Analytical Chem Soil HW Solutions 3

Analytical Chem Soil HW Solutions 3 - case, the soil...

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3 1. Mass of 50ml density bottle empty, g 25.07 2. Mass of 50ml density bottle + 20g of dry soil particles, g 45.07 3. Mass of 50ml density bottle + 20g of dry soil particles, with remainder of space in bottle filled with water, g 87.55 4. Mass of 50ml density bottle filled with water only, g 75.10 Calculate the relative density (specific gravity) of the soil particles. A 1 kg sample of the same soil taken from the ground has a natural water content of 27% and occupies a total volume of 0.52 litre. Determine the unit weight, the specific volume and the saturation ratio of the soil in this state. Calculate also the water content and the unit weight that the soil would have if saturated at the same specific volume, and the unit weight at the same specific volume but zero water content. Q1.3 Solution The particle relative density (grain specific gravity) G s is defined as the ratio of the mass density of the soil grains to the mass density of water. For a fixed volume of solid - in this
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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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This note was uploaded on 12/28/2011 for the course CHM 4302 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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