dU=δQ+SW R 8Q- pdv 0 igure 3.6 Reversible and irreversible processes. (A) The system reaches the state X from the standard state O through a path I involving irreversible processses. It is assumed that the same transformation can be achieved through a reversible transformation r. ( B) An example of an irreversible process is the sponta- neous expansion of a gas into vacuum. The same change can be achieved reversibly through an iso- thermal expansion of a gas that occurs infinitely Slowly so that the heat absorbed from the reservoir equals the f=constant work done on the piston. In a reversible isothermal 1)MR写 expansion the change in entropy can be calculated using ds=de/t
dU=Q+W = Q - pdV
Clausius deduction (1865 FIGURE 20-5 The Carnot cycle Heat In a cyo the cycle for the Carnot engine begins at point a on this Pl 几L diagram (1) The gas is first panded isothermally, with addition of heat @), along path ab at temperature TH. ( 2)Ne d→ the gas expands adiabatically Adiabatic Adiabatic from b to c-no heat is exchanged, compression lexpansion but the temperature drops to TL (3)The gas is then compressed at int temperature Ti, path c to d d 0=0 and heat e. flows out. (4)Finally he gas is compressed adiabatically path da, back to its original state. No Carnot engine actually exists, but as a theoretical engine it played c→d Isothermal Important role in the development of thermodynamics compression e= W QH 1-Q/QH=1-TLTH
e= W/ QH = 1-QL /QH =1-TL /TH Clausius’ deduction (1865)
doren 0 T 二 FIGURE 20-11 Any reversible cycle can be approximated as a series of Carnot cycles (The dashed lines represent isotherms
= 0 T dQrev
do ds T FIGURE 20-12 The integral, ds, of the entropy for a reversible cycle is zero. Hence the difference in entropy between states a and b ∫adS, is the same for path I as for path II
T dQ dS =
do SB- SA T FIGURE 20-12 The integral, ds, of the entropy for a reversible cycle is zero. Hence the difference in entropy between states a and b ∫adS, is the same for path I as for path II
− B A B A T dQ S S