第一次课后作业答案:选择题: 1 B 2 C 3 D 5 D3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 105页参考答案4-1:b 4-2: c 4-4: a 4-5: a3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 不同温度的S值也可以直接用饱和水表查得。计算结果是0.3363-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 6-43-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 在总压101.33kPa、温度350.8K下,苯(1)-正己烷(2)形成x1=0.525的恒沸混合物。此温度下两组分的蒸汽压分别是99.4kPa和97.27kPa,液相活度系数模型选用Margules方程,气相服从理想气体,求350.8K下的气液平衡关系3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 和3-1-|||-.Delta n+dfrac (1)(2)(a)^2+g-Delta x=q- . △z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)_(1)}(s) dfrac ({10)^4cdot (m)_(B)}(3600s)=2mgdfrac ({E)_(8)}(s) =sqrt ({120)^2-(50)^2}=100sqrt (dfrac {1)(lg )} .-|||-Delta h=-2583times (10)^6J Delta a=109sqrt (dfrac {1)({b)_(8)}}-|||-.dfrac (1)(2)(m)^2=1.65times (10)^4dfrac (1)(s) cdot Delta (I)_(2)=erasure 1.729dfrac (J)(s)-|||-.+dfrac (1)(2)-(m)^2-gcdot L=-2567times (10)^6dfrac (1)(8) =(2567-(10)^6-dfrac (1)(8))=2567-(10)^6-erasure -|||-=dfrac (2583-2567)(2567)-100% 的函数式。