Determine the approximate density of a Ti-6Al-4V titanium alloy that has a composition of 90 wt% Ti, 6 wt% Al, and 4 wt% V. [Hint: look in the book on how to use density data in composition calculations] (3 points)

ρ = 4407.03 kg/m³

Explanation:

The Density of a metal alloy is given by the following equation:

1/ρ = m₁/ρ₁ + m₂/ρ₂ + m₃/ρ₃

where,

ρ = density of allow = ?

ρ₁ = density of Titanium (Ti) = 4540 kg/m³

ρ₂ = density of Aluminum (Al) = 2710 kg/m³

ρ₃ = density of Vanadium (V) = 6110 kg/m³

m₁ = mass fraction of Titanium (Ti) = 90% = 0.9

m₂ = mass fraction of Aluminum (Ti) = 6% = 0.06

m₁ = mass fraction of Vanadium (V) = 4% = 0.04

Therefore,

1/ρ = 0.9/(4540 kg/m³) + (0.06)/(2710 kg/m³) + (0.04)/(6110 kg/m³)

1/ρ = 1.9823 x 10⁺⁴ m³/kg + 0.2214 x 10⁻⁴ m³/kg + 0.0654 x 10⁻⁴ m³/kg

1/ρ =  2.2691 x 10⁻⁴ m³/kg

ρ = 1/(2.2691 x 10⁻⁴ m³/kg)

ρ = 4407.03 kg/m³

answer

i think the answer is a) as air pressure in an area increases, the density of the gas particles in that area decreases.

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