A container with a volume of #8 L# contains a gas with a temperature of #180^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?
The new volume is 16L.
Let's identify our known and unknown variables.
The first volume we have is
The equation we use is:
The volume should have units of liters and the temperature must have units of Kelvins. In our case, both have good units!
Now rearrange the equation and plug in the given values
#V_2 = (T_2 * V_1)/(T_1)#
#V_2 = (360cancel("K") * "8 L") / (180 cancel("K"))#
#V_2 = "16 L"#
When using the Kelvin scale, you do not put the degree symbol. You just write K.