Question

a core ring with diameter 0f 4cm and a height of 5cm, the weight of the core ring with the undisturbed soil is 120g. the weight of the core ring with soil after oven drying for 24 hours is 105g. the volume occupied by air (Va)in the undisturbed soil sample is 16cm3. taking a particle density of the soil to be 2.65g/cm3, calculate the porosity and gravitational volume?

Answer 1

Answer on Question #82336, Chemistry/ General Chemistry

a core ring with diameter 0f 4cm and a height of 5cm, the weight of the core ring with the undisturbed soil is 120g. the weight of the core ring with soil after oven drying for 24 hours is 105g. the volume occupied by air (Va)in the undisturbed soil sample is 16cm3. taking a particle density of the soil to be $2.65\mathrm{g/cm^3}$ , calculate the porosity and gravitational volume?

Solution

Porosity, $n = \frac{V_{\mathrm{v}}}{V_{\mathrm{t}}}$ , where $V_{\mathrm{v}}$ - volume of voids $(V_{\mathrm{w}} + V_{\mathrm{a}})$ , $V_{\mathrm{t}}$ - total volume

$$V _ {t} = V _ {\text {c o r e r i n g}} = \pi r ^ {2} \times h = \pi (d / 2) ^ {2} \times h = 3. 1 4 \times (4 / 2) ^ {2} \times 5 = 6 2. 8 \mathrm {c m} ^ {3}$$ $$V _ {a} = 1 6 \mathrm {c m} ^ {3}$$ $$V _ {w} = \frac {M _ {w}}{\rho_ {w}}$$ $$M _ {w} = M _ {t} - M _ {s} = 1 2 0 \mathrm {g} - 1 0 5 \mathrm {g} = 1 5 \mathrm {g}$$ $$V _ {w} = \frac {1 5 g}{1 g / c m ^ {3}} = 1 5 c m ^ {3}$$ $$V _ {v} = V _ {w} + V _ {a} = 1 6 \mathrm {c m} ^ {3} + 1 5 \mathrm {c m} ^ {3} = 3 1 \mathrm {c m} ^ {3}$$ $$n = \frac {3 1}{6 2 . 8} = 0. 4 9$$

Gravitational volume :

$$\rho_ {s} = \frac {M _ {s}}{V _ {s}} = G _ {s} \rho_ {w}$$ $$\therefore V _ {s} = \frac {M _ {s}}{G _ {s} \rho_ {w}} = \frac {1 2 0}{2 . 6 5 \times 1} = 4 5. 2 8 c m ^ {3}$$

Answer: 0.49 ; 45.28 cm³

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