# A balloon with a volume of 5.3 L is taken from an indoor temperature of 24 C to the outdoors. The volume of the balloon outside is 4.9 L. How do you determine the C temperature outside?

Aug 31, 2016

The outside temperature is 0 °C.

#### Explanation:

Given

The volume ${V}_{1}$ of a gas at a temperature ${T}_{1}$.
A second volume ${V}_{2}$

Find

The second temperature ${T}_{2}$

Strategy

A problem involving two gas volumes and two temperatures must be a Charles' Law problem.

The formula for Charles' Law is

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {V}_{1} / {T}_{1} = {V}_{2} / {T}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

Solution

We can rearrange Charles' Law to get

T_2 = T_1 × V_2/V_1

${V}_{1} = \text{5.3 L}$; ${T}_{1} = \left(24 + 273.15\right) K = \text{297.15 K}$
${V}_{2} = \text{4.9 L}$; ${T}_{2} = \text{?}$
${T}_{2} = \text{297.15 K" × (4.9 cancel("L"))/(5.3 cancel("L")) = "274 K}$
$\text{200 K" = "(274 - 273.15) °C" = "0 °C}$