# Question 641bc

Apr 23, 2017

The volume of the bubble will be ${\text{1.45 cm}}^{3}$.

#### Explanation:

Given

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

Find

The second volume ${V}_{2}$

Strategy

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

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

We can rearrange this equation to get

V_2 = V_1 × T_2/T_1

Solution

${V}_{1} = {\text{1.15 cm}}^{3}$; ${T}_{1} = \text{(22 + 273.15) K" = "295.15 K}$
V_2 = ?#; $\textcolor{w h i t e}{m m m l} {T}_{2} = \text{(99 + 273.15) K" = "372.15 K}$
${T}_{2} = 1.15 {\text{cm"^3 × (372.15 cancel("K"))/(295.15 cancel("K")) = "1.45 cm}}^{3}$
The new volume is ${\text{1.45 cm}}^{3}$.