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At \(20^{\circ} \mathrm{C}\) , the surface tension of water is 72.8 dynes/cm and that of carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\) is 26.8 dynes/cm. If the gauge pressure is the same in two drops of these liquids, what is the ratio of the volume of the water drop to that of the \(\mathrm{CCl}_{4}\) drop?

Short Answer

Expert verified
The ratio of the volume of the water drop to the CCl4 drop is 20:1.

Step by step solution

01

Understand the Relation Between Surface Tension and Pressure

The pressure inside a drop due to surface tension follows the formula \( P = \frac{2T}{r} \), where \( T \) is the surface tension and \( r \) is the radius of the drop. This means that the pressure is inversely proportional to the radius when the surface tension is constant. Since the gauge pressure is the same in both drops, apply this formula to both liquids to relate radii.
02

Equate the Pressures of Both Drops

Since the pressure is the same in both drops,\[ \frac{2T_{\text{water}}}{r_{\text{water}}} = \frac{2T_{\text{CCl}_4}}{r_{\text{CCl}_4}} \]Simplify to get \( \frac{T_{\text{water}}}{r_{\text{water}}} = \frac{T_{\text{CCl}_4}}{r_{\text{CCl}_4}} \).
03

Express the Radius of Water in Terms of CCl4

From the equation \( \frac{T_{\text{water}}}{r_{\text{water}}} = \frac{T_{\text{CCl}_4}}{r_{\text{CCl}_4}} \), solve for \( r_{\text{water}} \):\[ r_{\text{water}} = r_{\text{CCl}_4} \cdot \frac{T_{\text{water}}}{T_{\text{CCl}_4}} \]Substitute the given surface tension values:\[ r_{\text{water}} = r_{\text{CCl}_4} \cdot \frac{72.8}{26.8} \].
04

Calculate the Ratio of Radii

Compute the ratio of the radii:\[ \frac{r_{\text{water}}}{r_{\text{CCl}_4}} = \frac{72.8}{26.8} = 2.716 \]
05

Relate Volume to Radius

The volume of a spherical drop is proportional to the cube of its radius, \( V = \frac{4}{3} \pi r^3 \). Therefore, the ratio of the volumes of water and \( \mathrm{CCl}_4 \) is the cube of the ratio of their radii:\[ \frac{V_{\text{water}}}{V_{\text{CCl}_4}} = \left( \frac{r_{\text{water}}}{r_{\text{CCl}_4}} \right)^3 = (2.716)^3 \].
06

