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Question

# Some helium occupies $1200 \mathrm{~ft}^3$ at $70{\degree} \mathrm{F}$. Find the temperature at which its volume will be $600 \mathrm{~ft}^3$.

Solution

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$\textbf{Given:}$

$V_1=1200 \text{ ft}^3$

$V_2=600 \text{ ft}^3$

$T_1=70 ^\circ \text{ F}=530 ^\circ \text{R}$

To solve the problem we use Charles's law which reads:

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

From it we express $T_2$ and by entering the data given in the task we get the result:

\begin{align*} T_2&=\dfrac{V_2 \cdot T_1}{V_1}\\ &=\dfrac{600 \cdot 530}{1200}\\ &\boxed{T_2=265^\circ \text{R} } \end{align*}

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