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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