Question

A 2.8-carat diamond is grown under a high pressure of 58 × 10 9 N / m 2 .

Answer 1

Answer:

a)  ΔV = 2,118 10⁻⁸ m³   b)  ΔR= 0.0143 cm

Explanation:

a) For this part we use the concept of density

    ρ = m / V

As we are told that 1 carat is 0.2g we can make a rule of proportions (three) to find the weight of 2.8 carats

    m = 2.8 Qt (0.2 g / 1 Qt) = 0.56 g = 0.56 10-3 kg

   

    V = m / ρ

    V = 0.56 / 3.52

    V = 0.159 cm3

We use the relation of the bulk module

    B = P / (Δv/V)

    ΔV = V P / B

    ΔV = 0.159 10⁻⁶ 58 10⁹ /4.43 10¹¹

    ΔV = 2,118 10⁻⁸ m³

b) indicates that we approximate the diamond to a sphere

    V = 4/3 π R³

For this part let's look for the initial radius

    R₀ = ∛ ¾ V /π

    R₀ = ∛ (¾ 0.159 /π)

    R₀ = 0.3361 cm

Now we look for the final volume and with this the final radius

    $V_{f}$ = V + ΔV

    $V_{f}$ = 0.159 + 2.118 10⁻²

    $V_{f}$ = 0.18018 cm3

    $R_{f}$ = ∛ (¾ 0.18018 /π)

    $R_{f}$ = 0.3504 cm

The radius increment is

    ΔR = $R_{f}$ - R₀

    ΔR = 0.3504 - 0.3361

    ΔR= 0.0143 cm