Chapter 13, Problem 55P

To determine

The diameter of bubble as its reaches the surface.

Answer to Problem 55P

The diameter is $2.1\text{mm}$.

Explanation of Solution

Depth of lake is $80.0\text{m}$, temperature at the bottom of lake is $4°C$, water temperature at the surface is $18°\text{C}$, and the initial diameter of bubble is $1.00\text{mm}$.

The number of moles will be same for at any point.

Write the ideal gas law in case of bubble.

$n_{i}R=n_{f}R$ (I)

Here, $n_i$ is the initial number of moles, $n_f$ is the final number of moles, and $R$ is the universal gas constant.

Write the equation for $n_{i}R$ in terms of pressure, temperature, and the volume.

$n_{i}R=\frac{P_i{V}_i}{T_i}$ (II)

Here, $P_i$ is the pressure at bottom, $V_i$ is the volume at bottom, and $T_i$ is the temperature at the bottom.

Write the equation for $n_{f}R$ in terms of pressure, temperature, and the volume.

$n_{f}R=\frac{P_f{V}_f}{T_f}$ (III)

Here, $P_f$ is the pressure at top, $V_f$ is the volume at top, and $T_f$ is the temperature at the top.

Rewrite equation (I) by substituting equations (II) and (III).

$\frac{P_i{V}_i}{T_i}=\frac{P_f{V}_f}{T_f}$ (IV)

Write the equation for $P_i$ in terms of $P_f$.

$P_i=P_f+\rho gh$ (V)

Here, $\rho$ is the density of water, $g$ is the gravitational acceleration, and $h$ is the depth of lake.

Write the equation for $V_f$.

$V_f=\frac{1}{6}\pi d_f^2$ (VI)

Here, $d_f$ is the diameter of bubble at the top.

Write the equation for $V_i$.

$V_i=\frac{1}{6}\pi d_i^2$ (VII)

Here, $d_i$ is the diameter of bubble at the bottom.

Rewrite equation (IV) by substituting equations (V), (VI), and (VII).

$\frac{\left(P_f+\rho gh\right)\left(\frac{1}{6}\pi d_i^2\right)}{T_i}=\frac{P_f\left(\frac{1}{6}\pi d_f^2\right)}{T_f}$

Rewrite the above relation in terms of $d_f$.

$d_f=d_i{\left(\frac{\left(P_f+\rho gd\right)T_f}{P_f{T}_i}\right)}^{1/3}$

Conclusion:

Substitute $1.00\text{mm}$ for $d_i$, $1.0\text{atm}$ for $P_f$, $1.0\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$, $80.0\text{m}$ for $h$, $18°\text{C}$ for $T_f$, and $4°\text{C}$ for $T_i$ in the above equation to find $d_f$.

$\begin{array}{c}{d}_f=\left(1.00\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right){\left(\frac{\left(\begin{array}{l}1.0\text{atm}\left(\frac{1.013\times {10}^5\text{Pa}/\text{atm}}{1.0\text{atm}}\right)+\\ \left(1.0\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\left(80.0\text{m}\right)\end{array}\right)\left(18°\text{C}\right)\left(\frac{273.15\text{K}+18\text{K}}{18°\text{C}}\right)}{1.0\text{atm}\left(\frac{1.013\times {10}^5\text{Pa}/\text{atm}}{1.0\text{atm}}\right)\left(4°\text{C}\right)\left(\frac{273.15\text{K}+4\text{K}}{4°\text{C}}\right)}\right)}^{1/3}\\ =\left(1.00\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right){\left(9.261\right)}^{1/3}\\ =2.1\times {10}^{-3}\text{m}\left(\frac{1\text{mm}}{{10}^{-3}\text{m}}\right)\\ =2.1\text{mm}\end{array}$

Therefore, the diameter is $2.1\text{mm}$.