# Question 0c095

Jun 24, 2017

${704}^{\text{o""C}}$

#### Explanation:

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

$\frac{{V}_{1}}{{T}_{1}} = \frac{{V}_{2}}{{T}_{2}}$ (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

$\text{K" = 20^"o""C} + 273 = 293$ $\text{K}$

The original volume (${V}_{1}$) is given as $1.50$ $\text{L}$, and the final volume (${V}_{2}$) is $5.00$ $\text{L}$.

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

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