Answer:
Given:
[tex]V_{1} = \text{400.0 mL}[/tex]
[tex]T_{1} = 225.0^{\circ}\text{C + 273.15 = 498.15 K}[/tex]
[tex]T_{2} = 127.0^{\circ}\text{C + 273.15 = 400.15 K}[/tex]
Unknown:
[tex]V_{2}[/tex]
Solution:
[tex]\dfrac{V_{1}}{T_{1}} = \dfrac{V_{2}}{T_{2}}[/tex]
[tex]V_{2} = V_{1} \times \dfrac{T_{2}}{T_{1}}[/tex]
[tex]V_{2} = \text{400.0 mL} \times \dfrac{\text{400.15 K}}{\text{498.15 K}}[/tex]
[tex]\boxed{V_{2} = \text{321.3 mL}}[/tex]
Therefore, the volume of the gas at 127.0°C is 321.3 mL
[tex]\begin{gathered}\\\end{gathered}[/tex]
