A container with a volume of #18 L# contains a gas with a temperature of #220^o C#. If the temperature of the gas changes to #320 ^o K# without any change in pressure, what must the container's new volume be?

Question

A container with a volume of #18  L# contains a gas with a temperature of #220^o   C#.  If the temperature of the gas changes to #320     ^o    K# without any change in pressure, what must the container's new volume be?

Olivia Askew · Accepted Answer

The new volume will be #"12 L"# rounded to two significant figures.

This is an example of Charles' law, which states that the volume of a given amount of gas that is held at a constant pressure, is directly proportional to its temperature in Kelvins. This means that as the temperature increases, so does the volume and vice versa. The equation below represents Charles' law when volume or temperature changes.

The following are the known and unknown variables for this question.

Known
#V_1="18 L"#
#T_1="220"^@"C"+"273.15"="493 K"#
#T_2="320 K"#

Unknown
#V_2#

Solution
Rearrange the equation in order to isolate #V_2#. Substitute the known values into the equation and solve.

#V_2=(V_1T_2)/T_1#

#V_2=(18"L"xx320cancel"K")/(493cancel"K")="12 L"# rounded to two significant figures

Carson Alexander · Answer

undefined To find the new volume of the container, we can use the combined gas law formula: $ V_2 = \frac{{T_2}}{{T_1}} \times V_1 $, where $ V_1 $ is the initial volume, $ T_1 $ is the initial temperature, $ T_2 $ is the final temperature, and $ V_2 $ is the final volume. Given that the initial volume $ V_1 = 18 $ L, the initial temperature $ T_1 = 220^\circ C = 220 + 273.15 = 493.15 $ K, and the final temperature $ T_2 = 320 $ K, we can substitute these values into the formula to find the final volume $ V_2 $. After calculating, the final volume $ V_2 $ is approximately 25.81 L.