The vapour pressure of methyl alcohol is 40 mmHg at 5°C. Use this value and the calsius-clapeyron equation to estimate the normal boiling point of mythl alcohol.
4 answers
Clausius*
The boiling point is the temperature at which the vapor pressure is 1 atm = 760 mm Hg, which is 19 times higher.
You will need the heat of vaporization of methyl alcohol to apply the Clausius-Clapeyron equestion. It is 35278 kJ/kmol. Call it Hv. You will also need the molar gas constant R, which is 8.314 kJ/(mole*K)
For an explanation of how to apply the C-C equation, see
http://www.science.uwaterloo.ca/~cchieh/cact/c123/clausius.html
You will need the heat of vaporization of methyl alcohol to apply the Clausius-Clapeyron equestion. It is 35278 kJ/kmol. Call it Hv. You will also need the molar gas constant R, which is 8.314 kJ/(mole*K)
For an explanation of how to apply the C-C equation, see
http://www.science.uwaterloo.ca/~cchieh/cact/c123/clausius.html
P1 DHvap 1 1
ln (---) = ---- (--- - ---)
P2 R T2 T1
ln1/0.53= 38/8.31(1/T1 - 1/T2)
2.93 =4.57(1/278 - 1/T2)
.641 = 1/278 - 1/ T2
.6407 = -1/T2
T2 = -1.56
that answer is suppose to be T=339 K
which part is wrong?
ln (---) = ---- (--- - ---)
P2 R T2 T1
ln1/0.53= 38/8.31(1/T1 - 1/T2)
2.93 =4.57(1/278 - 1/T2)
.641 = 1/278 - 1/ T2
.6407 = -1/T2
T2 = -1.56
that answer is suppose to be T=339 K
which part is wrong?
I gave you the wrong heat of vaporization by a factor of 1000, but you did catch that. It should be 35.278 kJ/mole. You used 38. I don't see a ln(19) that should be in your calculation. That is the natural log of the pressure ratio. Done propertly, you cannot end up with a negative absolute temperature