Question

Question #0c095

Answer 1

`704^"o""C"`

Explanation:

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"`