Question #8ced5

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Answer 1 · 2016-06-22T01:21:48

The new temperature is -73 °C.

Given

The volume $V_{1}$ $V_1$ of a gas at a temperature $T_{1}$ $T_1$ .
A second volume $V_{2}$ $V_2$

Find

The second temperature $T_{2}$ $T_2$

Strategy

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

The formula for Charles' Law is

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

Solution

We can rearrange Charles' Law to get

$T_2 = T_1 × V_2/V_1$

In your problem,

$V_1 = "15 L"$ ; $T_1 = (327 + 273.15) K = "600.15 K"$
$V_2 = color(white)(ll)"5 L"$ ; $T_2 = "?"$

∴ $T_2 = "600 K" × (5 cancel("L"))/(15 cancel("L")) = "200 K"$

$"200 K" = "(200 - 273.15) °C" = "-73 °C"$

The new temperature is -73 °C.