# An inflated balloon has a volume of 5.75 L at a temperature of 22°C. At what temperature in degrees Celsius will the volume of the balloon decrease to 5.38 L?

Mar 2, 2016

The volume will be at $\text{5.38 L}$ when the temperature is at $\textcolor{b l u e}{\text{2.85"^@"C}}$.

#### Explanation:

This is an example of Charles' law which states that at constant pressure, the volume is directly proportional to the temperature in Kelvins. The equation to use is ${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$.

Given/Known
${V}_{1} = \text{5.75 L}$
${T}_{1} = \text{22"^@"C""+273.15=295 K}$
${V}_{2} = \text{5.38 L}$

Unknown
${T}_{2}$

Solution
Rearrange the equation to isolate ${T}_{2}$. Substitute the known values into the equation and solve. Convert the temperature from Kelvins to degrees Celsius.

${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$

${T}_{2} = \frac{{T}_{1} {V}_{2}}{V} _ 1$

T_2=((295"K")*(5.38cancel"L"))/(5.75cancel"L")

${T}_{2} = \text{276 K}$

Convert Kelvins to $\text{^@"C}$.

$\text{^@"C""=} K - 273.15$

$\text{^@"C"=276-273.15="2.85"^@"C}$