Chapter 15, Problem 34P

(a)

To determine

To calculate:

Calculate the derivative of speed of the sound in air as respect to absolute temperature.

Answer to Problem 34P

Derivative of the speed of the sound in air as respect to absolute temperature is $\frac{\text{1}}{\text{2}}\frac{\text{dT}}{\text{d}}$ .

Explanation of Solution

Given:

Differentials $\text{dv}$ and ${\text{dF}}_{\text{T}}$ obey $\frac{\text{dv}}{\text{v}}\text{=}\frac{\text{1}}{\text{2}}{\text{dF}}_{\text{T}}{\text{/F}}_{\text{T}}$ .

Formula used:

$\text{dv/dT}$ is evaluated as,

  $\frac{\text{dv}}{\text{dT}}\text{=}\frac{\text{d}}{\text{dT}}\left[\sqrt{\frac{\text{γRT}}{\text{M}}}\right]$

Calculation:

The speed of sound in a gas is given by $\text{v=γRTM}$ .

Where,

  $\text{R=}$ Gas constant

  $\text{T=}$ Absolute temperature

  $\text{M=}$ Molecular mass of the gas

  $\text{γ=}$ Constant that is characteristic of the particular molecular structure of the gas.

To estimate the percentage change in the speed of sound if the temperature increases from $\text{0}\text{°C}$ to $\text{27}\text{°C}$ then $\frac{\text{dT}}{\text{ΔT}}$ is evaluate as $\frac{\Delta v}{v}$

For evaluating the $\text{dv/dT}$ is,

  $\begin{array}{l}\frac{\text{dv}}{\text{dT}}\text{=}\frac{\text{d}}{\text{dT}}\left[\sqrt{\frac{\text{γRT}}{\text{M}}}\right]\\ \text{=}\frac{\text{1}}{\text{2}}\sqrt{\frac{\text{M}}{\text{γRT}}}\left(\frac{\text{γR}}{\text{M}}\right)\\ \text{=}\frac{\text{1}}{\text{2}}\frac{\text{v}}{\text{T}}\end{array}$

Now, to separate the variables to obtain,

  $\frac{\text{dv}}{\text{v}}\text{=}\frac{\text{1}}{\text{2}}\frac{\text{dT}}{\text{d}}$

Conclusion:

Derivative of the speed of the sound in air as respect to absolute temperature is $\frac{\text{1}}{\text{2}}\frac{\text{dT}}{\text{d}}$ .

(b)

To determine

To calculate:

The percentage change in speed of the sound when temperature changes from $0$ to $\text{27}\text{°C}$

Answer to Problem 34P

The percentage change in speed of the sound when temperature changes from $0$ to $\text{27}\text{°C}$ $\text{5}\text{.0}\text{\%}\text{.}$

Explanation of Solution

Given:

Differentials $\text{dv}$ and ${\text{dF}}_{\text{T}}$ obey $\frac{\text{dv}}{\text{v}}\text{=}\frac{\text{1}}{\text{2}}{\text{dF}}_{\text{T}}{\text{/F}}_{\text{T}}$ .

Temperature $\text{1=0°}\text{C=273}\text{K}$

Temperature $\text{2=27°}\text{C=300}\text{K}$

Formula used:

  $\frac{\text{dv}}{\text{v}}\text{=}\frac{\text{1}}{\text{2}}\frac{\text{dT}}{\text{T}}$

Calculation:

The given equation is:

  $\frac{\text{dv}}{\text{v}}\text{=}\frac{\text{1}}{\text{2}}{\text{dF}}_{\text{T}}{\text{/F}}_{\text{T}}$

First differentiate the expression with respect to $\text{T}$ and then separate the variables to show that the differentials satisfy $\frac{\text{dv}}{\text{v}}\text{=}\frac{\text{1}}{\text{2}}\frac{\text{dT}}{\text{T}}$ .

To estimate the percentage change in the speed of sound if the temperature increases from $\text{0}\text{°C}$ to $\text{27}\text{°C}$ then $\frac{\text{dT}}{\text{ΔT}}$ is evaluate as $\frac{\Delta v}{v}$

Approximate the $\text{dT}$ with $\text{ΔT}$ and $\text{dv}$ with $\text{Δv}$ ,

Put the numerical values to get,

  $\begin{array}{l}\frac{\text{Δv}}{\text{v}}\text{=}\frac{\text{1}}{\text{2}}\left(\frac{\text{300}\text{K-273}\text{K}}{\text{273}\text{K}}\right)\\ \frac{\text{Δv}}{\text{v}}\text{=5}\text{.0}\text{\%}\text{.}\end{array}$

Conclusion:

Thus, the percentage change in speed of the sound when temperature changes from $0$ to $\text{27}\text{°C}$ $\text{5}\text{.0}\text{\%}\text{.}$

(c)

To determine

To calculate:

Calculate the value at $\text{27}\text{°C}$ using the differential approximation.

