Question

Show that 60.0 L of gasoline originally at 15.0ºC will expand to 61.1 L when it warms to 35.0ºC, as claimed in Example 13.4.

a) (60.0 L)
b) (60.6 L)
c) (61.1 L)
d) (61.7 L)

Answer 1

The volume of gasoline will expand from 60.0 L to 61.1 L when its temperature increases from 15.0°C to 35.0°C.

Step-by-step explanation:

In order to show that 60.0 L of gasoline originally at 15.0°C will expand to 61.1 L when it warms to 35.0°C, we can use the equation for volume expansion.

Let's assume that the coefficient of volume expansion of the gasoline is given by βs, and the initial volume of the gasoline is Vi.

Using the equation ΔV = βs * Vi * ΔT, where ΔV is the change in volume, βs is the coefficient of volume expansion, Vi is the initial volume, and ΔT is the change in temperature, we can calculate the change in volume:

ΔV = βs * Vi * ΔT = βs * 60.0 L * (35.0°C - 15.0°C) = βs * 60.0 L * 20.0°C

Based on Example 13.4, the change in volume is 1.1 L. Therefore, we can set up the following equation:

1.1 L = βs * 60.0 L * 20.0°C

Simplifying the equation, we find that βs = 1.1 L / (60.0 L * 20.0°C) = 0.000917 °C^-1.

Now, we can calculate the final volume of the gasoline:

Final volume = Vi + ΔV = 60.0 L + 0.000917 °C^-1 * 60.0 L * (35.0°C - 15.0°C) = 61.1 L.

Therefore, the correct answer is (c) 61.1 L.