Question

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?

Answer 1

The temperature needed to reduce the volume from `"8.62 L"` to `"2.61 L"` is `-123^@"C"`.

Explanation:

Your variables, `"L"` and `""^@"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="8.62 L"`

`T_1="221"^@"C" + 273.15="494 K"`

`V_2="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=(V_2T_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.

`"150. K"-273.15"="-123^@"C"`