Chapter 16, Problem 33SP

Air is trapped in the sealed lower end of a capillary tube by a mercury column as shown in Fig. 16-1. The top of the tube is open. The temperature is 14 °C, and atmospheric pressure is 740 mmHg. What length would the trapped air column have if the temperature were 30 °C and atmospheric pressure were 760 mmHg?

To determine

The length of the air column trapped into the capillary column shown in Fig. 16-1when the temperature is $30°\text{C}$ and the atmospheric pressure is $760\text{mmHg}$.

Answer to Problem 33SP

Solution:

$12.4\text{ cm}$

Explanation of Solution

Given data:

Air is trapped in the sealed lower end of the capillary tube at $14°\text{C}$ temperatureand $740\text{mmHg}$ atmospheric pressure. The length of air column at these conditions is as shown in figure 16-1.

The length of the air column is to be determined at $30°\text{C}$ temperatureand $760\text{mmHg}$ atmospheric pressure.

Formula used:

Write the expression for thepressure on a substance at a certain depth.

$P=P_0+\rho gh$

Here, $P_0$ is the atmospheric pressure, $\rho$ is the density of substance, $g$ is the acceleration due to gravity, and $h$ is the depth.

Write the expression forgas law.

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Here, $P_1$ is the pressure at volume $V_1$ and temperature $T_1$, and $P_2$ is the pressure at volume $V_2$ and temperature $T_2$.

The unit conversion of temperature from $°\text{C}$ to $\text{K}$ is,

$T_{\left(\text{K}\right)}=273+T_{\left(°\text{C}\right)}$

Here, $T_{\left(K\right)}$ is the temperature in kelvin and $T_{\left(°\text{C}\right)}$ is the temperature in $°\text{C}$.

Explanation:

The air inside the tube is compressed because of the pressure of mercury column in the tube.

The expression for thepressure on a substance at a certain depth is,

$P=P_0+\rho gh$

When atmospheric pressure is $740\text{ mmHg}$, substitute $740\text{ mmHg}$ for $P_0$, $13593\text{kg}/{\text{m}}^3$ for $\rho$(density of mercury), $9.8\text{m}/{\text{s}}^2$ for $g$, and $8\text{cm}$ for $h$.

$\begin{array}{c}{P}_1=\left(740\text{ mmHg}\left(\frac{133.32\text{ Pa}}{1\text{ mmHg}}\right)\right)+\left(13593\text{kg}/{\text{m}}^3\right)\left(9.8\text{m}/{\text{s}}^2\right)\left(8\text{ cm}\left(\frac{{10}^{-2}\text{m}}{1\text{ cm}}\right)\right)\\ =\left(9.8\times {10}^4\text{ Pa}\right)+\left(1.06\times {10}^4\text{ Pa}\right)\\ =1.086\times {10}^5\text{ Pa}\end{array}$

When atmospheric pressure is $760\text{ mmHg}$, substitute $760\text{ mmHg}$ for $P_0$, $13593\text{kg}/{\text{m}}^3$ for $\rho$, $9.8\text{m}/{\text{s}}^2$ for $g$, and $8\text{cm}$ for $h$.

$\begin{array}{c}{P}_2=\left(760\text{ mmHg}\left(\frac{133.32\text{ Pa}}{1\text{ mmHg}}\right)\right)+\left(13593\text{kg}/{\text{m}}^3\right)\left(9.8\text{m}/{\text{s}}^2\right)\left(8\text{ cm}\left(\frac{{10}^{-2}\text{m}}{1\text{ cm}}\right)\right)\\ =\left(1.01\times {10}^5\text{ Pa}\right)+\left(1.06\times {10}^4\text{ Pa}\right)\\ =1.12\times {10}^5\text{ Pa}\end{array}$

The expression forvolume is,

$V=Ah$

Here, $V$ is the volume, $A$ is the area of the tube, and $h$ is the height of the aircolumn.

For pressure $P_1$, substitute $12\text{cm}$ for $h$.

$V_1=A\left(12\text{ cm}\right)$

For pressure $P_2$, let the height of the air-column trapped be $l$.

$V_2=Al$

Convert the temperature at pressure $P_1$ from $°\text{C}$ to kelvin by using the following conversion.

$T_{\left(\text{K}\right)}=273+T_{\left(°\text{C}\right)}$

For pressure $P_1$, substitute $14°\text{C}$ for $T_{\left(°\text{C}\right)}$

$\begin{array}{c}{T}_1=273+14°\text{C}\\ \text{=287 K}\end{array}$

For pressure $P_2$, substitute $30°\text{C}$ for $T_{\left(°\text{C}\right)}$

$\begin{array}{c}{T}_2=273+30°\text{C}\\ \text{=303 K}\end{array}$

The expression forgas law is,

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Substitute $1.086\times {10}^5\text{Pa}$ for $P_1$, $A\left(12\text{ cm}\right)$ for $V_1$, $287\text{ K}$ for $T_1$, $1.12\times {10}^5\text{Pa}$ for $P_2$, $Al$ for $V_2$, and $303\text{ K}$ for $T_2$

$\frac{\left(1.086\times {10}^5\text{ Pa}\right)\left(A\left(12\text{ cm}\right)\right)}{\left(287\text{ K}\right)}=\frac{\left(1.12\times {10}^5\text{ Pa}\right)\left(Al\right)}{\left(303\text{ K}\right)}$

Solve for $l$.

$\begin{array}{c}l=\frac{\left(1.086\times {10}^5\text{ Pa}\right)\left(A\left(12\text{ cm}\right)\right)\left(303\text{ K}\right)}{\left(1.12\times {10}^5\text{ Pa}\right)\left(287\text{ K}\right)A}\\ =\frac{3.949\times {10}^8}{3.2144\times {10}^7}\text{cm}\\ \text{=12}\text{.4 cm}\end{array}$

Conclusion:

The length of the air column at temperature, $30°\text{C}$ and atmospheric pressure, $760\text{mmHg}$ is $12.4\text{ cm}$.