Question

Question: Air at 0.00 ⁰C and 1 atm pressure has a density of $\mathbf{1}\mathbf{.}\mathbf{29}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}\mathbf{}}\mathbf{}\mathit{g}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{3}}$, and the speed of sound is 331 m/s at that temperature. Compute the ratio $\lambda$of the molar specific heats of air. (Hint: See Problem 91)

Short answer

Answer

the ratio $\gamma$ of the two molar-specific heats ,$\gamma =1.40$ .

Step-by-step solution

Step 1: Given

1. The pressure is $p=1.00\text{atm}=1.01\times {10}^5\text{Pa}$.
2. The speed of sound is $v_s=331\text{m}/\text{s}$.
3. The air at temperature$0.000°\text{C}$ .

The density is $\rho =1.29\times {10}^{-3}\text{g}/{\text{cm}}^3=1.29\text{kg}/{\text{m}}^3$.

Determining the concept

By using the formula, from problems19-91,for the speed of the sound in gas, find the ratio of the molar-specific heat at constant pressure to that at constant volume, The speed with which the sound travels in gas is given as,

${\mathbf{v}}_{\mathbf{s}}\mathbf{=}\sqrt{\frac{\mathbf{\gamma p}}{\mathbf{\rho }}}$

where, ${\mathbf{v}}_{\mathbf{s}}$is velocity,$\rho$is the density, and p is pressure.

Determining the ratio 

From the problems19-91, The speed with which the sound travels in gas is given as,

${\text{v}}_s=\sqrt{\frac{\gamma \text{p}}{\rho }}$

Where $\gamma ={\text{C}}_p/{\text{C}}_V$

Squaring both sides,

${\text{v}}_s^2=\frac{\gamma \text{p}}{\rho }$

Therefore, $\lambda$ is given by,

$\gamma =\frac{\rho v_s^2}{p}$

$$\begin{gathered} \gamma =\frac{1.29\text{kg}/{\text{m}}^3\times {\left(331\text{m}/\text{s}\right)}^2}{1.01\times {10}^5\text{Pa}} \\ \gamma =1.40 \end{gathered}$$

Hence, the ratio of the molar-specific heat at constant pressure to that at constant volume, of air is $\gamma =1.40$.