Question 1.42: Refer to Example Problem 1.38, where for a 300-ft^3 oxygen t...

(a) To determine the resulting density of the oxygen in the tank, one must note that the mass of the oxygen in Equation 1.33 \rho =\frac{M}{V} , and thus, the weight of the oxygen Equation 1.44 \gamma =\frac{W}{V}=\frac{Mg}{V}=\rho g remains a constant as follows:

slug:=1lb\frac{sec^{2} }{ft}                P_{1} :=500psi=7.2\times10^{4} psf               V_{1} =300ft^{3}

g:=32.2\frac{ft}{sec^{2}}                P_{1} :=0.093 \frac{slug}{ft^{3} }                W_{1}:=897.385lb

V_{2} =100ft^{3}               W_{2}:=W_{1}=897.385lb               \gamma _{2} :=\frac{W_{2}}{V_{2}} =8.974\frac{lb}{ft^{3}}
\rho _{2} :=\frac{\gamma_{2} }{g} =0.279\frac{slug}{ft^{3} }

(b) Given compression under isothermal conditions (constant temperature), to determine the resulting pressure of the oxygen in the tank, one may apply Equation 1.155, \frac{p}{\rho } =C as follows:

Guess value:    P_{2}:=1psf

Given

(c) Given compression under isothermal conditions (constant temperature), the resulting bulk modulus of elasticity of the oxygen in the tank is equal to the absolute pressure of the oxygen in Equation 1.157 E_{\upsilon } =\frac{dp}{-\frac{dV}{V} } =\frac{dp}{\frac{d\rho }{\rho } } =\rho \frac{dp}{d\rho } =\rho \frac{d(C\rho )}{d\rho } =C\rho \frac{d\rho }{d\rho} =C\rho =p. Thus: