Unformatted text preview: case, the soil particles - the specific gravity is equal to the mass of the dry soil particles divided by the mass of water they displace. The mass of the dry soil particles is given by (m 2-m 1 ) = 20.00g The mass of water displaced by the soil particles is given by (m 4-m 1 ) - (m 3-m 2 ) = (50.03) - (42.48) = 7.55g G s = (m 2-m 1 )/[(m 4-m 1 )-(m 3-m 2 )] = (20.00g) ÷ (7.55g) = 2.65 For the sample of natural soil, the unit weight is equal to the actual weight divided by the total volume, γ = (1kg × 9.81N/kg 0.001kN/N) (0.52 10-3 m 3 ) ⇒ = 18.865 kN/m 3 The water content w = m w /m s = 0.27. For the 1kg sample, we know that m w +m s = 1kg, hence 1.27 m s = 1kg m s = 0.7874kg and m w = 0.2126kg The volume of water v w = m w / ρ w = 0.2126kg 1kg/litre = 0.2126litre The volume of solids v s = m s / s = 0.7874kg 2.65kg/litre = 0.2971litre The specific volume v is defined as the ratio v t /v s = 0.52litre/0.297litre v = 1.75...
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- Fall '11
- Analytical Chemistry, 0.52 litre, 0.52litre, 0.297litre, 0.2126litre, 0.2971litre