Heat and Work First Law of Thermodynamics 9.1 Heat Heat and Work Thermodynamic Processes Work Thermodynamics is the science of heat and work Heat is a form of energy Calorimetry Mechanical equivalent of heat Mechanical work done on a system produces a rise 1calorie =4.184 J in temperature like heat added to the system. Work First Law of Thermodynamics The change in the internal energy of a system U, is equal to the heat Heat input into a system can input Q minus the work done by the system. produce work ∆=∆WFx ∆UQW=− ∆=W PAxPV ∆=∆ Heat Q Internal For the case of Energy thermal expansion U Heat input Work work is done by Q W the system. System The internal energy is the energy stored in the system. For an ideal gas the internal energy is the kinetic energy of the gas 1 Thermodynamic processes in an PV diagram ideal gas T1 T2 U U1 2 P • State 2 State 1 (P1, V1, T1) State 2 (P2, V2, T2 ) • State 1 • The state of the ideal gas is determined by the three parameters, PVT. V • A thermodynamic process is the transition between states with input or output of heat and work with changes in internal energy. Each point in the diagram represents a State of the system • The internal energy U is a property of the state. ∆U determined by The temperature T is not plotted but is not an independent variable. the initial and final state and is independent of path T can be calculated from P and V using the ideal gas law. • The heat absorbed and work done in the process depend on the The points of constant temperature on the PV diagram are hyperbola path. nRT P = For constant T V Work can be calculated if the Thermodynamic processes process is reversible. x x State 2 (T , P , V ) State 2 (T , P , V ) Path A 2 2 2 Path A 2 2 2 P P Path B x x State 1 (T1, P1, V1) State 1 (T1, P1, V1) V V The change in U going from 1 to 2 is the same for the We can calculate the work done in going from 1 to 2 if two paths. ∆UA =∆UB the change is a reversible process. i.e. it goes slowly The change in Q and W going from 1 to 2 is different through well defined states that are in quasi-equilibrium. between the two paths. QA ≠ QB , WA ≠ WB Work cannot be calculated for Reversible process an irreversible process. x State 2 (T , P , V ) Path A 2 2 2 T is increased slowly from 0 to 100o C P ? x State 1 (T1, P1, V1) V We cannot calculate the work done in going from 1 to 2 if The temperature is changed very slowly so that the gas sample is at a uniform temperature. the process is irreversible. i.e. goes rapidly through states that are not well characterized and not in quasi-equilibrium. However, we can determine ∆U, since this is a state property. 2 Irreversible process Work done in a process V2 dW = PdV WPdv= ∫ V1 The gas is not in a well defined thermodynamic The work is the area under the state as it is heated. PV curve. Some thermodynamic processes Question expansion of ideal gas T constant Find the work done in going from A to B along the 1. Constant Temperature process- heat line. Isothermal Constant T bath P constant 2. Constant Pressure process – 2P1 B Isobaric. e.g heating a balloon T changes at constant P P1 A 3. Constant Volume process. Heating V constant a gas thermometer T changes at constant V 4. Adiabatic process – no heat 2V1 Q=0 V1 exchange. Rapid compression or expansion of a gas T changes No contact with V with heat bath Thermodynamic processes Isothermal volume change T is constant U is constant T T2 1 Increasing T ∆U0= All the heat goes W=Q into work Constant T • Constant P Q =W Constant V Work VV22nRT Adiabatic W== PdV dV • P ∫∫V (Q=0) VV11 nRT P = V W= nRTlnV21− nRTlnV V work is positive Expansion 2 > 1 V V V work is done by the gas WnRTln= 2 1 V2 work is negative V1 Compression < 1 V1 work is done on the gas 3 Constant Volume Constant Pressure At constant volume W=0 Constant Pressure P2 , T2 ∆U = Q Work V P Q 2 no work is done all the heat goes WPdV= ∫ = P(V21−=∆ V ) P V into internal energy V P1 , T1 1 We can use this to find ∆U Heat input V QUPV= ∆+∆ QncT=∆V QncTPVncTnRT= ∆+∆= ∆+ ∆ ∆=UncT ∆ This is true for all processes since U is a State VV V function and only dependent on T. QncT= P∆ specific heat at constant volume Specific heat at constant pressure, c ∆U P the specific heat of the gas cV depends cV = nT∆ on the structure of the gas molecule. ccR= + the specific heat at constant P is larger Units JmoleK/ ⋅ PV than the specific heat at constant V. because work is done by the gas Adiabatic Process Adiabatic compression Heat input Q=0 The air (γ=1.40) in an automobile engine is compressed quickly so that appreciable heat exchange does not occur. For a compression ratio Work is done at the ∆U= -W V /V =10. Find the temperature of the gas compressed from an initial expense of internal energy 1 2 temperature of 20o C. Q=0 Define “Adiabatic exponent” γ ccR+ R γ ==PV =+1 ccVV c V γdepends on the nature of the gas, γ =1.40 for air. γ γ γ PV relationship PV11= PV 2 2 PV= constan t γ −−11γ TV relationship TV= TV γ −1 11 2 2 TV= constan t PV− PV W = 11 2 2 γ −1 Adiabatic compression Santa Anna Conditions T= 20o C The air (γ=1.40) in an automobile engine is compressed quickly so that appreciable heat exchange does not occur. For a compression ratio P= 85 kPa V1/V2 =10. Find the work done in compressing a gas from 1atm if the capacity is 2.8 l. P = 100 kPa T=? High Desert. Coast Santa Ana winds originate in the high desert region of California and is heated by adiabatic compress during the rapid descent to the coast. For the above conditions what would the temperature at the coast be? 4 Thermodynamic processes Work done in expansion One mole of air (γ=1.40) at P1 = 0.25 atm, T1 = 300 K, 3 V1=0.1m expands by a factor of 2. Find the work done if the expansion is a) isobaric,b) isothermal c) adiabatic. Which process does the most work? T1 P Defining T constant V constant P constant No Heat input Characteristic ∆U=0 W=0 Q=0 First Law Q=W ∆U=Q ∆U=Q-W ∆U=-W T2 V1 V2 Work done by nRTln(V2/V1) W=0 P(V2-V1) PV11− PV 2 2 V the gas γ −1 γ Other PV=constant Q=mCV∆T Q=ncP∆T PV =constant relations γ-1 cP=cP +R TV =constant Temperature change in expansion Heat input on expansion 3 One mole of air (γ=1.40) at P1 = 0.25 atm, T1 = 300 K, One mole of air (γ=1.40) at P1 = 0.25 atm, T1 = 300 K, V1=0.1m 3 expands by a factor of 2. Find the temperature change if the V1=0.1m expands by a factor of 2. Find the heat input if expansion is a) isobaric, b) isothermal c) adiabatic. Which process the expansion is a) isobaric, b) isothermal c) adiabatic. produces the highest temperature rise? What process results in the largest heat input? cV (air) =5/2R T1 T1 P P T2 T2 V1 V2 V1 V2 V V 5.