Calculation of specific weight, specific volume, and density, determination of gas constant R

Below are some of the examples worked out to calculate specific weight, specific volume, specific gravity, and density of substance(matter) at given conditions:
Problem 1:
At a temperature of 100° F and 120 psi (pressure per square inch) absolute, calculation of the above properties for methane (CH4):
solution: 
By using equation of state  =   PV= NRT  ( R= 96.3 ft/°R)
specific weight = P/RT = n/V = 120 × 144 / [96.3 (460+100)] = 0.320 lb/ft³  (note  ° R= 460 + ° F)

density = specific weight / gravity = 0.320/32.2 = 0.00994 slug/ft³

specific volume = 1/density = 1/0.00994 = 101 ft³/slug
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Problem 2:
For oil of volume 6 m³ and weighs 47 kN
solution:
specific weight = 47 kN / 6 m³ = 7.833 kN/m³

density =7833/9.81 = 798 kg/m³

specific gravity = specific weight of oil /specific weight of water = 7.833/9.79 =0.800

Note: specific gravity of a liquid is the ratio of the specific weight of a particular liquid to the specific of water under the same conditions, and its specific gravity is as no unit it has only a numerical value.
specific gravity of substance is calculated as: = weight of substance / weight of equal volume of water
= specific weight of substance / specific weight of water
= density of substance / density of water
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Problem 3:
At 90° F and 300psi absolute, find the gas constant and density for a gas that has 11.4 ft³/ lb.
solution:
Data given is the condition of gas i.e temperature and pressure and specific volume
R= P/(specific volume . T) = [(30.0 × 144)11.4]/460+90 = 89.5 ft/° R

density = specific weight / gravity = 1 / ( volume × gravity ) = 1 / (11.4 × 32.2) = 0.00272 slugs/ft³

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