Question

Gasoline is primarily a mixture of hydrocarbons and is sold with an octane rating that is based on a comparison with the combustion properties of isooctane. Gasoline usually contains an isomer of isooctane called tetramethylbutane (C8H18), which has an enthalpy of vaporization of 43.3 kJ/mol and a boiling point of 106.5°C. Determine the vapor pressure of tetramethylbutane on a very hot summer day when the temperature is 38°C.

A. 47.9 torr
B. 67.8 torr
C. 89.3 torr
D. 80.3 torr
E. 36.7 torr

Air (consisting mostly of nitrogen (N2), oxygen (O2), and argon (Ar)) dissolves in water due to?

A. dipole-induced dipole forces.
B. ion–dipole forces.
C. dipole–dipole forces.
D. hydrogen bonding.
E. dispersion forces.

Thanks!

Answer 1

Gasoline is primarily a mixture of hydrocarbons and is sold with an octane rating that is based on a comparison with the combustion properties of isooctane.

Gasoline usually contains an isomer of isooctane called tetramethylbutane (C8H18), which has an enthalpy of vaporization of 43.3 kJ/mol and a boiling point of 106.5°C.

Determine the vapor pressure of tetramethylbutane on a very hot summer day when the temperature is 38°C.

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(P2/760) = 43300/8.314*(1/(106.5+273)-1/(38+273))

ln(P2/760)  = -3.0227

P2 = 760* exp(-3.0227)

P2 = 36.9889 torr

Question 2.

Air (consisting mostly of nitrogen (N2), oxygen (O2), and argon (Ar)) dissolves in water due to?

A. dipole-induced dipole forces. --> TRUE, the dipole in water, re-arranges the poles in nonpolar air, thereofre, they re-arrange to form a dipole + induced dipole
B. ion–dipole forces. --> no ion forms
C. dipole–dipole forces. --> false, air has just nonpolar species
D. hydrogen bonding. --> no H bonds present in air
E. dispersion forces. --> this will not help in dissolving