A container with a volume of #18 L# contains a gas with a temperature of #170^o C#. If the temperature of the gas changes to #640 ^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 #170^o   C#.  If the temperature of the gas changes to #640     ^o    K# without any change in pressure, what must the container's new volume be?

Nora Arzate · Accepted Answer

The new volume is #=26L#

Utilize Charles' Law

#"volume"/"Temperature"="Constant"#, #"At constant pressure"#

#V_1/T_1=V_2/T_2#

The initial volume is #=V_1=18L#

The initial temperature is #T_1=170+273=443K#

The final temperature is #T_2=640K#

The last book is

#V_2=T_2/T_1*v_1=640/443*18=26L#

Sebastian Acosta · Answer

undefined The ideal gas law states that $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles of gas, $R$ is the gas constant, and $T$ is temperature in Kelvin. Since pressure remains constant, we can rearrange the equation to find the new volume using the initial and final temperatures:

$$
\frac{{V_1}}{{T_1}} = \frac{{V_2}}{{T_2}}
$$

Substituting the given values:

$$
\frac{{18}}{{443 \, \text{K}}} = \frac{{V_2}}{{913 \, \text{K}}}
$$

Solving for $V_2$:

$$
V_2 = \frac{{18 \times 913 \, \text{K}}}{{443 \, \text{K}}}
$$

$$
V_2 \approx 37.08 \, \text{L}
$$

Therefore, the container's new volume must be approximately 37.08 liters.