Question 1.39: A 69-m^3 helium tank is compressed under a pressure of 800 k...

(a) To determine the density of the helium in the tank, one may apply either Equation
1.104 p=\rho RT or Equation 1.117 pV=NR_{u} T. Equation 1.117 pV=NR_{u} T is applied in this Example Problem as follows:

which is solved for N, the number of moles of helium, as follows:

N:=kg\frac{m}{sec^{2} }                   p:=800\frac{kN}{m^{2} }                  V:=69m^{3}

T:=10^{\circ } C=283.15K                     R_{u}:=8314\frac{Nm}{kgmolK}

Guess value:                    N_{m} :=1kgmol

Given

p V=N_{m}R_{u}T
N_{m}:=Finf(N_{m})=23.448kgmol

Note that the number of moles, N is called N_{m} in Mathcad in order to distinguish between it and the force units of Newton, N. Then, the density, ρ of the helium is 78 Fluid Mechanics for Civil and Environmental Engineers computed by applying Equation 1.33 \rho =\frac{M}{V} , as follows:

where the mass, M of the helium is computed by applying Equation 1.107 m=\frac{M}{N} as follows:

where the molar mass, m for helium is read from Table A.5 in Appendix A as follows:

mm:=4.003 \frac{kg}{kgmol}                   M:=mm       N_{m}=93.864 kg
\rho :=\frac{M}{V} =1.36\frac{kg}{m^{3} }

Note that the molar mass, m is called mm in Mathcad in order to distinguish between it and the mass units of meters, m.
(b) To determine the weight of the helium in the tank, Equation 1.44 \gamma =\frac{W}{V}=\frac{Mg}{V}=\rho g is applied as follows:

\gamma =\frac{W}{V} =\frac{Mg}{V} \rho g
g:=9.81\frac{m}{sec^{2} }                  \gamma :=\rho g=13.345\frac{N}{m^{3} }                   W:=\gamma           V=920.804N