Question #641bc

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Answer 1 · 2017-04-23T15:35:07

The volume of the bubble will be $"1.45 cm"^3$ .

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 $"1.45 cm"^3$ .