Answer
the gage pressure of oxygen contained in the tank is $57.8408 \mathrm{kPa}$
Work Step by Step
Calculate the absolute pressure of oxygen contained in a tank considering ideal gas law.
$$
p_{a b s}=\rho R T
$$
Here, $\rho$ is the density of oxygen in the tank, $R$ is the specific gas constant of oxygen, and $T$ is the temperature of compressed air in tank.
Substitute $2 \mathrm{~kg} / \mathrm{m}^{3}$ for $\rho, 259.8 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}$ for $R, 298 \mathrm{~K}\left(25^{\circ} \mathrm{C}\right)$ for $T$.
$$
\begin{aligned}
p_{a b s} &=(2)(259.8)(298) \\
&=154840.8 \mathrm{~Pa} \\
&=154.8408 \mathrm{kPa}
\end{aligned}
$$
Calculate the gage pressure of oxygen in the tank.
$$
p_{g}=p_{a b s}-p_{a t m}
$$
Here, $p_{\text {atm }}$ is the atmospheric pressure.
Substitute $154.8408 \mathrm{kPa}$ for $p_{a b s}$ and $97 \mathrm{kPa}$ for $p_{\text {atm }}$.
$$
\begin{aligned}
p_{g} &=154.8408-97 \\
&=57.8408 \mathrm{kPa}
\end{aligned}
$$
Thus, the gage pressure of oxygen contained in the tank is $57.8408 \mathrm{kPa}$
