则 2:0=p:p0→z′=p=-133×10- ×542×103=071s-1 1013×105 7-14(1)如图 又∵B→>C等温过程,故TB=TC 由p=nRTP=2PH=B 题714图 则TB=2T KT 2nd- p a: a-t t B→>C等温过程PcVc=PB1B→Pc×2A=2pHA→PC=P4 c:λ4=2:1 715(1) VM =1.738.31×4000 =7.0×10°(ms-) (2)d=4+2=1 =-(1+3)×10 2×10-0m (3)z=√2m2n2v=√2z×4x1020×40×1035×7×103 5×10 716(1)69 则 8 1 5 4 0 0 0 0 5.42 10 0.71 1.013 10 1.33 10 : : − − = = = = z s p p z z p p z 7-14 (1)如图 M RT v 2 3 = ∴ c A Tc TA v : v : 2 2 = 又 B →C 等温过程,故 TB = TC . 由 RT PB PA VA VB M m pV = = 2 = 则 TB = 2TA ∴ : 2 :1 2 2 Vc VA = (2) A A c c c A P T P T p KT : : 2 d 2 = = B →C 等温过程 pCVC = pBVB pC 2VA = 2pAVA pC = pA C : A = 2 :1 7-15 (1) M RT v 1.73 2 = 7.0 10 (ms ) 2 10 8.31 4000 1.73 3 1 3 − − = = (2) (1 3) 10 2 10 m 2 1 2 2 1 2 −10 −10 = + = + = d d d (3) 20 25 3 2 2 = 2 d = 2 410 4010 710 − z n v 10 1 5 10 s − = 7-16 (1) 题 7-14 图