Answer to Problem 34P

The value at $\text{27}\text{°C}$ using the differential approximation is $\text{347}\text{m/s}$ .

Explanation of Solution

Given:

Speed of the sound $\text{=331}\text{m/s}$ .

Temperature $\text{=0°}\text{C}$

Formula used:

  ${\text{v}}_{\text{FT}}{\text{=v}}_{\text{IT+}}{\text{v}}_{\text{IT}}\frac{\text{Δv}}{\text{v}}$

Calculation:

According to the question, ${\text{ΔF}}_{\text{T}}$ is evaluated as,

  ${\text{v}}_{\text{FT}}{\text{=v}}_{\text{IT+}}{\text{v}}_{\text{IT}}\frac{\text{Δv}}{\text{v}}$

Using the differential approximation, approximate the speed of sound at $\text{300}\text{K}$ .

  $\begin{array}{l}{\text{v}}_{\text{FT}}{\text{=v}}_{\text{IT+}}{\text{v}}_{\text{IT}}\frac{\text{Δv}}{\text{v}}\\ {\text{v}}_{\text{300}\text{K}}\approx {\text{v}}_{\text{273}\text{K}}{\text{+v}}_{\text{273}\text{K}}\frac{\text{Δv}}{\text{v}}\\ {\text{v}}_{\text{300}\text{K}}{\text{=v}}_{\text{273}\text{K}}\left(\text{1+}\frac{\text{Δv}}{\text{v}}\right)\end{array}$

Now, put the numerical values and evaluate the ${\text{v}}_{\text{300}\text{K}}$ ,

  $\begin{array}{l}{\text{v}}_{\text{300}\text{K}}\text{=}\left(\text{331}\text{m/s}\right)\left(\text{1+0}\text{.0495}\right)\\ {\text{v}}_{\text{300}\text{K}}\text{=347}\text{m/s}\end{array}$

Conclusion:

Thus, the value at $\text{27}\text{°C}$ using the differential approximation is $\text{347}\text{m/s}$ .

(d)

To determine

To explain:

Calculate an approximation comparison with result of an exact calculation.

Answer to Problem 34P

Approximation comparison with result of an exact calculation $\left({\text{v}}_{\text{300}\text{K}}\right)$ is $\text{347}\text{m/s}$ .

Explanation of Solution

Given:

Speed of the sound $\text{=331}\text{m/s}$ .

Temperature $\text{=0°}\text{C}$

Formula used:

The speed of sound wave at the absolute temperature is:

  $\text{v =}\sqrt{\frac{\text{γRT}}{\text{M}}}$

Here,

Molecular mass of hydrogen: $\text{M}$

Constant (hydrogen is diatomic gas): $\text{γ}$

Absolute temperature: $\text{T}$

Gas constant: $\text{R}$

Calculation:

The speed of sound wave at the temperature $\text{300}\text{K}$ is:

  $\begin{array}{l}\text{v =}\sqrt{\frac{\text{γRT}}{\text{M}}}\\ {\text{v}}_{\text{300}\text{K}}\text{=}\sqrt{\frac{\text{γR}\left(\text{300}\text{K}\right)}{\text{M}}}\end{array}$

The speed of sound wave at the temperature $\text{273}\text{K}$ is:

  $\begin{array}{l}\text{v =}\sqrt{\frac{\text{γRT}}{\text{M}}}\\ {\text{v}}_{\text{273}\text{K}}\text{=}\sqrt{\frac{\text{γR}\left(\text{273}\text{K}\right)}{\text{M}}}\end{array}$

Now, divide the first of these equations by the second and solve for ${\text{v}}_{\text{300}\text{K}}$

  $\begin{array}{l}\frac{{\text{v}}_{\text{300}\text{K}}}{{\text{v}}_{\text{273}\text{K}}}\text{=}\frac{\sqrt{\frac{\text{γR}\left(\text{300}\text{K}\right)}{\text{M}}}}{\sqrt{\frac{\text{γR}\left(\text{273}\text{K}\right)}{\text{M}}}}\\ \frac{{\text{v}}_{\text{300}\text{K}}}{{\text{v}}_{\text{273}\text{K}}}\text{=}\sqrt{\frac{\left(\text{300}\text{K}\right)}{\left(\text{273}\text{K}\right)}}\end{array}$

And,

  $\begin{array}{l}{\text{v}}_{\text{300}\text{K}}\text{=}\left(\text{331}\text{m/s}\right)\sqrt{\frac{\text{300}}{\text{273}}}\\ {\text{v}}_{\text{300}\text{K}}\text{=347}\text{m/s}\end{array}$

Conclusion:

Approximation comparison with result of an exact calculation $\left({\text{v}}_{\text{300}\text{K}}\right)$ is $\text{347}\text{m/s}$ .