A driver with a nearly empty fuel tank may say she is "running on fumes."

If a 15.0 gallon automobile gas tank had only gasoline vapor remaining in it, what is the farthest the vehicle could travel if it gets 20.0 miles per gallon on liquid gasoline?

Assume the average molar mass of molecules in gasoline is 101 g/mol, the density of liquid gasoline is 0.75 g/mL, the pressure is 747 mmHg, and the temperature is 25 ∘C.

The farthest the vehicle could travel (if it gets 20.0 miles per gallon on liquid gasoline) is 1.62 miles.

Explanation:

The automobile gas tank has a volume capacity of 15 gallons which can be converted to liters: 15 * 3.7854 = 56.781 liters

We can find the moles of gasoline by using the ideal gas equation: PV = nRT.

Make n (number of moles) the subject of the formula: n = PV/RT, where:

P = 747 mmHg

V = 56.781 liters

R (universal gas constant) = 0.0821 liter·atm/mol·K

T = 25 ∘C = (273 + 25) K = 298 K

1 atm (in the unit of R) = 760 mmHg

Therefore n = 747 * 56.781 / (0.0821 * 760 * 298) = 2.281 mol.

Given that the molar mass of the gasoline = 101 g/mol,

the mass of gasoline = n * molar mass of gasoline = 2.281 mol * 101 g/mol = 230.38 g

the density of the liquid gasoline = 0.75 g/mL

In order to calculate the distance the vehicle can travel, we have to calculate volume of gasoline available = mass of the liquid gasoline : density of liquid gasoline

= 230.38 g : 0.75 g/mL = 307.17 mL = 0.3071 liters = 0.3071 : 3.7854 = 0.0811 gallons

since the vehicle gets 20.0 miles per gallon on liquid gasoline, the distance traveled by the car = gallons available * miles per gallon = 0.0811 * 20 = 1.62 miles.