理想气体pvt过程计算.ppt

1、00-7-10,1,理想气体恒温,例 物质的量为 n 的理想气体由始态 p1, V1 , T 恒温变化压力, 体积, 到达末态 p2, V2 , T, 求过程的焓变 H.,H = U + (pV) = U + (nRT),理想气体恒温的单纯 pVT 变化中, U = 0, 所以 H = U + nR (T) = 0,计算表明, 理想气体作单纯 pVT 变化时其焓不随系统的压力和体积等变化, 或者说这种情况下系统的焓只是温度的函数.,00-7-10,2,理想气体恒容升温+恒压压缩，始末温度相同,例 1mol理想气体由202.65kPa, 10dm3 恒容升温, 使压力升高到2026.5kPa,

2、再恒压压缩至体积为1dm3. 求整个过程的W, Q, U, 及 H.,因 p1V1 = p3V3, 所以T3 = T1; 又因是理想气体, 故 U = 0, H = 0.,Q =W =W2 = p2(V3 V2) = 202.65 103Pa (110) 103m3 =18.239kJ,00-7-10,3,理想气体恒外压恒温,例 3 mol单原子理想气体, 从始态T1 = 300 K, p1 = 100 kPa, 反抗恒外压50 kPa作不可逆膨胀, 至终态T2 = 300 K, p2 = 50 kPa, 求这一过程的Q, W, U, H.,U = 0, H = 0 W = pamb (V2V

3、1) = pamb (nRT/ p2nRT/ p1) = 3.741 kJ Q = W = 3.741 kJ,00-7-10,4,理想气体恒压升温,例 2 mol H2 从400 K, 100 kPa恒压加热到1000 K, 已知Cp,m(H2) = 29.2 Jmol1K1, 求U, H, Q, W各为多少?,Qp = = 2 mol29.2 Jmol1K1(1000400)K = 35.04 kJ U = H (pV) = HnR(T2T1) = 25.06 kJ W = UQ =9.98 kJ,00-7-10,5,理想气体恒压升温，已知定容热容,例 3 mol某理想气体, 在101.3

4、kPa下恒压由20加热到80, 计算此过程的Q, W, U, H. 已知该气体 CV, m = (31.3813.4103 T / K)Jmol1K1.,U = QW = (7.921.50)kJ = 6.42 kJ,W = pamb(V2V1)= nR(T2T1) = 38.314(8020)J = 1497 J = 1.50 kJ,00-7-10,6,理想气体恒容，已知定压热容,例 3 mol某理想气体由409 K, 0.15 MPa经恒容变化到 p2 = 0.10 MPa, 求过程的Q, W, U及H. 该气体Cp, m = 29.4 Jmol1K1.,T2 = p2T1 / p1 =

5、(0.10409 / 0.15)K = 273 K QV = U = n CV, m T = 3 mol(29.48.31)Jmol1K1(409273)K = 8.635 kJ W = 0,H = U + (pV) = UnRT = 8635 J3 mol8.314 Jmol1K1(409273)K = 12.040 kJ,00-7-10,7,理想气体绝热压缩,例 某压缩机气缸吸入101.325kPa, 25的空气, 经压缩后压力提高至192.5kPa, 相应使温度上升到79. 已知该温度范围内空气的CV,m近似为25.29Jmol1 K1, 试求1mol空气压缩过程的W, Q 及系统的U.

6、,W = U = 1366J,U = U1 = nCV,m(T2T1) = 1 25.29(352.15298.15)J = 1366J,因是非恒容过程, Q不是恒容热, 故Q U, 也不能用CV,m直接求非恒容过程的热.,(空气视为理想气体),00-7-10,8,理想气体恒温+恒容升温,例 1mol理想气体由27, 101.325kPa受某恒定外压恒温压缩到平衡, 再恒容升温至97, 则压力升至1013.25kPa. 求整个过程的W, Q, U, 及H. 已知该气体的CV,m恒定为20.92Jmol1 K1.,W = W1 = pamb(V2V1) = p2(V2V1) = (p2V2 p3

7、 T2 V1 /T3) = nRT2 (1p3 T2 /p1T3 ) =1 8.314 300.1511013.23 300.15/(101.325 370.15 )J = 17.74kJ,pamb = p2 = p3 T2 /T3,U = nCV,m(T2T1) = 1 20.92(9727)J = 1.464kJ,H = nCp,m(T2T1) = 1 (20.92 + 8.314)(9727)J = 2.046kJ,Q = UW = (1.4617.74)kJ =16.276 kJ,00-7-10,9,理想气体3种条件下的升温,(1) dV = 0, W = 0 QV = U = nCV, m (T2T1) = (1020283)J = 56.6 kJ H = nCp, m(T2T1) = 10(208.314)283J =80.129 kJ,(2) Q = 0, U2 = U1 = 56.6 kJ, H2 = H1 = 80.129 kJ W = U2 = 56.6 kJ,(3) dp = 0, U3 = U1 = 56.6 kJ Qp = H3 = H1 = 80.129 kJ,例 某理想气体, 其CV