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