Chapter 19, Problem 49PQ

An ideal gas is trapped inside a tube of uniform cross-sectional area sealed at one end as shown in Figure P19.49. A column of mercury separates the gas from the outside. The tube can be turned in a vertical plane. In Figure P19.49A, the column of air in the tube has length L₁, whereas in Figure P19.49B, the column of air has length L₂. Find an expression (in terms of the parameters given) for the length L₃ of the column of air in Figure P19.49C, when the tube is inclined at an angle θ with respect to the vertical.

FIGURE P19.49

To determine

The expression for the length $L_3$ of the column of air.

Answer to Problem 49PQ

The expression for the length $L_3$ of the column of air is $\underset{\_}{L_3=\frac{L_1{L}_2}{L_2-\left(L_2-L_1\right)\cos \theta }}$.

Explanation of Solution

Three cases are depicted here. The first case in which the length of the air column is $L_1$, the second case in which the length of the air column is $L_2$, and the third case in which the length of air column is $L_3$. All the three cases are sketched in Figure 1, Figure 2, and Figure 3 below.

In all the three cases mercury separates the air from outside. In all the three cases mercury should be in static equilibrium. The forces experienced by the mercury are the force due to the pressure inside the tube, the force from the atmospheric pressure, and the force due to the weight of the mercury. Here the tube is maintained at constant pressure. So apply Boyle’s law.

Consider Figure 1.

Weight of mercury acts perpendicular to the orientation of the tube. Thus the mercury is in equilibrium whenever the atmospheric pressure is equal to the pressure inside the tube.

  $P_1^{\text{gas}}=P_{\text{atm}}$                                                                                                             (I)

Here, $P_1^{\text{gas}}$ is the pressure on mercury inside the tube in position A, and $P_{\text{atm}}$ is the atmospheric pressure.

Consider Figure 2.

Here the outside atmospheric pressure is balanced by the sum of pressure in the tube due to the air column and the pressure due the weight of mercury.

Write the expression for the pressure due to the weight of mercury in position B.

  $P_2^{\text{mercury}}=\frac{mg}{A}$                                                                                               (II)

Here, $P_2^{\text{mercury}}$ is the pressure due to the weight of mercury in position B, $m$ is the mass of mercury, $g$ is the acceleration due to gravity, and $A$ is the area of cross section of tube.

Write the expression for the density of mercury.

  ${\rho }_{\text{mercury}}=\frac{m}{V}$                                                                                                    (III)

Here, ${\rho }_{\text{mercury}}$ is the density of mercury, $m$ is the mass of mercury, and $V$ is the volume of mercury.

Solve equation (III) for $m$.

  $m={\rho }_{\text{mercury}}V$                                                                                               (IV)

Use expression (IV) in (II).

  $P_2^{\text{mercury}}=\frac{{\rho }_{\text{mercury}}Vg}{A}$                                                                                      (V)

$\frac{V}{A}$ is equal to the length of the mercury in the tube, denoted by $l$.

  $P_2^{\text{mercury}}={\rho }_{\text{mercury}}gl$ (VI)

Here, $l$ is the length of the mercury column.

The atmospheric pressure at position B is balanced by the sum of pressures due to the weight of mercury, and pressure due to the column of air in the tube.

  $P_2^{\text{gas}}+\left({\rho }_{\text{mercury}}gl\right)=P_{\text{atm}}$                                                                         (VII)

Here, $P_2^{\text{gas}}$ is the pressure due to the air column in position 2.

Consider the position 3.

Here the pressure due to atmosphere is balanced by pressure inside the tube due to gas column denoted by $P_3^{\text{gas}}$ and the vertical component of mercury column.

Write the expression for the vertical component of pressure due to the weight of mercury column.

  $P_3^{\text{mercury}}={\rho }_{\text{mercury}}gl\cos \theta$                                                                       (VIII)

Here, $P_3^{\text{mercury}}$ is the pressure due to the mercury column in position 3 of tube, and $\theta$ is the angle made by the tube with the vertical.

Write the expression for the balance of pressure in the tube kept in position 3.

  $P_3^{\text{mercury}}+{\rho }_{\text{mercury}}gl\cos \theta =P_{\text{atm}}$                                                                     (IX)

Since the temperature is constant, apply Boyle’s law. Boyle’s law states that the volume of a gas is directly proportional to the pressure of the gas at constant temperature.

Write the expression for Boyle’s law for case A and case B.

  $P_1^{\text{gas}}{V}_1=P_2^{\text{gas}}{V}_2$                                                                                          (X)

Write the expression for volume of air in tube 1.

  $V_1=A{L}_1$                                                                                                       (XI)

Here, $L_1$ is the length of air column in tube 1.

Write the expression for volume of air in tube 2.

  $V_2=A{L}_2$                                                                                               (XII)

Here, $L_2$ is the length of air column in tube 2.

Use expression (XI), (VII), (XII) and (I) in expression (X).

  $P_{\text{atm}}A{L}_1=\left(P_{\text{atm}}-{\rho }_{\text{mercury}}gl\right)L_{2}A$                                                                  (XIII)

Solve expression for $P_{\text{atm}}$.

  $P_{\text{atm}}=\left({\rho }_{\text{mercury}}gl\right)\frac{L_2}{L_2-L_1}$                                                                           (XIV)

Write the expression for Boyle’s law for case A and case C.

  $P_1^{\text{gas}}{V}_1=P_3^{\text{gas}}{V}_3$                                                                                       (XV)

Write the expression for volume of air in tube 3.

  $V_3=A{L}_3$                                                                                                    (XVI)

Here, $L_3$ is the length of air column in tube 3.

Use expression (XI), (XVI), (IX) and (I) in expression (XV).

  $P_{\text{atm}}A{L}_1=\left(P_{\text{atm}}-{\rho }_{\text{mercury}}gl\cos \theta \right)L_{3}A$                                                   (XVII)

Solve expression (XVII) for $P_{\text{atm}}$.

  $P_{\text{atm}}=\left({\rho }_{\text{mercury}}gl\cos \theta \right)\frac{L_3}{L_3-L_1}$                                                             (XVIII)

Equate the right hand sides of equations (XIV) and (XVIII) and solve for $L_3$.

  $\begin{array}{c}\left({\rho }_{\text{mercury}}gl\right)\frac{L_2}{L_2-L_1}=\left({\rho }_{\text{mercury}}gl\cos \theta \right)\frac{L_3}{L_3-L_1}\\ \frac{L_2}{L_2-L_1}=\frac{L_3\cos \theta }{L_3-L_1}\\ L_2{L}_3-L_2{L}_1=L_3\left(L_2-L_1\right)\cos \theta \end{array}$                                          (XIX)

Solve expression (XIX) for $L_3$.

  $\begin{array}{c}{L}_3\left(L_2-\left(L_2-L_1\right)\cos \theta \right)=L_1{L}_2\\ L_3=\frac{L_1{L}_2}{L_2-\left(L_2-L_1\right)\cos \theta }\end{array}$

Conclusion:

Therefore, the expression for the length $L_3$ of the column of air is $\underset{\_}{L_3=\frac{L_1{L}_2}{L_2-\left(L_2-L_1\right)\cos \theta }}$.