A sample of gas occupies a volume of 325 mL and has a temperature of 434.9 kelvin. What will the volume of the sample of gas be if the temperature is increased to 812 K? Be sure to include units with your answer. DO NOT use scientific notation for your answer. DO NOT use a space between the numerical value and the units (i.e. a volume of 1.35 L would be recorded as 1.35L, not 1.35 L).

To get the final volume of the gas sample, we will use Charles law. According to the law, the volume of a given amount of gas is directly proportional to its temperature. Mathematically;

[tex]\begin{gathered} V\alpha T \\ V=kT \\ k=\frac{V}{T} \\ \frac{V_1}{T_1}=\frac{V_2}{T_2}_{} \end{gathered}[/tex]

V1 and V2 are the initial and final volume

T1 and T2 are the initial and final temperatures.

Given the following parameters:

[tex]\begin{gathered} V_1=325mL \\ T_1=434.9K \\ T_2=812K \\ V_2_{}=\text{?} \end{gathered}[/tex]

Substitute the given parameters into the formula to have:

[tex]\begin{gathered} V_2=\frac{V_1T_2}{T_1} \\ V_2=\frac{325mL\times812\cancel{K}}{434.9\cancel{K}} \\ V_2=\frac{263,900}{434.9} \\ V_2=606.81mL \end{gathered}[/tex]

Hence the volume of the sample of gas if the temperature is increased to 812 K is 606.81mL