Question 1.38: A 300-ft^3 oxygen tank is compressed under a pressure of 500...

(a) To determine the density of the oxygen in the tank, one may apply either Equation 1.104 p=\rho RT or Equation 1.117 pV=NR_{u} T. Equation 1.104 is applied in this Example Problem as follows:
p=\rho RT
where the gas constant, R for oxygen is read from Table A.5 in Appendix A as follows:

slug:=1lb\frac{sec^{2} }{ft}                 P:=500psi=7.2\times 10^{4} psf

T:=40^{\circ } =499.67^{\circ } R                R:=1554\frac{ft lb}{slug^{\circ } R}

Guess value:                          \rho :=1\frac{slug}{ft^{3} }

Given

P=\rho RT
\rho :=Find (\rho )0.093\frac{slug}{ft^{3} }

(b) To determine the weight of the oxygen in the tank, Equation 1.44 \gamma =\frac{W}{V}=\frac{Mg}{V}=\rho g is applied as follows:

V:=300ft^{3}                 g:=32.2\frac{ft}{sec^{2} }

\gamma :=\rho         g=2.991 \frac{lb}{ft^{3} }                 W:=\gamma V=897.385lb