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Physics 231 Lecture 33 • • Main M i points i off today’s d ’ lecture: l Work in thermodynamic processes: Wsystem = PΔV • First Law of Thermodynamics: ΔU = ΔQ − PΔV • • Specific heat at constant pressure Processes – cyclic: ΔP = ΔV = ΔT = 0 – isobaric: ΔP = 0 – isovolumetric: ΔV = 0 – isothermal: ΔT = 0 – adiabatic: ΔQ = 0 Conceptual question • Which Whi h gives i th the llargestt average radiation di ti absorbed b b d on a 1 m2 area at that distance? – 1. a 50-W source at a distance R. – 2. a 100-W source at a distance 2R. – 3. a 200-W source at a distance 4R. Pabsorbed b b d = fraction of area covered eemitted power Pabsorbed 1m 2 = eP 2 4π r 1m 2 P1 = e50 2 4π R 1m 2 e100 P2 = 2 4π (2R) 1m 2 P2 = e200 2 4π (4R) Work and the 1st Law of thermodynamics • • • Consider the ideal gas contained in the volume under the cylinder at right. The piston compresses the gas very slowly, moving downwards with a constant velocity. The gas exerts a force Fggas=PA upward on the piston and conversely, the piston exerts a force Fpiston=-PA downward on the gas. The work done by the piston on the gas is: W piston = Fpiston Δy = PA Δy = − PΔV • The work done by the gas on the piston is: Wgas = − Fgas Δy = − PA Δy = PΔV • If U denotes the internal energy of the system, the conservation of energy gy dictates: ΔU = ΔE th = ΔQ + Wpiston = ΔQ − Wgas = ΔQ − PΔV 1st Law of Thermodynamics Here I have changed notation to U, because other systems, besides the ideal gas, can have potential energy. Why does work increase thermal energy • • Recall R ll that th t the th thermal th l energy U is i given i by b U=3/2Nk U 3/2NkBT If work done on the gas increases U it must increase T. • Positive work done on the gas increases U because the collisions with the piston moving downward increases the velocity of the molecules. Negative work done on the gas decreases U because the collisions with the piston moving upward decreases the velocity of the molecules. • Example • The workk done Th d to t compress one mole l off a monatomic t i ideal id l gas is i 6200 J. The temperature of the gas changes from 350 K to 550 K. How much heat flows between the gas and its surroundings? Determine whether the heat flows into or out of the gas. gas Work done by the gas is : Wgas = −6200J 3 Internal energy change in the gas is: ΔU = nR ( ΔT ) 2 3 3 Q = ΔU + Wgas = nR ( ΔT ) − 6200J = (8.31J / K )( 550 − 350K ) − 6200J 2 2 Q = −3700J Computation of work: • • • Whatt is Wh i the th workk done d b the by th gas going i ffrom (Pi ,V Vi) tto (Pf ,V Vf) ? – a) Pi(Vi-Vf) – b) Pi(Vf-Vi) – c) 0 – d) Vi(Pi-Pf) – e) Vf(Pi-P Pf) The work done by the gas is the mathematical area under the curve Pi(Vf-Vi). It is negative if the volume is decreasing. It is pushing to increase the volume, but the volume is being g decreased. No work is done on the isovolumetric (constant volume) part of the path! Computation of work: • • • Whatt is Wh i the th workk done d on the th gas going i ffrom (Pi ,V Vi) tto (Pf ,V Vf) ? – a) Pi(Vi-Vf) – b) Pi(Vf-Vi) – c) 0 – d) Vi(Pi-Pf) – e) Vf(Pi-P Pf) Note that the work done on the gas is (-1) times the mathematical area under the curve, -Pi(Vf -Vi). The mathematical area is negative if the volume is decreasing. The work done on the g gas is positive, p , you y are compressing p g it and thereby y exerting a force in the direction of that it is being compressed. No work is done on the isovolumetric (constant volume) part of the path! Conceptual quiz • A gas is taken from an initial state of pressure and volume, PA VA,to a final, different, state, PB VB. As it changes work is done on the gas. g – a) The amount of work depends on the path taken from A to B – b) The amount of work does not depend on the path taken from A to B quiz: • • • Whatt is Wh i the th workk done d on the th gas going i ffrom (Pi ,V Vi) tto (Pf ,V Vf) ? – a) Pf(Vi-Vf) – b) Pf(Vf-Vi) – c) 0 – d) Vi(Pi-Pf) – e) Vf(Pi-P Pf) Note that the work done on the gas is the negative of the mathematical area under the curve. The mathematical area is negative if the volume is decreasing. We are compressing the gas and the volume is decreasing and we are pushing in the direction to decrease it. No work is done on the isovolumetric part of the path! Computation of work: • Whatt is Wh i the th workk done d on the th gas going i ffrom (Pi ,V Vi) tto (Pf ,V Vf) ? – a) Pi(Vi-Vf) – b) Pf (Vi-Vf) – c) 0 – d) (Pi+ Pf)(Vi-Vf)/2 – e)) -(P ( i+ Pf)( )(Vi-Vf))/2 Heat capacity at constant pressure • Consider a cylinder y filled with a n=2 moles of monatomic gas and plugged at the top by a frictionless piston. Above the piston is atmospheric pressure. This ensures the piston will be at a constant pressure P=299kPa. The bottom of the cylinder piston is a heat source that can heat the gas. The sides of the cylinder and the piston have zero thermal conductivity. What is th heat the h t Q required i d to t raise i the th temperature t t off the th gas from initial temperature Ti =300 K to final temperature Tf =400K? Heat source Q = ΔU + W 3 ΔU = n RΔT 2 W = pΔV 3 = n R ( Tf − Ti ) 2 = p ( Vf − Vi ) = pVf − pVi = nRTf − nRTi = nR ( Tf − Ti ) 3 5 Q = n R ( Tf − Ti ) + nR ( Tf − Ti ) = n R ( Tf − Ti ) = 28.31*(400 − 300)J = 1.66kJ 2 2 3 5 c p = c v + R = R + R = R is the molar heat capacity at constant pressure 2 2 Example • A person takes in a breath of 0°C air and holds it until it warms to 37 0°C 37.0 C. The air has an initial volume of 0 0.600 600 L and a mass of 7 7.70 70 x 10–4 kg. Determine (a) the work done by the air on the lungs if the pressure remains constant at atmospheric pressure, (b) the change in internal energy gy of the air,, and (c) ( ) the energy gy added to the air by y heat. Model the air as if it were a diatomic gas: cV =5/2R., cp=7/2R. Pf = Pi = Patm = 1.01x105 Pa Wgas = Patm ( Vf − Vi ) = nR ( Tf − Ti ) nRTf Tf Patm Vf − 1 = Patm Vi − 1 = Patm Vi − 1 = Patm Vi nRTi Ti Patm Vi 310 Wgas = 1.01x105 Pa .0006m3 − 1 = 8.2J 273 Assume diatomic gas: Tf 5 5 5 Wgas = 20.5J = ΔU = nR ( Tf − Ti ) = nRTi − 1 2 2 2 Ti 7 ΔQ = ΔU + Wgas = nR ( Tf − Ti ) = 28.7J 2 ( )( )