Chapter 13, Problem 106P

To determine

The decrease in the rms speed of the air molecules in lungs.

Answer to Problem 106P

The decrease in the rms speed of the air molecules in lungs is $-25\text{m/s}$.

Explanation of Solution

The temperature change of ground squirrel during hibernation is from $40.0\text{°C}$ to $10.0\text{°C}$. The air in the lungs of the squirrel is composed of $75.0\%$ nitrogen and $25.0\%$ of oxygen.

Write the formula for the rms speed.

$v_{\text{rms}}=\sqrt{\frac{3kT}{m}}$ (I)

Here, $v_{\text{rms}}$ is the rms speed, $k$ is the Boltzmann’s constant, $T$ is the temperature, $m$ is the mass.

Refer equation (I) and write an equation for the change in rms speed.

$\Delta v_{\text{rms}}=\sqrt{\frac{3k{T}_f}{m}}-\sqrt{\frac{3k{T}_i}{m}}$ (II)

Here, $\Delta v_{\text{rms}}$ is the change in rms speed, $T_f$ is the final temperature, $T_i$ is the initial temperature.

Write the formula for the mass of air.

$M=\left(0.75\right)\left(2\times M_N\right)+\left(0.25\right)\left(2\times M_O\right)$ (III)

Here, $M_N$ is the molar mass of nitrogen, $M_O$ is the molar mass of oxygen.

Substitute equation (III) in equation (II).$\Delta v_{\text{rms}}=\sqrt{\frac{3k{T}_f}{\left(0.75\right)\left(2\times M_N\right)+\left(0.25\right)\left(2\times M_O\right)}}-\sqrt{\frac{3k{T}_i}{\left(0.75\right)\left(2\times M_N\right)+\left(0.25\right)\left(2\times M_O\right)}}$ (IV)

Conclusion:

Substitute $1.38\times {10}^{-23}\text{J/K}$ for $k$, $40.0\text{°C}$ for $T_i$, $10.0\text{°C}$ for $T_f$, $14.0067\text{u}$ for $M_N$, $15.9994\text{u}$ for $M_O$.

$\begin{array}{c}\Delta v_{\text{rms}}=\left\{\begin{array}{l}\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{J/K}\right)\left(10.0\text{°C}\right)}{\left(0.75\right)\left(2\times 14.0067\text{u}\right)+\left(0.25\right)\left(2\times 15.9994\text{u}\right)}}\\ -\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{J/K}\right)\left(40.0\text{°C}\right)}{\left(0.75\right)\left(2\times 14.0067\text{u}\right)+\left(0.25\right)\left(2\times 15.9994\text{u}\right)}}\end{array}\right\}\\ =\left\{\begin{array}{l}\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{J/K}\right)\left(10.0+273\text{K}\right)}{\left[\left(0.75\right)\left(2\times 14.0067\text{u}\right)+\left(0.25\right)\left(2\times 15.9994\text{u}\right)\right]\left(1.66\times {10}^{-27}\text{kg/u}\right)}}\\ -\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{J/K}\right)\left(40.0+273\text{K}\right)}{\left(0.75\right)\left(2\times 14.0067\text{u}\right)+\left(0.25\right)\left(2\times 15.9994\text{u}\right)\left(1.66\times {10}^{-27}\text{kg/u}\right)}}\end{array}\right\}\\ =493\text{m/s}-518\text{m/s}\\ =-25\text{m/s}\end{array}$

The decrease in the rms speed of the air molecules in lungs is $-25\text{m/s}$.