Question 11.13: Using the compressibility charts, find the pressure exerted ...

From Table C.12a, we find that

T_c = 240.  \text{R}

p_c = 507  \text{psia}

and

v_c = \frac{1.49  \text{ft³/lbmole}}{28.011  \text{lbm/lbmole}} = 0.053  \text{ft³/lbm}

Also, from Table C.13a, we find that R = 0.0709  \text{Btu/lbm.R}. Then, we have

T_R = \frac{T}{Tc} = \frac{−78.0 + 460.}{240.} = 1.60

and

v =\frac{1.00  \text{ ft³}}{8.20  \text{lbm}} = 0.122  \text{ft³/lbm}

with

v′_c = \frac{RT_c}{p_c} = \frac{[0.0709  \text{Btu/(lbm.R)]}(240.  \text{R})(778.16  \text{ft.lbf/Btu})}{(507  \text{lbf /in²})(144  \text{in²/ft²})} = 0.181  \text{ ft³/lbm}

so that

v′_R = \frac{v}{v′_c} = \frac{0.122}{0.181} = 0.67 (notice that we do not use the actual critical specific volume v_c here)

Using T_R = T/T_c = 1.60 and v′_R = v/v′_ c = 0.67, we find from Figure 11.6 that

p_R = \frac{p}{p_c} = 2.10    and    Z = 0.850

Then, we can calculate

p = p_cp_R = 507 (2.10) = 1070  \text{psia}