Question

The normal boiling point of methanol is 64.70 °C. At 21.21 °C, the vapor pressure of methanol is 0.1316 atm. What is the vapor pressure of methanol when the temperature is 53.00 °C?

Answer 1

Heat of vaporisation for methanol = 36600 J / mol

R= 8.314 JK-1mol-1 ( from data book)

Given,

vapour pressure P1 = 0.1316 atm , temperature T1= (21.21 + 273) = 294.21 K

Vapour pressure P2 = ? , Temperature T2 = (53+ 273) = 326 K

Now using Clausius- Clapeyron equation:

Log (P2/P1) = ∆H/2.303R{ (T2 - T1)/T1T2}

Log ( P2/.1316)

= 36600/2.303 × 8.314{(326-294.21)/ (294.21 × 326)}

= .6335

Log (P2/.1316) = .6335

P2/.1316= 10^(.6335)

P2/.1316 = 4.3

P2 =0 .5659 atm

So vapour pressure at 53°C will be 0.5659 atm