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Vapour mixture calculation
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Methanol (CH3OH) - ethanol (CH3CH2OH) solutions are almost ideal. If the vapour pressures of methanol and ethanol at 335K are (8.1 x 10*4) and (4.5 x 10*4)Nm -2 respectively, calculate the composition (by mole fraction) of the vapour over a mixture of 64g of methanol and 46 g of ethanol at... 顯示更多 Methanol (CH3OH) – ethanol (CH3CH2OH) solutions are almost ideal. If the vapour pressures of methanol and ethanol at 335K are (8.1 x 10*4) and (4.5 x 10*4)Nm -2 respectively, calculate the composition (by mole fraction) of the vapour over a mixture of 64g of methanol and 46 g of ethanol at 335K (Relative atomic masses: H:=1.0, C=12, O=6)
最佳解答:
PM: Partial pressure of methanol (CH3OH) in vapour mixture XM: Mole fraction of methanol (CH3OH) in liquid mixture PMo = vapour pressure of methanol (CH3OH) = 8.1 * 104 N m-2 PE: Partial pressure of ethanol (CH3CH2OH) in vapour mixture XE: Mole fraction of ethanol (CH3OH) in liquid mixture PEo = vapour pressure of ethanol (CH3CH2OH) = 4.5 * 104 N m-2 Molar mass of CH3OH = 12 + 1*4 + 16 = 32 g mol-1 Molar mass of CH3CH2OH = 12*2 + 1*6 + 16 = 46 g mol-1 No. of moles of CH3OH used = 64/32 = 2 mol No. of moles of CH3CH2OH = 46/46 = 1 mol In vapour: PM = XM*PMo = (2/3) * (8.1 * 104) = 5.4 * 104 N m-2 PE = XE*PEo = (1/3) * (4.5 * 104) = 1.5 * 104 N m-2 Total pressure P = 5.4 * 104 + 1.5 * 104 = 6.9 * 104 N m-2 Mole fraction of CH3OH in vapour = (5.4 * 104)/( 6.9 * 104) = 18/23 (or 0.783) Mole fraction of CH3CH2OH in vapour = (1.5 * 104)/( 6.9 * 104) = 5/23 (or 0.217) =
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