※ 引述《ppppp367 (小p)》之銘言： : 1.A gas at 250k and 15 bar has a molar volume 12 per cent : larger than that calculated from the perfect : gas law calculate (a) the compression factor under these conditions and : (b) the molar volume of the gas 15bar = 15/1.01325 atm = 14.8 atm n = 1 PV = nRT (for ideal gas) 14.8 * V = 1 * 0.082 * 250 ==> V = 1.385 L real gas => 1.385 * 1.12 = 1.55 L (for part b) PV = jnRT 14.8 * 1.55 = j * 1 * 0.082 * 250 ===> j = 1.12 (for part a) : 2.I2 has a heat capacity Cp of 80j/mol k and a boiling point at 458k : When 1 mol of I2 358k to 458k in the liquid state at 1atm, calculate the : (a)三角形h (b)三角形s H S (a) (b) set Cp independent of T (liquid state) (a) △H = n Cp dT = 1 * 80 * ( 458 - 358 ) = 8000 J (b) △S = (Cp / T) dT = 80 ln (458/358) = ... : an electric current of 2.0A from a 12V supply is passed : for 1 hour through a resistance the liquid vaporized completely at 458k : and 1 atm calculate(c) Q(d) W (e)三角形U (f)三角形H (g)三角形S Q由電提供 我不會XD Q得到後 Qp = △H 再帶入 △G = 0 = △H - T△S 得S .....不太會XD : the vapor at 458K and 1 atm was then heated to 558K and 2 atm : calculate (h)三角形U (i)三角形H (j)三角形S : (for the I2 vapor Cp=30+0.005T J/(mol K) : 麻煩大大幫忙 這兩題真想很久還是卡關 : 大大解出來 一定P幣奉上 感謝 (h) (i) (j) 458k 1atm --> 558k 2atm 這邊就不是很確定了.. △H = n Cp dT = (30+0.005T)dT 由458積分到558 △U = n Cv dT = n (Cp - R) dT 同上 △S = (30+0.005T)/T dT 458積分到558 ...........以上沒考慮壓力的影響 不確定可不可以 很久沒碰了 不知道有沒有錯 有錯請指正 --

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