Q. 2.117 - An Introduction to General, Organic, and Biological Chemistry

Question

A car travels at $55 \mathrm{mi}/\mathrm{h}$ and gets $11 \mathrm{km}/\mathrm{L}$ of gasoline. How many gallons of gasoline are needed for a $3.0$-h trip? (2,6)

Step-by-Step Solution

Answer

The number of gallons of gasoline required are determined as $6.4 gal$.

Step 1 : Given Information

The given is the distance travelled by a car with gasoline.

The objective is to determine the number of gallons of gasoline.

The top speed of the car is 55 mi/hr.

Amount of gasoline it gets as : 11 km/L

Step 2 : Explanation

The car's top speed is $55 mi / h$. The rate of use of gasoline is $11 km / L$.

We need to figure out how many gallons of gasoline are needed for a $3.0$-hour drive.

We'll utilize the provided data to calculate the conversion factors. We must also convert between miles and kilometers, as well as liters and gallons.

Step 3 : Explanation

The speed of the car is $55\mathrm{mi}/\mathrm{h}$. So, the conversion factors are

$\frac{55 mi}{1 h} and \frac{1 h}{55 mi}$

The consumption of gasoline is increasing. $11 \mathrm{km}/\mathrm{L}$. So, the conversion factors are $\frac{11 km}{1 L} and \frac{1 L}{ 11 km}$

Step 4 : Explanation

The relationship between miles and kilometers is equal $1 \mathrm{km}=0.621\mathrm{mi}$ 

Hence, the conversion factors are $\frac{1 \mathrm{km}}{0.621\mathrm{mi}}$ and $\frac{0.621\mathrm{mi}}{1 \mathrm{km}}$

The relationship between liters and gallons is equal.

$1 \mathrm{L}=0.265 \mathrm{gal}$

Hence, the conversion factors are,

$\frac{1 \mathrm{L}}{0.265\mathrm{gal}}\text{ and }\frac{0.265\mathrm{gal}}{1 \mathrm{L}}$

Step 5 : Explanation

The relevant conversion factors are employed, and each factor is set up to cancel the unit before it until the desired unit is produced.

Consequently, we have

$3.0 \mathrm{h}\times \frac{55\mathrm{mi}}{1 \mathrm{h}}\times \frac{1 \mathrm{km}}{0.621\mathrm{mi}}\times \frac{1\mathrm{t}}{11 \mathrm{km}}\times \frac{0.265\mathrm{gal}}{1\mathrm{t}}=6.4\mathrm{gal}$

Hence, a $3.0 h$ trip requires $6.4 \mathrm{gal}$ of gasoline.