Question

A gasoline spill is approximately 4 liters of liquid. What volume of vapor is created at 1 bar and 293 K when the liquid evaporates? The density of regular gasoline can be estimated by treating it as pure isooctane (2,2,4-trimethylpentane rhoL = 0.692 g/cm3) at 298 K and 1 bar.

Answer 1

Answer:

$V=591.748 L$

Explanation:

Assumption:

Ideal Vapors/Ideal gas

Formula for ideal Gas:

$PV=nR_uT$

Where:

P is the pressure

V is the Volume

n is the number of moles = m/M

R_u is Universal Gas Constant=0.08314 L*bar/(K*mol)

T is the temperature in Kelvin

Calculating Number of moles n:

n=Mass/Molar Mass

$Mass=\rho_L*Volume\\Mass=0.692*(4000 cm^3)........... (4 liter * 1000cm^3/Liters =4000 cm^3)\\Mass=2768 g$

Molar Mass of gasoline=114g/mol

$n=\frac{2768}{114} \\n=24.2807 moles$

Now:

$PV=nR_uT$

$V=\frac{nR_uT}{P}\\V=\frac{24.2807*0.08314*293}{1 bar}\\V=591.748 L$