# 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?

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.

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.

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