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热力学 习题解答(部分)

17-2 , 17-3 , 17-4 , 17-5 , 17-6 , 17-10. 热力学 习题解答(部分). P. D. A. O. C. B. V. 17-2 如图所示, AB 、 DC 是绝热过程, COA 是等温过程,在此过程中放热 100 J ,图中 OAB 面积为 30 J , ODC 面积为 70 J ,那么在 BOD 这个过程中吸收的热量为多少 ?. 解: Q AB + Q BOD + Q DC + Q COA = 0 + Q BOD + 0 + Q COA = S ODC - S OAB

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热力学 习题解答(部分)

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  1. 17-2,17-3,17-4,17-5,17-6,17-10 热力学习题解答(部分)

  2. P D A O C B V 17-2如图所示,AB、DC是绝热过程,COA是等温过程,在此过程中放热 100 J,图中OAB面积为 30 J,ODC面积为 70 J,那么在BOD这个过程中吸收的热量为多少? 解:QAB + QBOD + QDC + QCOA = 0 + QBOD + 0 + QCOA = SODC- SOAB = 70 - 30 = 40 J QBOD = 40 - QCOA = 40 - (- 100) = 140 J

  3. 17-3理想气体从状态a出发经图中不同过程膨胀一微小体积,其 ac为等温线,ae 为绝热线 ,试问经历图中哪些过程 (1)会导致体系的温度升高, (2)会导致体系的温度下降, (3)体系 吸热, (4)体系放热, (5)体系热容量为负值。 解:(1)等体过程 bc P ,T   Tab ; (2) Tad,Tae,Taf  P a b c d e f V

  4. P a b c d e f V (3) 考察 acea 循环过程 等体过程 ce:P  T ; A = 0, Qce = v Cv T < 0, Qac + Qce + Qea = A > 0 Qac + Qce + 0 = A  Qac = A - Qce > 0  吸热: ab , ac , ad 过程 (4)放热: af 过程

  5. 升温过程: ab; 降温过程: ad , ae , af . 吸热过程: ab , ac , ad 放热过程: af (5) ab : 吸热 Q > 0 , 升温T > 0 , C > 0 ; ac : 吸热 Q > 0 , 等温T = 0 , C =  ; ad : 吸热 Q > 0 , 降温T < 0 , C < 0 ; ae: 绝热 Q = 0 , 降温T < 0 , C = 0 ; af : 放热 Q < 0 , 降温T < 0 , C > 0 ;

  6. P ( 105 Pa ) A 1.5 B 0.5 V(10-3 m3) 1 3 17-40.1 mol 的理想气体(假设摩尔等体热容 CV = 3 R / 2 经历如图所示 的过程,由状态 A 经一直线到达状态 B , (1)试证所示状态 A、B 温度相同; 解: PAVA = 1.5  102 (Pa m3) PBVB = 1.5  102 (Pa m3) 因 PAVA = PBVB  TA = TB

  7. P ( 105 Pa ) A 1.5 B 0.5 V(10-3 m3) 1 3 (2)在此过程中,气体吸热多少 ? Q = E + A = 0 + A = ( 0.5 + 1.5 ) ( 3 - 1 )  102 / 2 = 2  102 J

  8. P ( 105 Pa ) A 1.5 B 0.5 V(10-3 m3) 1 3 (3)在 A 到 B 的过程中,哪一点温度最高 ? AB 过程方程: ( P - PA ) / ( V - VA ) = ( PB - PA ) / ( VB - VA )  P = 2  105 - 5  107 V 状态方程: PV = v RT T = PV / v R = ( 2105 - 5107 V ) V/ vR

  9. P ( 105 Pa ) A 1.5 H B 0.5 V(10-3 m3) 2 1 3 T = ( 2  105 - 5  107 V ) V / v R 求极值: dT/dV = 0  ( 2  105 - 10  107 V ) / v R = 0  VH = 2  10-3 m3 PH = 2  105 - 5  107 Vh = 1  105 Pa TH = 240.6 K 状态 H恰好在 AB 的中点

  10. (4)在这过程中经历每一微小变化时, 气体是否总是吸热 ? ( CV = 3 R / 2 ) P = 2  105 - 5  107 V T = ( 2  105 - 5  107 V ) V / v R dT = ( 2  105 - 10  107 V ) / v R dQ = dU + P dV = v CV dT + P dV = ( 5  105 - 2  108 V ) dV dV > 0 , dQ > 0 吸热:  V < 2.5  10-3 m3 dQ < 0 放热:  V > 2.5  10-3 m3

  11. P ( 105 Pa ) A 1.5 E B 0.5 V(10-3 m3) 1 3 即存在 E状态: VE = 2.5  10-3 m3 PE = 0.75  10-5 Pa , TE = 225.6 K 在AE 过程中, 始终吸热 ; 在EB 过程中, 始终放热 .

  12. 17-5:1 mol 理想气体 ( 设 CV = 5R /2 ) 以状态 A ( P1,V1 ) 沿 P-V 图所示直线变化到状态 B ( P2,V2 ),试求: (1)气体的内能的增量; (2)气体对外界所作的功; (3)气体吸收的热量; (4)此过程的摩尔热容。 P P2 B P1 A V V1 V2 解:(1)  E= CV ( T2 -T1 ) = 5R( T2 -T1 )/2= 5( P2V2 - P1V1 )/2 (2) A = S = ( P2V2 - P1V1 )/2 (3)Q= E+A=5(P2V2 -P1V1 )/2+(P2V2 -P1V1 )/2 = 3(P2V2 -P1V1 )

  13. (4)、(3)结论对于微元状态变化也成立,所以 dQ = 3d(PV) = 3d(RT) = 3R dT C = dQ/dT = 3R

  14. 另一种解法: 状态方程:PV = RT 过程方程:P =  V   V2 = RT 摩尔热容:C = dQ/dT = dE /dT + PdV/dT = CV + P dV/dT 对关系式  V2 = RT 两边求微分: 2 VdV = RdT  dV/dT = R/2V = R/2P C = CV+ P dV/dT = 5R /2 + R/2 = 3R

  15. 17-6设一摩尔固体的状态方程可写作 v = v0 + aT + bp,内能可表示为 u = cT-apT,其中 a、 b、c 和 v0 均是常数。试求: (1) 摩尔焓的表达式; (2) 摩尔热容量 CP 和 CV。 解: h = u + pv = cT-apT + p (v0 + aT + bp) = cT + p v0 + bp2 (2) CP = ( h /  T)P = c u = cT - a ( v - v0 - aT ) T/b = [c - a ( v - v0 ) /b ]T + a2 T2 /b CV = ( u /  T)V = c - a ( v - v0 ) /b + 2 a2 T /b

  16. 17-10一物体在恒压强下被可逆地加热 (设热容保持恒定 ) 试证明这物体的熵的变化为△S= υCP ln ( T2 / T1 )。 解: △S =  dQ/T =  υCP dT/ T = υCP ln ( T2 / T1 )

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