91Ó°ÊÓ

The temperature $T=0°\text{C}=273\text{K}$.

1. The increase in the temperature is $∆T=1K$.

By differentiating the equation of speed of sound in air with respect to temperature, from problem 19-91, find the speed of sound increase for each increase in air temperature by 1 ⁰C.

From problem 19-91, the speed of sound in air is,

$v_s=\sqrt{\frac{\mathrm{γ}RT}{M}}…………………………………….\left(1\right)$

where,$\mathrm{γ}=C_p/C_V$ , R is the gas constant, T is the temperature, vis velocity and M is the molar mass.

Differentiating Eq. (1) with respect to T,

$$\begin{gathered} \frac{d{v}_s}{dT}=\frac{1}{2}\sqrt{\frac{\mathrm{γ}R}{M}}{T}^{\left(-\frac{1}{2}\right)} \\ \frac{d{v}_s}{dT}=\frac{1}{2T}\sqrt{\frac{\mathrm{γ}RT}{M}} \end{gathered}$$

From Eq. (1),

$\frac{d{v}_s}{dT}=\frac{v_s}{2T}$

Therefore,

$d{v}_s=\frac{dT}{2T}{v}_s$

Therefore, sound’s speed is changed by-,

$∆v_s=\frac{∆T}{2T}{v}_s$

At ,$T=0°\text{C}=273\text{K}$ the speeds of the sound is $v_s=331\text{m}/\text{s}$. With$∆T=1$

For the given values the above equation becomes.

$$\begin{gathered} \mathrm{Δ}{v}_s=\frac{1\text{K}}{2×273\text{K}}×331\text{m}/\text{s} \\ =0.606\text{m}/\text{s} \\ =0.61\text{m}/\text{s} \end{gathered}$$

Hence, the speed of sound increases for each increase in air temperature by 1 ⁰Cis$\mathrm{Δ}{v}_s=0.61\text{m}/\text{s}$.