Question

RS ITem 26 ,

Gasoline is a mixture of hydrocarbons, a major component of which is octane, CH3CH2CH2CH2CH2CH2CH2CH3. Octane has a vapor pressure of 1.860kPa at 25.0 ∘Cand a vapor pressure of 19.302kPa at 75.0 ∘C.

Part A

Calculate the heat of vaporization of octane.

Part B

Calculate the normal boiling point of octane.

Part C

Calculate the vapor pressure of octane at 40.0 ∘C .

Answer 1

A)

Hoctane

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

ln(19.30/1.86) = H/8.314*(1/(298) - 1/(75+273))

H = (8.314*ln(19.30/1.86))/(1/(298) - 1/(75+273))

H = 40342.599987 J/mol

H= 40.34 kJ/mol

B)

Tb of octane is at P = 101.3 KPa

then

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

ln(101.3/1.86) = 40342.59/8.314*(1/(298) - 1/(T))

Solve for T

3.9975 = 4852.36(1/(298) - 1/(T))

(1/(298) - 1/(T)) = 3.9975/4852.36

T = 394.96 K

T = 121.96°C

C)

ln(101.3/P1) = 40342.59/8.314*(1/(298) - 1/(40+273))

solve for P

ln(101.3/P) = 0.78034

P = (exp(0.78034)/101.3)^-1

P = 46.42074 kPa