Question

Calculate the ratio of the speed of sound in neon to that in water vapour at any temperature. Given that the molecular weight of neon gas $=2.02\times 10^{- 2} \, kg \, mole^{- 1}$ and molecular weight of water vapour $=1.8\times 10^{- 2} \, kg \, mole^{- 1}$ . $\gamma _{\text{neon}}=\frac{5}{3}$ and $\gamma $ for water vapour $=\frac{4}{3}$ .

• A. $1.1$
• B. $1.7$
• C. $1.9$
• D. $1.3$

Answer 1

Answer: (A)

Let v₁ and v₂ be the speeds of sound in neon and water vapour respectively. Suppose that M₁ and M₂ are the molecular weights of neon and water vapour respectively. If γ is the ratio of the two specific heats, then
$\frac{\upsilon_{1}}{\upsilon_{2}} = \sqrt{\frac{\text{RT } \gamma _{\text{neon}}}{\text{M}_{1}}} \times \sqrt{\frac{\text{M}_{2}}{\text{RT } \gamma _{\text{water vapour}}}} = \sqrt{\frac{\text{M}_{2}}{\text{M}_{1}}} \times \sqrt{\frac{\gamma _{\text{neon}}}{\gamma _{\text{water vapour}}}}$
Here, $M_{1}=2.02\times 10^{- 2}kgmole^{- 1}$ ;
$M_{2}=1.8\times 10^{- 2}kgmole^{- 1}$
$\therefore \frac{\upsilon_{1}}{\upsilon_{2}} = \sqrt{\frac{1 \cdot 8 \times 1 0^{- 2} \times 5 / 3}{2 \cdot 0 2 \times 1 0^{- 2} \times 4 / 3}} = 1 \text{.} 1$