Question

A storage tank contains ethanol. As liquid ethanol is taken from the lower part of the tank, nitrogen is fed into the upper part of the tank so that the pressure above the liquid in the tank is always 780 mmHg. Currently the depth of the liquid ethanol in the tank is 1.2 m. It is a summer day. The temperature in the tank is 50 OC.

(a) If you took a sample of the vapor in the tank, what would its composition be?

Answer 1

a)

At 50 °C; liquid ethanol will vaporise and these vapors occupy the space above the liquid ethanol. The liquid and vapor phases are in equilibrium and the partial pressure exerted by ethanol vapors is equal to the vapor pressure of ethanol at 50 °C.

Mathematically;

From Dalton's Law of Partial Pressures:

P_e = y P

where;

P_e = Partial pressure of ethanol vapor which is equal to its vapor pressure

y = Mole fraction of ethanol in the gas phase

P = Total pressure = 780 mm Hg

The vapor pressure of ethanol can be estimated from the Antoine's Equation:

log₁₀ P_e = 8.04494 - 1554.3 / (222.65 + T)

Here T = 50 °C

log₁₀ P_e = 8.04494 - 1554.3 / (222.65 + 50)

log₁₀ P_e = 8.04494 - 5.70072

log₁₀ P_e = 2.3442248

P_e = 10 ^(2.3442248)

P_e = 221 mm Hg

Given: P = 780 mm Hg

Therefore:

221 = 780 y

y = 221 / 780

y = 0.2833

Therefore;

The molar composition of the sample is:

Ethanol = 0.2833 mol / mol

Nitrogen = 0.7167 mol / mol