Given the following, solve for the new temperature: $\begin{array} { l l }
\text { Initial volume } & 9 \mathrm {~L} \
\text { Initial temperature } & 24 \text { degrees Celsius } \
\text { New volume } & 3 \mathrm {~L}
\end{array}$

Question

Given the following, solve for the new temperature: $\begin{array} { l l } 
\text { Initial volume } & 9 \mathrm {~L} \
\text { Initial temperature } & 24 \text { degrees Celsius } \
\text { New volume } & 3 \mathrm {~L}
\end{array}$

Quizplus · Accepted Answer

Using Charles's Law, which states that volume is directly proportional to temperature (in Kelvin), we can set up the equation: $ \frac{V_1}{T_1} = \frac{V_2}{T_2} $. Convert the initial temperature to Kelvin (24°C = 297K). Solve for the new temperature: $ \frac{9}{297} = \frac{3}{T_2} $. This gives $ T_2 = 99K $, which is -174°C.

Given the Following, Solve for the New Temperature A) 174 Degrees Celsius B) 0 Degrees Celsius C) -50

Given the Following, Solve for the New Temperature A) 174 Degrees Celsius
B) 0 Degrees Celsius
C) -50