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Question

Some oxygen has a density of 1.75 kg/m31.75 \mathrm{~kg} / \mathrm{m}^3 at normal atmospheric pressure. Find its pressure (in kPa\mathrm{kPa} ) when the density is changed to 1.45 kg/m31.45 \mathrm{~kg} / \mathrm{m}^3 .

Solution

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Answered 2 years ago
Answered 2 years ago
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Given:\textbf{Given:}

D1=1.75 kg m3D_1=1.75 \text{ } \dfrac{ \text{kg}}{\text{ m}^3}

D2=1.45 kg m3D_2=1.45 \text{ } \dfrac{ \text{kg}}{\text{ m}^3}

P1=101 kPaP_1=101 \text{ kPa}

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

D1D2=P1P2\dfrac{D_1}{D_2}=\dfrac{P_1}{P_2}

From it we express P2P_2 and include the data given in the problem to get the solution:

P2=D2P1D1P2=1.451011.75P2=83.7 kPa\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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