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Question

# Some oxygen has a density of $1.75 \mathrm{~kg} / \mathrm{m}^3$ at normal atmospheric pressure. Find its pressure (in $\mathrm{kPa}$ ) when the density is changed to $1.45 \mathrm{~kg} / \mathrm{m}^3$ .

Solution

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$\textbf{Given:}$

$D_1=1.75 \text{ } \dfrac{ \text{kg}}{\text{ m}^3}$

$D_2=1.45 \text{ } \dfrac{ \text{kg}}{\text{ m}^3}$

$P_1=101 \text{ kPa}$

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

$\dfrac{D_1}{D_2}=\dfrac{P_1}{P_2}$

From it we express $P_2$ and include the data given in the problem to get the solution:

\begin{align*} P_2&=\dfrac{D_2 \cdot P_1}{D_1}\\ P_2&=\dfrac{1.45 \cdot 101}{1.75}\\ &\boxed{P_2=83.7 \text{ kPa}} \end{align*}

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