Data:
\(V_{initial} = 4.9\:L\)
\(T_{initial} = 30^0C\)
converting to Kelvin
TK = TC + 273
TK = 30 + 273 → TK = 303 → \(T_{initial} = 303\:K\)
\(V_{final} = ? (in\:liters)\)
\(T_{final} = 199^0C\)
TK = TC + 273
TK = 199 + 273 → TK = 472 → \(T_{final} = 472\:K\)
By the first Law of Charles and Gay-Lussac, we have:
\(\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }\)
Solving:
\(\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }\)
\(\frac{ 4.9 }{ 303 } = \frac{ V_{f} }{ 472 }\)
Product of extremes equals product of means:
\(303* V_{f} = 4.9*472\)
\(303 V_{f} = 2312.8\)
\(V_{f} = \frac{2312.8}{303}\)
\(\boxed{\boxed{V_{f} \approx 7.63\:L}}\end{array}}\qquad\quad\checkmark\)
