A certain car tire has a volume 9.5 L at 25C. What would it’s volume be if the temperature drops to 5.0 C?1) name the law first 2) identify the given, write down the equation (formula)

From the given information, you will observe that the pressure of the tire remains constant, so the law governing the situation is called the Charle's law

Step 1: State Charle's law

Charle's law state that the volume of a given is directly proportional to the temperature of the mass, provided that the pressure remains constant.

The above law can be expressed mathematically below

[tex]\begin{gathered} \text{ V }\propto\text{ T} \\ \text{ V = kT} \\ \text{ Divide both sides by T} \\ \frac{V}{T}\text{ = }\frac{kT}{T} \\ \text{ k = }\frac{V}{T} \\ k\text{ is the constant \lparen Pressure\rparen} \end{gathered}[/tex]

Step 2: Convert the temperature to kelvin from degrees Celcius

[tex]\begin{gathered} \text{ T1= 25}\degree C\text{ } \\ T\text{ K= t + 273.15} \\ \text{ T k = 25 + 273.15} \\ \text{ Tk = 298.15K} \\ \\ \text{ T2 = t + 273.15} \\ \text{ T2 = 5 +273.15} \\ \text{ T2 = 278.15K} \end{gathered}[/tex]

Step 3: Substitute the given data into the formula in step 1

[tex]\begin{gathered} \frac{V1}{T1}\text{ = }\frac{V2}{T2} \\ \\ \text{ }\frac{9.5}{298.15}\text{ = }\frac{V2}{278.15} \\ \text{ Cross multiply} \\ \text{ 9.5}\times\text{ 278.15 = V2 }\times298.15 \\ 2642.425\text{ = 298.15V2} \\ \text{ Divide both sides by 298.15} \\ \text{ }\frac{2642.425}{298.15}\text{ = }\frac{298.15V2}{298.15} \\ \text{ } \\ \text{ V2 = 8.86L} \\ V2\approx8.9L \end{gathered}[/tex]

Hence, the final volume of the car tire is 8.9L