Compute the Final Ratio

Calculate \((2.716)^3 \) to get the ratio of their volumes:\[ \frac{V_{\text{water}}}{V_{\text{CCl}_4}} = 20.0 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure in Liquids
Pressure in liquids is a compelling topic that involves understanding how force acts over an area within a liquid. In the context of drops, the pressure within is influenced by the surface tension of the liquid forming the drop. Surface tension, essentially the skin-like surface layer of a liquid forming drops, contributes to pressure inside the drop. When we talk about pressure in this context, it is the pressure exerted by the liquid's surface tension that matters. This pressure can be described using the formula:\[ P = \frac{2T}{r} \]where:
  • \( P \) is the internal pressure,
  • \( T \) is the surface tension, and
  • \( r \) is the radius of the drop.
A fascinating property here is the relationship between pressure and radius where, given constant surface tension, the pressure is inversely proportional to the radius. Smaller radii mean higher internal pressure for a liquid drop. This principle is crucial when comparing droplets of different substances like water and carbon tetrachloride, where the pressures must be equal.
Volume of a Drop
The volume of a drop is a fundamental concept in understanding the size and capacity of a spherical drop. When discussing the volume of liquid drops, we generally refer to its geometrical calculation. For any spherical object, the volume is determined by the formula:\[ V = \frac{4}{3} \pi r^3 \]Knowing this formula helps in understanding how changes in the radius of a drop affect its volume tremendously. Since the formula includes the cube of the radius, even a slight increase or decrease in the drop’s radius results in a significant change in its volume.
In the exercise, when drops of two different liquids have the same internal pressure, their radii vary because of different surface tensions. While the radii and surface tensions affect pressure equivalently, they in turn, also alter the volumes in a cubically proportional manner.
Spherical Drops
A drop of liquid naturally forms into a spherical shape due to surface tension. The spherical shape minimizes the surface area for a given volume, which is an inherently energy-efficient configuration. The reason behind this behavior stems from physics principles where liquids assume a shape with the least surface tension possible. This concept simplifies calculations for engineers and scientists by allowing easy prediction of drop behavior under varying conditions.
When analyzing different liquid drops such as those in our exercise, it's essential to consider their spherical nature. This forms the basis of their pressure and volume relationship assessments. Understanding spherical drops aids in recognizing how the inter-play of surface tension and gravitational forces create their shape and influence apparent behaviors.
Radius-Volume Relationship
In spherical drops, there is a vivid and noteworthy connection between radius and volume. This radius-volume relationship is pivotal for understanding how changes in drop size result in variations in volume.For a spherical drop, the volume \( V \) is given by:\[ V = \frac{4}{3} \pi r^3 \]This equation indicates that volume changes with the cube of the change in radius. Thus, if the radius of a drop increases or decreases, the volume will increase or decrease by the cube of that change. For instance, if one drop has a radius twice that of another, its volume becomes eight times larger.This cubical relationship becomes crucial in exercises where you need to compare volumes of drops from two different substances with various surface tensions while maintaining equal internal pressure. It emphasizes how even minor changes in radius due to diverse surface tensions result in significant volume differences, aiding in precise calculations and predictions. When working on such problems, always pay attention to how the radius affects the drop's volume.

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Most popular questions from this chapter

On the earth, the pressure generated by the heart is sufficient to pump blood to a height of 1.3 \(\mathrm{m} .\) The density of blood is 1.04 \(\mathrm{g} / \mathrm{cm}^{3} . )\) If the top of your head were 0.5 \(\mathrm{m}\) above your heart, what would be the strongest gravitational acceleration that you could endure on another planet before your heart would be unable to pump blood to your brain? $$\begin{array}{l}{\text { A. } 2 g} \\ {\text { B. } 3 g} \\ {\text { C. } 5 g} \\ {\text { D. } 10 g}\end{array}$$

Water discharges from a horizontal cylindrical pipe at the rate of 465 \(\mathrm{cm}^{3} / \mathrm{s}\) . At a point in the pipe where the radius is \(2.05 \mathrm{cm},\) the absolute pressure is \(1.60 \times 10^{5} \mathrm{Pa} .\) What is the pipe's radius at a constriction if the pressure there is reduced to \(1.20 \times 10^{5} \mathrm{Pa}\) ?

By how many newtons do you increase the weight of your car when you fill up your 11.5 gal gas tank with gasoline? A gallon is equal to 3.788 \(\mathrm{L}\) and the density of gasoline is 737 \(\mathrm{kg} / \mathrm{m}^{3}\) .

A copper sphere with a mass of 0.20 \(\mathrm{g}\) and a density of 8900 \(\mathrm{kg} / \mathrm{m}^{3}\) is observed to fall with a terminal speed of 6.0 \(\mathrm{cm} / \mathrm{s}\) in an unknown liquid. Find the viscosity of the unknown liquid if its buoyancy can be neglected.

Artery blockage. A medical technician is trying to determine what percentage of a patient's artery is blocked by plaque. To do this, she measures the blood pressure just before the region of blockage and finds that it is \(1.20 \times 10^{4}\) Pa, while in the region of blockage it is \(1.15 \times 10^{4}\) Pa. Furthermore, she knows that blood flowing through the normal artery just before the point of blockage is traveling at 30.0 \(\mathrm{cm} / \mathrm{s}\) , and the specific gravity of this patient's blood is \(1.06 .\) What percentage of the cross-sectional area of the patient's artery is blocked by the plaque?

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