# A sample of sulfur hexafluoride gas occupies 8.62 L at 221 degrees Celsius. Assuming that the pressure remains constant, what temperature (in Celsius) is needed to reduce the volume to 2.61 L?

##### 1 Answer
Jul 5, 2017

The temperature needed to reduce the volume from $\text{8.62 L}$ to $\text{2.61 L}$ is $- {123}^{\circ} \text{C}$.

#### Explanation:

Your variables, $\text{L}$ and $\text{^@"C}$ represent volume and temperature. The gas law that involves the relationship between the volume of a gas and its temperature is Charles' law. http://chemistry.bd.psu.edu/jircitano/gases.html

Charles' law states that the volume of a given amount of gas held at constant pressure is directly proportional to its Kelvin temperature. This means that as the temperature increases, the volume also increases, and vice-versa. The equation used to solve gas problems involving Charles' law is:

${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$,

where the temperatures must be in Kelvins. Add $273.15$ to the Celsius temperature to get the Kelvin temperature.

Organize your data:

Known

${V}_{1} = \text{8.62 L}$

${T}_{1} = \text{221"^@"C" + 273.15="494 K}$

${V}_{2} = \text{2.61 L}$

Unknown

T_2=?

Solution

Rearrange the equation given above to isolate "T_2. Insert your data into the new equation and solve.

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

T_2=(2.61color(red)cancel(color(black)("L"))xx494"K")/(8.62color(red)cancel(color(black)("L")))="150. K"

To convert from Kelvins to degrees Celsius, subtract $273.15$ from the Kelvin temperature.

$\text{150. K"-273.15"="-123^@"C}$