Question

# 1a.  From the following vapor pressure data for octane, an estimate of the molar heat of vaporization...

1a.  From the following vapor pressure data for octane, an estimate of the molar heat of vaporization of C8H18 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 CHCl3 when the external pressure is 1.34 atm is  K.

1c.

The vapor pressure of liquid butanol, C4H9OH, is 100. mm Hg at 343 K.

A sample of C4H9OH is placed in a closed, evacuated 528 mL container at a temperature of 343 K. It is found that all of the C4H9OHis in the vapor phase and that the pressure is 67.0 mm Hg. If the volume of the container is reduced to 381 mL at constant temperature, which of the following statements are correct?

Choose all that apply.

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.

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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