Question

**1a**. From the following vapor
pressure data for **octane**, an estimate of the molar
heat of vaporization of
**C _{8}H_{18}**
is kJ/mol.

P, mm Hg | T, Kelvins |

100 |
339 |

400 |
377 |

**1b.**

The normal boiling point of liquid **chloroform**
is **334** K. Assuming that its molar heat of
vaporization is constant at **29.9** kJ/mol, the
boiling point of **CHCl _{3}** when the
external pressure is

**1c.**

The vapor pressure of liquid **butanol**,
**C _{4}H_{9}OH**, is

A sample of

Only butanol vapor will be present.

Some of the vapor initially present will condense.

The pressure in the container will be 100. mm Hg.

Liquid butanol will be present.

No condensation will occur.

Answer #1

1a.

Recall that in equilibrium; especially in vapor-liquid equilibriums, we can use Clasius Clapyeron combination equation in order to relate two points in the same equilibrium line.

The equation is given as:

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

Where

P2,P1 = vapor pressure at point 1 and 2

dH = Enthalpy of vaporization, typically reported in kJ/mol, but we need to use J/mol

R = 8.314 J/mol K

T1,T2 = Saturation temperature at point 1 and 2

Therefore, we need at least 4 variables in order to solve this.

Substitute all known data:

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

Change negative signs

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

ln(400/100) = dHvap/8.314*(1/(339) - 1/(377))

dHvap =38763.495 = 38.76 kJ/mol

1b

again, use clasius equation

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

ln(1.34/1) = 29900/8.314*(1/(334 - 1/T)

T = (-ln(1.34/1) *8.314/29900+1/(334) )^-1

T = 343 K = 70 °C

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