A container with a volume of #"16 L"# contains a gas with a temperature of #"180 K"#. If the temperature of the gas changes to #"360 K"# without any change in pressure, what must the container's new volume be?

Question

A container with a volume of #"16 L"# contains a gas with a temperature of #"180 K"#.  If the temperature of the gas changes to #"360 K"# without any change in pressure, what must the container's new volume be?

Elena Albarran · Accepted Answer

The final volume will be exactly #"32 L"#.

This is an illustration of Charles' law, which states that the Kelvin temperature is directly proportional to the volume of a given amount of a gas at constant pressure. You should use the following equation: #V_1/T_1=V_2/T_2#, where #V# is volume and #T# is temperature in Kelvins. Known #V_1="16 L"# #T_1="180 K"# #T_2="360 K"# Unknown #V_2# Solution Rearrange the equation to isolate #V_2#. Insert the known values and solve. #V_2=(V_1T_2)/T_1# #V_2=(16"L"xx360cancel"K")/(180cancel"K")="32 L"#

Sebastian Acosta · Answer

undefined Using the ideal gas law equation $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles of gas, $R$ is the ideal gas constant, and $T$ is temperature in Kelvin, we can rearrange the equation to solve for $V$. Since the pressure is constant, we have:

$V_1 / T_1 = V_2 / T_2$

Substituting the given values, we get:

$16 \, \text{L} / 180 \, \text{K} = V_2 / 360 \, \text{K}$

Solving for $V_2$, we find:

$V_2 = (16 \, \text{L} \times 360 \, \text{K}) / 180 \, \text{K} = 32 \, \text{L}$

Therefore, the container's new volume must be $32 \, \text{L}$.