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.

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zainsubhani zainsubhani · Answer 1 · 2020-01-23T06:51:25+01:00

Answer:

[tex]V=591.748 L[/tex]

Explanation:

Assumption:

Ideal Vapors/Ideal gas

Formula for ideal Gas:

[tex]PV=nR_uT[/tex]

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

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

Molar Mass of gasoline=114g/mol

[tex]n=\frac{2768}{114} \\n=24.2807 moles[/tex]

Now:

[tex]PV=nR_uT[/tex]

[tex]V=\frac{nR_uT}{P}\\V=\frac{24.2807*0.08314*293}{1 bar}\\V=591.748 L[/tex]