Answer
Solution:
Volume of the tank $=2 \mathrm{ft}^{3}$
Weight of gas $=0.3 \mathrm{lb}$
Pressure exerted by the gas, $P_{\text {gauge }}=12 \mathrm{psi}$
Considering atmospheric pressure, $P_{\text {atom }}=14.7$ psi
Absolute pressure, $P_{\text {abs }}=P_{\text {atm }}+P_{\text {gauge }}$
$P_{\text {abs }}=12+14.7$
$P_{\text {abs }}=26.7 \mathrm{psi}$
Gas temperature $=80 \mathrm{~F}$
Converting Fahrenheit to Celsius,
$$
\begin{array}{l}
T_{c}=\left(\frac{5}{9}\right)\left(T_{f}-32^{\circ}\right) \\
T_{c}=\left(\frac{5}{9}\right)\left(80^{\circ}-32^{\circ}\right)
\end{array}
$$
$T_{c}=26.66^{\circ} \mathrm{C}$
Converting $\mathrm{C}$ to $\mathrm{K}$,
$T=273+26.66$
$T=299.66 \mathrm{~K}$
Converting Kelvin to Rankin,
$1 \mathrm{~K}=1.8^{\circ} \mathrm{R}$
But,
$299.66 \mathrm{~K}=299.66 \times 1.8^{\circ} \mathrm{R}$
$T=539.388^{\circ} \mathrm{R}$
Mass of the gas $=\frac{\text { Weight }}{\text { Acceleration due to gravity }}$
$m=\frac{0.3}{32.2}$
$m=9.316 \times 10^{-3}$ slug
Density of gas $=\frac{\text { Mass }}{\text { Volume }}$
$\rho=\frac{9.316 \times 10^{-3}}{2}$
Density of gas $(\rho)=4.658 \times 10^{-3}$ slug $/ \mathrm{ft}^{3}$
From the relation,
$P=R T$
$26.7=4.658 \times 10^{-3} \times R \times 539.388$
$$
R=\frac{26.7}{4.658 \times 10^{-3} \times 539.388}
$$
$R=10.626$
Converting inches to feet,
$R=10.626 \times(12)^{2}$
$R=1530.28 \frac{\mathrm{lb}-\mathrm{ft}}{\text { slug }-{ }^{\circ} \mathrm{R}}$
So, from the table we can observe that the gas constant that we obtained is nearer to the gas
constant of Oxygen rather than Helium.
Hence, the gas in the closed tank is oxygen.
Work Step by Step
Solution:
Volume of the tank $=2 \mathrm{ft}^{3}$
Weight of gas $=0.3 \mathrm{lb}$
Pressure exerted by the gas, $P_{\text {gauge }}=12 \mathrm{psi}$
Considering atmospheric pressure, $P_{\text {atom }}=14.7$ psi
Absolute pressure, $P_{\text {abs }}=P_{\text {atm }}+P_{\text {gauge }}$
$P_{\text {abs }}=12+14.7$
$P_{\text {abs }}=26.7 \mathrm{psi}$
Gas temperature $=80 \mathrm{~F}$
Converting Fahrenheit to Celsius,
$$
\begin{array}{l}
T_{c}=\left(\frac{5}{9}\right)\left(T_{f}-32^{\circ}\right) \\
T_{c}=\left(\frac{5}{9}\right)\left(80^{\circ}-32^{\circ}\right)
\end{array}
$$
$T_{c}=26.66^{\circ} \mathrm{C}$
Converting $\mathrm{C}$ to $\mathrm{K}$,
$T=273+26.66$
$T=299.66 \mathrm{~K}$
Converting Kelvin to Rankin,
$1 \mathrm{~K}=1.8^{\circ} \mathrm{R}$
But,
$299.66 \mathrm{~K}=299.66 \times 1.8^{\circ} \mathrm{R}$
$T=539.388^{\circ} \mathrm{R}$
Mass of the gas $=\frac{\text { Weight }}{\text { Acceleration due to gravity }}$
$m=\frac{0.3}{32.2}$
$m=9.316 \times 10^{-3}$ slug
Density of gas $=\frac{\text { Mass }}{\text { Volume }}$
$\rho=\frac{9.316 \times 10^{-3}}{2}$
Density of gas $(\rho)=4.658 \times 10^{-3}$ slug $/ \mathrm{ft}^{3}$
From the relation,
$P=R T$
$26.7=4.658 \times 10^{-3} \times R \times 539.388$
$$
R=\frac{26.7}{4.658 \times 10^{-3} \times 539.388}
$$
$R=10.626$
Converting inches to feet,
$R=10.626 \times(12)^{2}$
$R=1530.28 \frac{\mathrm{lb}-\mathrm{ft}}{\text { slug }-{ }^{\circ} \mathrm{R}}$
So, from the table we can observe that the gas constant that we obtained is nearer to the gas
constant of Oxygen rather than Helium.
Hence, the gas in the closed tank is oxygen.
