Answer:
a. 0.32 b. 1448 m/s
Explanation:
We know v ∝ √T where v = velocity of sound and T = absolute temperature.
Let v₁ = velocity of sound at 17°, v₁ = fλ where f = frequency of sound = 512 Hz and λ = 66.5 cm = 0.665 m
So, v₁ = fλ = 512 Hz × 0.665 m = 340.48 m/s
T₁ = 17 + 273 = 290 K
Let v₂ = velocity of sound in air at NTP = unknown and T₂ = temperature at NTP = 0°C + 273 = 273 K
Now v₁/v₂ = √T₁/√T₂
So, v₂ = (√T₂/√T₁)v₁
= [√(T₂/T₁)]v₁
substituting the values of the variables, we have
v₂ = [√(273 K/290 K)]340.48 m/s
v₂ = [√0.9413]340.48 m/s
v₂ = (0.9702)340.48 m/s
v₂ = 330.35 m/s
Also v = √(γP/ρ) where v = velocity of sound in air at NTP = 330.35 m/s, γ = ratio of molar heat capacities, P = pressure at NTP = 1.013 × 10⁵ Pa and ρ = density of air = 1.293 kg/m³
Since, v = √(γP/ρ)
making γ subject of the formula, we have
γ = v²ρ/P
substituting the values of the variables, we have
γ = (330.35 m/s)² × 1.293 kg/m³/1.013 × 10⁵ Pa
= 31975.36 kg/m²s² ÷ 1.013 × 10⁵ Pa
= 0.32
b. Speed of sound in mercury v₃ = √(B/ρ) where B = Bulk modulus of mercury = 28.5 × 10⁹ Pa and ρ = density of mercury = 13600 kg/m³
v₃ = √(B/ρ)
= √(28.5 × 10⁹ Pa/13600 kg/m³)
= √(28.5 × 10⁹ Pa/13.6 × 10³ kg/m³)
= √(2.096 × 10⁶) m/s
= 1.448 × 10³ m/s
= 1448 m/s
