Question

A driver fills an 18.9L steel gasoline can with gasoline at 15.0°C right up to the top. He forgets to replace the cap and leaves the can in the back of his truck. The temperature climbs to 30.0°C by 1pm. How much gasoline spills out of the can​

Answer 1

Answer:

ΔV = 0.98 L

Explanation:

First, we will calculate the increased volume using Charles' Law:

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$

where,

V₁ =initial volume = 18.9 L

V₂ = final volume = ?

T₁ = initial temperature = 15°C + 273 = 288 k

T₂ = final temperature = 30°C + 273 = 303 k

Therefore,

$\frac{18.9\ L}{288\ k} = \frac{V_2}{303\ k}$

V₂ = 19.88 L

Now, we calculate the change in volume:

ΔV = V₂ - V₁ =  19.88L - 18.9 L

ΔV = 0.98 L

This is the volume of gasoline that will spill out.