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

Answer:

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"#