Explain with the aid of a P- V diagram why the work is not the same in the two processes. (b) In which process is the work done greater? (c) Find the work done on each if the compressibility of the copper is 0.7 x 10-6 atm-1. (d) Calculate the change in volume in each case. (a) Vgas changes significantly Vcopper almost no change (b) Wgas > Wcopper Thursday, March 20, 2014 Phys235 : Thermodynamics 3 6 (c) For gas V2 P1 nRT Wgas PdV dV nRTn P1V1n V V1 P2 5 N 1 3 Wgas 10 2 0.5m n 8.15x10 4 J m 5 1 V V P T For copper W copper V PdV P dP V PdP P T 6 1 2 0.7x10 V P2 P12 0.5 5 2 12 x1.013x10 5 2 2 0.42J Thursday, March 20, 2014 Phys235 : Thermodynamics 3 7 (d) Volume Change For gas P1V1 V2 0.1m 3 P2 Vgas = (0.5-0.1)m3=0.4 m3 For copper V =V0[1-(P2-P1)] = 0.5[1-0.7x10-6(5-1)] = 0.5[1-2.8x10-6] Thursday, March 20, 2014 ≈V0 Phys235 : Thermodynamics 3 8 Example 2: A volume of 10 m3 contains 8 kg of oxygen at a temperature of 300 K. Find the work necessary to decrease the volume to 5 m3, (a) at a constant pressure and (b) at constant temperature. (c) What is the temperature at the end of the process in (a)? (d) What is the pressure at the end of the process in (b) ? (e) Show both processes in the P- V plane. Thursday, March 20, 2014 Phys235 : Thermodynamics 3 9 8x10 3 mass n 0.25x10 3 mole molecular weight 32 nRT P V 0.25x10 3 8.314 300 10 6.23x10 4 Nm2 (a) P = const W = ∫PdV = P(V2-V1)=6.23x104(5-10)=-3.11x105 Joules (b) T = const V2 1 nRT W PdV dV nRTn nRTn 4.32x10 5 J 2 V V1 Thursday, March 20, 2014 Phys235 : Thermodynamics 3 10 (c) T2 = ? For (a) T2 T1 V2 V1 T2 T1 1 V2 300 150 K V1 2 (d) P2 = ? For (b) P1V1 P2V2 P1V1 4 2 P2 6.23x10 12.46x10 4 Nm 2 1 V2 Thursday, March 20, 2014 Phys235 : Thermodynamics 3 11 (e) 160x10 3 140 Pressure (Pa) 120 100 (b) 80 60 (a) 40 20 0 4 Thursday, March 20, 2014 6 8 3 Volume (m ) Phys235 : Thermodynamics 3 10 12 12 Heat Flow Q=Wtotal-Wadiabatic (i.e. non-adiabatic work) Q -ve : heat flow out of the system Q +ve : heat flow into the system Using these quantities we can state the 1st law of thermodynamics as ∆U=Ub-Ua=Q - W Increase in internal energy = heat flow into the system - work done by the system. dU=d’Q-pdV Note that d’Q depends on the path. dU depends on the state of the system not the path taken. Thursday, March 20, 2014 Phys235 : Thermodynamics 3 13 Mechanical Equivalent of Heat 1 Calorie = 1 g of H2O temperature increased by 1 K at 14.5 ˚C = 4.158 Joules Heat Capacity lim Q d'Q Q C , C T 0 T dT T Cp : P = const Cv : V = cont Cp and equation of state is sufficient to determine Cv Thursday, March 20, 2014 Phys235 : Thermodynamics 3 14 Heat of transformation (latent heat) : A change of phase change of volume work is involved in the process (1st order phase change). At T=const. and P= const. w = P(v2- v1) (specific work). From the 1st law (u2- u1) = - P(v2- v1) = (u2+ Pv2) - (u1+ Pv1). If we define specific enthalpy h = u + Pv then = h2 - h1. h is a state function whose value only depends on the state of the system Thursday, March 20, 2014 Phys235 : Thermodynamics 3 15 Heat flow in any reversible isobaric process is equal to the change in enthalpy. Terms in thermodynamics. 12 (solid-liquid) Heat of Fusion 23 (liquid-vapour) Heat of Vaporisation 13 (solid-vapour) Heat of sublimation Thursday, March 20, 2014 Phys235 : Thermodynamics 3 16 for H2O 23 = h'''- h''= 22.6x105 J kg-1 w = P(v'''- v'') = 1.8x105 J kg-1 (u'''- u'') = 23 - w = 20.8x105 J kg-1 So the enthalpy change is due to 92% change of internal energy and 8% work done in expanding the H2O. Thursday, March 20, 2014 Phys235 : Thermodynamics 3 17 Now dh 0 so if we have a cyclic process near the triple point in which the substances goes from solid to vapor then back to liquid and then to solid then ∆h1 + ∆h2 + ∆h3 = 0. then 13 - 23 - 12 = 0 thus 13 = 23 + 12 Thursday, March 20, 2014 Phys235 : Thermodynamics 3 18 General Form of the First Law ∆U + ∆Ek = Q - W Where ∆U : Change of internal energy ∆Ek : Change of kinetic energy and W =W* + Wc where Wc is the work done by conservative force. Wc = ∆Ep (potential energy) So ∆U + ∆Ek + ∆Ep = Q - W* ∆E = Q - W* Thursday, March 20, 2014 Phys235 : Thermodynamics 3 19

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