Data:
[tex]V_{initial} = 590\:mL[/tex]
[tex]T_{initial} = -55.0^0C[/tex]
converting to Kelvin
TK = TC + 273
TK = -55.0 + 273 → TK = 218.0 → [tex]T_{initial} = 218.0\:K[/tex]
[tex]V_{final} = ? (in\:milliliters)[/tex]
[tex]T_{final} = 30.0^0C[/tex]
TK = TC + 273
TK = 30.0 + 273 → TK = 303.0 → [tex]T_{final} = 303.0\:K[/tex]
By the first Law of Charles and Gay-Lussac, we have:
[tex] \frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }[/tex]
Solving:
[tex] \frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }[/tex]
[tex]\frac{ 590 }{ 218.0 } = \frac{ V_{f} }{ 303.0 }[/tex]
Product of extremes equals product of means:
[tex]218.0* V_{f} = 590*303.0[/tex]
[tex]218.0 V_{f} = 178770[/tex]
[tex]V_{f} = \frac{178770}{218.0} [/tex]
[tex]\boxed{\boxed{V_{f} \approx 820.04\:mL}}\end{array}}\qquad\quad\checkmark[/tex]
