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標題:

Vapor Pressure problem求解

發問:

Ammonia, NH3, has a normal boiling point of -33 °C, and its heat of vaporization is 25.1 kJ/mol. The liquid and gas heat capacities are 2.2 J/(g °C) and 4.7 J/(g °C), respectively. How much heat must be supplied to 236 g of liquid ammonia to raise its temperature from -60.0 °C to 5.0 °C? (Hint:... 顯示更多 Ammonia, NH3, has a normal boiling point of -33 °C, and its heat of vaporization is 25.1 kJ/mol. The liquid and gas heat capacities are 2.2 J/(g °C) and 4.7 J/(g °C), respectively. How much heat must be supplied to 236 g of liquid ammonia to raise its temperature from -60.0 °C to 5.0 °C? (Hint: Draw heating curve T versus heat added for such problems). help me please :(

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最佳解答:

no. of mole of 236 g NH? = 236/(14+1*3) = 13.88 mole Heat required from -60?C to -33?C : = 2.2*236*[(-33)-(-60)] = 14018.4 J Heat of vaporization : = 25.1*1000*13.88 = 348388 J Heat required from -33?C to -5?C : = 4.7*236*[(-5)-(-33)] = 31058 J Total amount of heat required = 14018.4 J + 348388 J + 31058 J = 393 kJ

其他解答:

氨，NH 3，有一個正常的沸點為-33℃，其汽化熱25.1千焦/摩爾。液體和氣體的熱容量為2.2 J / J /（G°C）和4.7（G°C）。多少熱量必須提供液氨236克，以提高其溫度-60.0°C至5.0°C？ （提示：繪製加熱曲線T與熱添加等問題）。5AEEC82B53E1B405