Methanol is stored at the temperature of 60 degree C. The vapor pressure of meth

Question

Methanol is stored at the temperature of 60 degree C. The vapor pressure of methanol is 13 .2 kPa at 20 degree C and 347 kPa at 100 degree C. Determine the vapor pressure of methanol at 60 degree C. Liquid methanol at 25 degree C is heated and vaporized for use in a chemical reaction. How much heat is required to heat and vaporize 10 mol/s of methanol to 600 degree C. At 25 degree C, liquid (Reference): C_p, liq = 75.86 + 0.1683 T C_p, vap = 42.93 + 0.083 T Heat of vaporization of methanol = 35 .27 kJ/mol Normal boiling point of methanol = 64.7 degree C

Explanation / Answer

2)

From Clausius-Clapeyron equation

ln(P1/P2) = (Hvap/R)((1/T2) - (1/T1))

T1=20 0c = 283 K                          P1 = 13.2 kpa

T2=100 0c = 373 K                        P2 = 347 kPa

ln(13.2 / 347) = (Hvap/8.314)((1/373) - (1/283))

-3.269 = (Hvap/8.314) * (-8.52*10-4)

3.269 = Hvap * 1.025 *10-4

Hvap = 31877 J/mol = 31.877 kJ/mol

Vapour pressure at 60 0c

ln(P1/P2) = (Hvap/R)((1/T2) - (1/T1))

T1=20 0c = 283 K                          P1 = 13.2 kpa

T2=60 0c = 333 K                          P2 = ?

ln(13.2 /P2) = (31877/8.314)((1/333) - (1/283))

ln(13.2 /P2) = -2.034

(13.2 /P2) = exp(-2.034) = 0.1308

P2 = 100.90 kPa

3)

Q = (nCpT)liq + Heat of vapourisation + (nCpT) vap

Q = (10*(75.86+(0.1683*25)*(64.7-25)) +(35.27) + (10*(42.93+(0.083*600)*(600-64.7))

Q = 30283.45+ 35270+ 496383=561937.14 j/sec

Q= 561937.14 watt

Note: unit of cp is not specified .Assume the unit of Cp is j/mol K

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