Question #0c095

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Answer 1 · 2017-06-24T21:27:32

$704^"o""C"$

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

$(V_1)/(T_1) = (V_2)/(T_2)$ (constant pressure and quantity)

Remember, when using any gas equation, the value for temperature must always be in Kelvin, so let's convert the temperature to Kelvin

$"K" = 20^"o""C" + 273 = 293$ $"K"$

The original volume ( $V_1$ ) is given as $1.50$ $"L"$ , and the final volume ( $V_2$ ) is $5.00$ $"L"$ .

Plugging in the values into the equation, and solving for the final temperature ( $T_2$ ), we have

$T_2 = (V_2T_1)/(V_1) = ((5.00cancel("L"))(293color(white)(l)"K"))/(1.50cancel("L")) = color(red)(977$ $color(red)("K"$

Which, in Celsius temperature is

$""^"o""C" = 977color(white)(l)"K" - 273 = color(red)(704^"o""C"$