第二章新答案

第二章新答案

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1 , 5 1. 理想气体等温可逆膨胀,体积从 V 1 膨胀到 10V 1 ,对外做功 41.85kJ ,体系的起始压力 202.65kPa ,试求: 1 )求 V 1 2 )若气体的量为 2mol ,那么体系的温度为多少? (1) 2 11 1 33 1 1 1 23 1 ln 10 41.85*10 202.65*10 * *ln 8.97*10 (m ) V Wp V V V V V V = = = (2) 32 202.65*10 *8.97*10 1093(K) R2 * 8 . 3 1 4 pV T n == = 2. 计算 1mol 理想气体在下列 4 个过程所做的体积功。已知始态体积为 50L ,终态体积 100L ;始态和终态温度为 373K 。试求: 1 )等温可逆膨胀。 2 )向真空膨胀。 3 )向外压恒定为气体终态的压力下膨胀。 4 )先在外压恒定为体积等于 75L 时气体的平衡压力下膨胀,当膨胀到 75L (此时温 度仍为 100 )以后,再在外压等于 100L 时的气体的平衡压力下膨胀。 (1) 2 1 100 R ln 1*8.314*373*ln 2150(J) 50 V WnT V = (2) 0(J) d V (3) 2 3 2 3 2 R 1*8.314*373 31011(Pa) 100*10 31011*(100 50)*10 1550(J) nT p V WpV = = + (4) 2 3 2 3 12 3 22 R 1*8.314*373 ' 41348(Pa) ' 75*10 ' 41348*(75 50)*10 1030(J) 31011*(100 75)*10 780(J) 1030 780 1810(J) p V V V WWW = = = =+= + = + + 3. 根据第 212 页所给数据 , 计算反应 CaCO 3 (s) = CaO(s) + CO 2 (g) (1) 298.15K 时的标准摩尔熵变 , (2) 400K 时的标准摩尔熵变 . [ 已知恒压热容 C p,m ( 单位 J.mol -1 .K -1 ) 分别为 ,CaCO 3 (s) 81.88, CaO(s) 42.80, CO 2 (g) 37 .13] Δ r S θ m = Δ r S θ m (CO 2 )+ Δ r S θ m (CaO)+ Δ r S θ m (CaCO 3 ) =213.64+39.7-91.4=160.44J/mol K (1) (2) Δ r S θ m (400K)= Δ r S θ m (298.15K)+ ∫Δ rCp.m/TdT =160.44+ =160.44+(-1.95)ln400/298.15 =159.87 J/mol K 400 298.15 (42.80+37.13-81.88)/TdT 2 , 5 4. 273K 、压力为 5×101 325kPa N 2 2L ,在外压 101 325kPa 下等温膨胀,直到 N 2 的压力也等于 101 325kPa 时为止。求反应的
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第二章新答案

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