Question

Question #641bc

Answer 1

The volume of the bubble will be `"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.

`color(blue)(bar(ul(|color(white)(a/a)V_1/T_1 = V_2/T_2color(white)(a/a)|)))" "`

We can rearrange this equation to get

`V_2 = V_1 × T_2/T_1`

Solution

In your problem,

`V_1 = "1.15 cm"^3`; `T_1 = "(22 + 273.15) K" = "295.15 K"`
`V_2 = ?`; `color(white)(mmml)T_2 = "(99 + 273.15) K" = "372.15 K"`

`T_2 =1.15 "cm"^3 × (372.15 cancel("K"))/(295.15 cancel("K")) = "1.45 cm"^3`

The new volume is `"1.45 cm"^3`.