A container with a volume of #6 L# contains a gas with a temperature of #280^o K#. If the temperature of the gas changes to #320 ^o K# without any change in pressure, what must the container's new volume be?

1 Answer
Apr 11, 2016

Answer:

The new volume will be #6.86L#

Explanation:

We start with the ideal gas law:

#PV = nRT#

and move all of the constant terms to one side. We know from the question that the temperature and the volume are changing - everything else is constant, in other words:

#V/T = (nR)/P = "constant"#

This means that we can set the two different sets of volumes and temperatures equal to each other using this equation, since we know they are constant:

#V_2/T_2 = V_1/T_1 = "constant"#

We can now solve for the new volume we are trying to find and plug in the numbers from the question:

#V_2 = (V_1*T_2)/T_1 = (6L*320K)/(280K) = 6.86L#