Chapter 19, Problem 38P

To determine

The greatest depth to which the bird dived.

Answer to Problem 38P

The greatest depth to which the bird dived is $\underset{\_}{7.13\text{m}}$.

Explanation of Solution

Write the equation from ideal gas equation for the top portion.

  $P_t{V}_t=nRT$                                                                                                     (I)

Here, $V_t$ is the volume of top portion, $n$ is the amount of gas in mole, $R$ is the gas constant, $T$ is the temperature, and $P_t$ is the pressure at top portion.

Write the expression for the top portion volume.

  $V_t=Ad$                                                                                                         (II)

Here, $A$ is the cross-sectional area and $d$ is the distance.

Rewrite the expression for the ideal gas equation for top portion.

  $P_{t}Ad=nRT$                                                                                                 (III)

Write the equation from ideal gas equation for the bottom portion.

  $P_b{V}_b=nRT$                                                                                                    (IV)

Here, $V_b$ is the volume of bottom portion, $n$ is the amount of gas in mole, $R$ is the gas constant, $T$ is the temperature, and $P_b$ is the pressure at bottom portion.

Write the expression for the bottom portion volume.

  $V_b=A\left(d-d^{\prime }\right)$                                                                                                 (V)

Here, $d^{\prime }$ is the distance.

Rewrite the expression for the ideal gas equation for bottom portion.

  $P_{b}A\left(d-d^{\prime }\right)=nRT$                                                                                        (VI)

Write the expression for the bottom pressure.

  $P_b=P_t+\rho gh$                                                                                                  (VII)

Here, $\rho$ is the density, $g$ is the gravitational acceleration and $h$ is the height.

Rearrange the expression for the height from equation (VII).

  $h=\frac{1}{\rho g}\left(P_b-P_t\right)$                                                                                            (VIII)

Conclusion:

Substitute $1\text{atm}$ for $P_t$, $6.50\text{cm}$ for $d$ in Equation (III).

  $\left(1\text{atm}\right)A\left(6.50\text{cm}\right)=nRT$                                                            (IX)

Substitute $2.70\text{cm}$ for $d^{\prime }$, $6.50\text{cm}$ for $d$ in Equation (V).

  $P_{b}A\left(6.50\text{cm}-2.70\text{cm}\right)=nRT$                                                       (X)

Divide (X) by (IX).

  $\begin{array}{c}\frac{P_{b}A\left(6.50\text{cm}-2.70\text{cm}\right)}{\left(1\text{atm}\right)A\left(6.50\text{cm}\right)}=\frac{nRT}{nRT}\\ \frac{P_b\left(3.80\text{cm}\right)}{\left(1\text{atm}\right)\left(6.50\text{cm}\right)}=1\\ P_b=1.73\times {10}^5{\text{N/m}}^{\text{2}}\end{array}$

Substitute $1.73\times {10}^5{\text{N/m}}^{\text{2}}$ for $P_b$, $1.013\times {10}^5{\text{N/m}}^{\text{2}}$ for $P_t$, $1030{\text{kg/m}}^{\text{3}}$ for $\rho$ and $9.8m/s^2$ for $g$ in Equation (VIII) to find $h$.

  $\begin{array}{c}h=\frac{1}{\left(1030{\text{kg/m}}^{\text{3}}\right)\left(9.8m/s^2\right)}\left(\left(1.73\times {10}^5{\text{N/m}}^{\text{2}}\right)-\left(1.013\times {10}^5{\text{N/m}}^{\text{2}}\right)\right)\\ =\frac{1}{\left(1030{\text{kg/m}}^{\text{3}}\right)\left(9.8m/s^2\right)}\left(0.717\times {10}^5{\text{N/m}}^{\text{2}}\right)\\ =7.13\text{m}\end{array}$

Thus, the greatest depth to which the bird dived is $\underset{\_}{7.13\text{m}}$.