Question #03e92

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Answer 1 · 2017-07-03T01:32:14

$V_{2} = 33.0$ $V_2 = 33.0$ $L$ $"L"$

To solve this problem, we can use the temperature-volume relationship of gases, illustrated by Charles's law:

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ $(V_1)/(T_1) = (V_2)/(T_2)$

Remember that for all gas equation problems, the temperature must always be in units of Kelvin.

Let's convert both temperatures:

$T_{1} = 19^{o} C + 273 = 292$ $T_1 = 19^"o""C" + 273 = 292$ $K$ $"K"$

$T_{2} = 122^{o} C + 273 = 395$ $T_2 = 122^"o""C" + 273 = 395$ $K$ $"K"$

And the initial volume $V_{1}$ $V_1$ is given as $24.4$ $24.4$ $L$ $"L"$ .

Plugging in known values, and solving for $V_{2}$ $V_2$ , we have

$V_2 = (T_2V_1)/(T_1) = ((395cancel("K"))(24.4color(white)(l)"L"))/(292cancel("K")) = color(blue)(33.0$ $color(blue)("L"$