Question

Estimate the liquid density (g/cm3) of propane at 298 K and 10 bar. Compare the price per kilogram of propane to the price of regular gasoline assuming the cost of 5 gal of propane for typical gas grills is roughly $30. The density of regular gasoline can be estimated by treating it as pure isooctane (2,2,4 trimethylpentane ! = 0.692 g/cm3 ) at 298 K and 1 bar.

Answer 1

Answer:

Density of propane = 17.8 g/L

Propane is more priced than gasoline

Explanation:

Given:

Temperature, T = 298 K

Pressure, P = 10 bar = 0.987 × 10 = 9.87 atm

now,

Molar mass of propane, M = 44.1 g/mol

From ideal gas law

⇒ PV = nRT

here,

n is the number of moles

R is the ideal gas constant = 0.0821 L.atm/mol.K

also,

Density, D = $\frac{\textup{Mass}}{\textup{Volume(V)}}$

or

V = $\frac{\textup{Mass}}{\textup{D}}$

and,

nM = mass

thus,

V = $\frac{\textup{nM}}{\textup{D}}$

substituting in the ideal gas relation

we have

P = $\frac{\textup{DRT}}{\textup{M}}$

or

D = $\frac{\textup{PM}}{\textup{RT}}$

or

D = $\frac{9.87\times44.1}{0.0821\times298}$

or

D = 17.8 g/L

Now,

1 gallon = 3.78 Liter

Therefore,

5 gallon = 5 × 3.78 Liter = 18.9 Liter

Thus,

mass of 5 gallon propane = Volume × Density

= 18.9 Liter × 17.8 g/L

= 336.42 g

or

= 0.336 kg

also it is given that Price of 5 gallon propane i.e 0.336 kg = $30

Therefore,

Price per kg = $\frac{30}{0.336}$

= $89.28

and,

Mass of 5 gallons i.e 18.9 Liter gasoline = Density × Volume

= 0.692 g/cm³ × 18.9 Liter

also,

1 L = 1000 cm³

thus,

= 0.692 g/cm³ × 18.9 × 1000 cm³

= 13078.8 g

or

= 13.078 kg

Therefore,

Price per kg of gasoline = $\frac{\ 30}{13.078}$

= $2.29

hence, propane is more priced than gasoline