A gas at 362 K occupies a volume of 0.67 L. At what temperature will the volume increase to 1.12 L?

1 Answer
Apr 11, 2017

Answer:

The temperature at which the volume increases to #"1.12 L"# is #6.1xx10^2##color(white)(.)"K"#.

Explanation:

This is an example of Charles' law , which states that the volume of a gas, held at constant pressure and amount, varies directly with the temperature in Kelvins. This means that when the volume increases, the temperature increases and vice-versa.

The equation to use is:

#V_1/T_1=V_2/T_2#

where #V_1# is the initial volume, #T_1# is the initial temperature, #V_2# is the final volume, and #"T_2"# is the final temperature.

Given/Known
#T_1="362 K"#
#V_1="0.67 L"#
#V_2="1.12 L"#

Unknown: #T_2#

Solution
Rearrange the equation to isolate #T_2#, then substitute the known values into the equation and solve.

#T_2=(V_2T_1)/V_1#

#T_2=(1.12"L"xx362"K")/(0.67"L")=6.1xx10^2##color(white)(.) "K"#
(#"605 K"# rounded to two significant figures)