3 Ch. 19 First Law of Thermodynamics 1. Thermodynamic systems 2. Work done during volume changes 3. Paths between thermodyn. states 4. Internal energy and the First Law 5. Kinds of thermodynamic processes 6. Internal energy of an ideal gas 7. Heat capacities of an ideal gas 8. Adiabatic processes for ideal gas
4 Sign conventions: a) Q is positive b) Q is negative c) W is positive d) W is negative e) In a steam engine the heat in and the work out (done) are both defined to be POSITIVE.
5 (a) Molecule hits a piston moving away from it; speed and KE of molecule decrease. (expan. -> cooling)(b) Molecule hits a piston moving toward it; speed and KE of molecule increase. (comp. -> heating)
6 WORK-ENERGY THEOREMWhen a molecule hits a wall moving away from it, the molecule does work on the wall; the molecule’s speed and kinetic energy decrease. The gas does positive work (on the piston); and the internal energy of the gas decreases.Molecule hits a wall moving toward it, the wall does work on the molecule; the molecule’s speed and kinetic energy increase. The gas does negative work (on the piston); and the internal energy of the gas increases.
7 The infinitesimal work done by the system (gas) during the small expansion dx is dW = p A dx = F dx so W = ò p dV
8 The work done equals the area under the curve on a p-V diagram The work done equals the area under the curve on a p-V diagram. (a) Volume increases: work and area are positive. (b) Volume decreases: work and area are negative. (c) A constant-pressure: Volume increases, W > 0.
9 (a) Three different paths between state 1 and state 2 (a) Three different paths between state 1 and state 2. (b)-(d) The work done by the system during a transition between two states depends on the PATH (or PROCESSES) chosen.
10 The final states are the same The final states are the same. The intermediate states (p and V) during the transition from state 1 to 2 are entirely different. Heat, like work, depends on the initial and final states AND the path between the states. (a) Heat is added slowly so as to keep the temperature constant. (b) Rapid, free expansion does no work and there is no heat transfer.HEAT ADDED
11 The First Law of Thermodynamics DU = Q - WIn a thermodynamic process, the internal energy of a system may increase, decrease, or remain the same: (a) DU > 0 (b) DU < 0 (c) DU = 0
12 Daily thermodynamic process of your body (a thermodynamic system)
13 The net work done by the system in the CYCLIC PROCESS aba is –500 J.
15 The First Law of Thermodynamics KINDS OF THERMODYNAMIC PROCESSES ADIABATIC – NO HEAT TRANSFER; Q=0; DU=-W ISOCHORIC – CONSTANT VOLUME; W=0; DU = Q ISOBARIC – CONSTANT PRESSURE; W=p (V2 – V1) ISOTHERMAL – CONSTANT TEMPERATURE; only for ideal gas DU = 0, Q = WThe First Law of ThermodynamicsDU = Q - W or Q = DU + W
16 Processes for an ideal gas: adiabatic Q=0 isochoric W=0 isothermal DU=0
17 The internal energy of an ideal gas depends only on its temperature, not on its pressure or volume. HANDOUT p1
20 Raising the temperature of an ideal gas by various processes: Q = DU + W and DU = f(T) Isochoric W = 0 so Q = DU Isobaric Q = DU + p(V2-V1)HANDOUT p2
21 A p-V diagram of an adiabatic process for an ideal gas from state a to state b. pV g = constant (adiabat)pV = constant (isotherm)
22 ADIABATIC PROCESSES for IDEAL GAS Q = 0 so DU = Q – W = 0 – p DV and DU = n Cv dT for an ideal gas DU = n Cv dT = – p DV = -(n R T/V) dV dT/T + (R/ Cv) dV/V = 0 dT/T + (g - 1) dV/V = 0 where g – 1 = (Cp/Cv ) –1 ò dT/T + (g - 1)ò dV/V = 0 ; lnT+(g-1) lnV = constln (TV g-1 ) = constant and T1V1 g-1 = T2V2 g-1And since pV = nRT, (pV/nR) V g-1 = constant’(p/nR) V g = constant’ p1V1 g = p2V2 g
23 Adiabatic compression of air in a cylinder of a diesel engine.
24 Carnot Cycle for an ideal gas pCarnot Cycle for an ideal gasHOTCOLD
25 The Carnot cycle for an ideal gas. Isotherms in light-blue. Adiabats in dark-blue.
26 Along which path is a) work done and b) heat transferred by the system greatest? Q = DU + W
27 The absolute value of heat transfer during one cycle is 7200 J. Assigned for HW
28 Given values for all p’s and V’s, calculate the work. A quantity of air is taken from state a to state b along the straight line.Is Tb > Ta?Given values for all p’s and V’s, calculate the work.
29 Q = 90 J W = 60 JQ = ? |W| = 35 JQ = ? W = 15 JQ bda = +45 J and Q ba (blue) = -65 J
30 Ub = 240 J. Uc = 680 J. Qbc = ? Qab = ? Qdc = ? Wadc = 120 J Qad = ? Wabc = 450 JQab = ?Qdc = ?Wadc = 120 JQad = ?Ud = 330 J.Ua = 150 J.