温: 4433VpVp? 3334 3410 4.24mVp pV ???????????? 10 t =0.25 st =0.5 s x3 2? t =1 s t =0 s (2)JVpVpV VTV VTRM mQQW 333114 321 212110 1.22 ln)( ln ln????????????????(3)%30 1 1 2???T T? 6.(1)2 31 2 23 12 ln ln 32 21T TCT TCT dT CT dT CSSS Vp TT V TT p???????????由于 31TT?,RCC Vp??,所以可得 1 21 2 ln lnV VRT TRS???(2)1 21 2111 ln ln 1V VRV V RT TT QS????(3)4 14 3 43 14 ln ln 0 34T TCT TCT dT CSSS pp TT p?????????? 2311VpVp??和???? 11441)/(/ppTT ???????? 121 113)/()/(VVpp 1 22 14 1 ln ln 1 lnV VRV VCT TCS pp????????三次计算的 S?都相等,说明熵变只与始末状态有关。练习十五 1.s1 ,?3 2 ,?3 14 ,s5 2 .见右图 3. [3] 4. [2] 5.(1)m x4.02 sin 4.0 0???????????,02 cos 2 0??????????v (2 )在振动方程中,令?3 4?t 得,m x2.06 37 sin 4.0?????????又sm v/73 .16 37 cos 2?????????,2/56 37 sin 10sm a???????????(3 )由2.02 5 sin 4.0???????????tx ,02 5 cos 2??????????tv
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