A container with a volume of #7 L# contains a gas with a temperature of #120^o K#. If the temperature of the gas changes to #360 ^o K# without any change in pressure, what must the container's new volume be?

Question

A container with a volume of #7 L# contains a gas with a temperature of #120^o   K#.  If the temperature of the gas changes to #360     ^o    K# without any change in pressure, what must the container's new volume be?

Olivia Askew · Accepted Answer

The new volume, #V_2#, is #color(blue)"21 L"#.

This is an example of Charles' law, which states that the volume of a given amount of a gas kept at constant pressure varies directly with the temperature in Kelvins. This means that if the volume increases, so will the temperature, and vice versa. The equation used to solve Boyle's law problems is #V_1/T_1=V_2/T_2#. Known #V_1="7 L"# #T_1="120 K"# #T_2="360 K"# Unknown #V_2="?"# Solution Rearrange the equation to isolate #V_2#. Substitute the known values into the equation and solve. #V_1/T_1=V_2/T_2# #V_2=(V_1T_2)/T_1# #V_2=(7"L"*360cancel"K")/(120cancel"K")# #V_2=color(blue)("21 L")#

Charles Aeschliman · Answer

undefined Using the combined gas law, we can find the new volume of the gas using the formula:

V2 = V1 * (T2 / T1)

Where:
V1 = initial volume = 7 L
T1 = initial temperature = 120 K
T2 = final temperature = 360 K

Plugging in the values:

V2 = 7 L * (360 K / 120 K) = 7 L * 3 = 21 L

So, the container's new volume must be 21 liters.