Chapter 19, Problem 53PQ

An air bubble starts rising from the bottom of a lake. Its diameter is 3.60 mm at the bottom and 4.00 mm at the surface. The depth of the lake is 2.50 m, and the temperature at the surface is 40.0°C. What is the temperature at the bottom of the lake? Consider the atmospheric pressure to be 1.01 × 10⁵ Pa and the density of water to be 1.00 × 10³ kg/m³. Model the air as an ideal gas.

53. Use the ideal gas law for the bottom (point 1) and the surface (point 2) of the lake. At the surface, the pressure is atmospheric pressure. However, at the bottom it is equal to to the sum of the atmospheric pressure and the pressure due to 2.50 m column of water.

$\begin{array}{l}{P}_2=1.01\times {10}^5\text{ Pa}\\ P_1=P_2+{\rho }_{W}g{h}_W\\ P_1=1.01\times {10}^5\text{ Pa}+(1.00\times {10}^3{\text{ kg/m}}^3)(9.81{\text{ m/s}}^2)(2.50\text{ m})\end{array}$

Use the ideal gas law (Eq. 19.17).

$T_1=\frac{P_1{V}_1}{P_2{V}_2}{T}_2$

The volume ratio at the bottom and top of the lake can be calculated with the diameters given.

$\frac{V_1}{V_2}=\frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3}={\left(\frac{1.8}{2.0}\right)}^3$

$\begin{array}{l}{T}_1=\frac{P_1}{P_2}\left(\frac{V_1}{V_2}\right)T_2\\ T_1=\frac{1.01\times {10}^5\text{ Pa}+(1.00\times {10}^3{\text{ kg/m}}^3)(9.81{\text{ m/s}}^2)(2.50\text{ m})}{1.01\times {10}^5\text{ Pa}}{\left(\frac{1.80}{2.00}\right)}^3(40.0+273.15\text{ K})\\ T_1=\boxed{284\text{ K}}\end{array}$

To determine

The temperature at the bottom of the lake.

Answer to Problem 53PQ

The temperature at the bottom of the lake is $284\text{K}$.

Explanation of Solution

The pressure at the surface of the lake is the atmospheric pressure and the pressure at the bottom of the lake is the sum of atmospheric pressure and pressure due to the water column.

Write an expression for the pressure at the bottom of the lake.

    $P_1=P_{atm}+P_w$                                                                                                          (I)

Here, $P_1$ is the pressure at the bottom of the lake, $P_{atm}$ is the atmospheric pressure and $P_w$ is the pressure due to the water column.

Write an expression for the pressure due to the water column.

    $P_w={\rho }_{w}g{h}_w$                                                                                                           (II)

Here, ${\rho }_w$ is the density of water, $g$ is the acceleration due to gravity and $h_w$ is the height of the water coumn.

Substitute equation (II) in equation (I).

    $P_1=P_{atm}+{\rho }_{w}g{h}_w$                                                                                                 (III)

Write an expression for the pressure at the top of the lake.

    $P_2=P_{atm}$                                                                                                               (IV)

Here, $P_2$ is the pressure at the top of the lake.

Write an expression for the temperature at the bottom of the lake.

    $T_1=\frac{P_1{V}_1}{P_2{V}_2}{T}_2$                                                                                                           (V)

Here, $T_1$ is the temperature at the bottom of the lake, $T_2$ is the temperature at the surface of the lake, $V_1$ is the volume at the bottom of the lake and $V_2$ is the volume at the top of the lake.

Write an expression for the volume at the bottom of the lake.

    $V_1=\frac{4}{3}\pi {\left(\frac{d_1}{2}\right)}^3$                                                                                                      (VI)

Here, $d_1$ is the diameter at the bottom.

Write an expression for the volume at the surface of the lake.

  $V_2=\frac{4}{3}\pi {\left(\frac{d_2}{2}\right)}^3$                                                                                                     (VII)

Here, $d_2$ is the diameter at the top.

Substitute equation (III), (IV), (VI) and (VII) in equation (V).

      $\begin{array}{c}{T}_1=\frac{\left(P_{atm}+{\rho }_{w}g{h}_w\right)\left(\frac{4}{3}\pi {\left(\frac{d_1}{2}\right)}^2\right)}{\left(P_{atm}\right)\left(\frac{4}{3}\pi {\left(\frac{d_2}{2}\right)}^2\right)}{T}_2\\ =\frac{\left(P_{atm}+{\rho }_{w}g{h}_w\right)d_1^3}{\left(P_{atm}\right)d_2^3}{T}_2\end{array}$                                                                            (VIII)

Conclusion:

Substitute $1.01\times {10}^5\text{Pa}$ for $P_{atm}$, $1.00\times {10}^3{\text{kg/m}}^3$ for ${\rho }_w$, $2.50\text{m}$ for $h_w$, $9.81{\text{m/s}}^2$ for g, $3.60\text{mm}$ for $d_1$, $4.00\text{mm}$ for $d_2$ and $40°\text{C}$ for $T_2$ in equation (VIII) to find  $T_1$.

  $\begin{array}{c}{T}_1=\left(\begin{array}{c}\frac{\left(1.01\times {10}^5\text{Pa}+\left(1.00\times {10}^3{\text{kg/m}}^3\right)\left(9.81{\text{m/s}}^2\right)\left(2.50\text{m}\right)\right){\left(\left(3.60\text{mm}\right)\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}^2}{\left(1.01\times {10}^5\text{Pa}\right){\left(\left(4.00\text{mm}\right)\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}^2}\\ \left(\left(40.0°\text{C+273}\text{.15}°\text{C}\right)\left(\frac{1\text{K}}{1°\text{C}}\right)\right)\end{array}\right)\\ =\frac{\left(1.01\times {10}^5\text{Pa}+\left(1.00\times {10}^3{\text{kg/m}}^3\right)\left(9.81{\text{m/s}}^2\right)\left(2.50\text{m}\right)\right){\left(3.60\times {10}^{-3}\text{m}\right)}^2}{\left(1.01\times {10}^5\text{Pa}\right){\left(4.00\times {10}^{-3}\text{m}\right)}^2}\left(313.15\text{K}\right)\\ =284\text{K}\end{array}$

Thus, the temperature at the bottom of the lake is $284\text{K}$.