Question

The density of air at NTP is $1.293kg{m}^{-3}$ and density of mercury at $0^{\circ }C$ is $13.6\times {10}^{3}kg{m}^{-3}$. If $C_p=0.2417calk{g}^{-10}{C}^{-1}$ and $C_v=0.1715$, the speed of sound in air at ${100}^{\circ }C$ will be $(g=9.8Nk{g}^{-1})$

• A. $260m{s}^{-1}$
• B. $332m{s}^{-1}$
• C. $350.2m{s}^{-1}$
• D. $369.4m{s}^{-1}$

Answer 1

Answer: (D)

$369.4m{s}^{-1}$

NTP conditions: $T={25}^{o}C=298.15KP=1bar={10}^{5}Pa$

Given: Density of air at NTP $\rho =1.293kg/m^3$

$\gamma =\frac{C_p}{C_v}=\frac{0.2417}{0.1715}=1.4$

Speed of sound in air at NTP, $v_{{25}^{o}C}=\sqrt{\frac{\gamma P}{\rho }}=\sqrt{\frac{1.4\times {10}^5}{1.293}}=330.15m/s$

Let speed of sound in air at ${100}^{o}C$ be $v_{{100}^{o}C}$

As $v\propto \sqrt{T}$

Thus $\frac{v_{{100}^{o}C}}{v_{{25}^{o}C}}=\sqrt{\frac{373.15}{298.15}}=1.118$

$⟹v_{{100}^{o}C}=1.118\times 330.15=369.35m/s$