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What radius must a water drop have for the difference between the inside and outside pressures to be 0.0200 atm? The surface tension of water is 72.8 dynes/cm.

Short Answer

Expert verified
The required radius is approximately 0.00718 cm.

Step by step solution

01

Identify Given Variables and Constants

Identify the known quantities in this problem: the pressure difference between inside and outside the water drop is \( \Delta P = 0.0200 \) atm, and the surface tension of water is \( \gamma = 72.8 \) dynes/cm (converted to N/m as 0.0728 N/m). We need to find the radius \( r \) of the water drop.
02

Understand the Core Concept

This problem involves the concept of surface tension affecting pressure inside a droplet. Liquid droplets are governed by the Young-Laplace equation, which relates surface tension to the pressure difference across the surface of a droplet of radius \( r \). The equation is \( \Delta P = \frac{2\gamma}{r} \).
03

Convert Pressure Units

Convert the pressure difference from atm to pascals, since the unit of surface tension is in N/m. We know that 1 atm = 101325 Pa, so \( \Delta P = 0.0200 \times 101325 \) Pa = 2026.5 Pa.
04

Rearrange Young-Laplace Equation

Rearrange the Young-Laplace equation to solve for the radius \( r \): \[ r = \frac{2\gamma}{\Delta P} \]. This formula gives you the radius in meters when \( \gamma \) is in N/m and \( \Delta P \) is in Pa.
05

Substitute Values and Calculate

Substitute the known values into the equation: \( \gamma = 0.0728 \) N/m and \( \Delta P = 2026.5 \) Pa. Calculate: \[ r = \frac{2 \times 0.0728}{2026.5} \approx 7.18 \times 10^{-5} \text{ m} \]. Convert to cm: \( 7.18 \times 10^{-5} \text{ m} = 0.00718 \text{ cm} \).
06

Conclusion

The radius of the water drop required to achieve the given pressure difference is approximately 0.00718 cm.

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Key Concepts

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

Young-Laplace equation
One of the fundamental principles in the study of surface tension is the Young-Laplace equation. This equation describes how the pressure difference (\(\Delta P\)) across the interface of a droplet is related to its surface tension (\(\gamma\)) and its radius (\(r\)). It is given by the formula:
  • \(\Delta P = \frac{2\gamma}{r}\)
This equation is crucial to understanding the physics of droplets, bubbles, and similar systems with curved surfaces.
The pressure inside a droplet is greater than outside, due to surface tension. This inward force creates a pressure imbalance which is necessary to keep the droplet intact.
This formula helps us determine one of the unknown variables, such as the pressure difference or the radius of the droplet, if the other quantities are known. It elegantly demonstrates the link between the physical characteristics of liquid and the forces at work.
pressure difference
Pressure difference is a key concept when analyzing the behavior of liquid droplets. It refers to the difference in pressure between the inside and outside of a droplet.
In the context of this exercise, we're interested in a specific pressure difference of \(0.0200\) atm, which is equivalent to \(2026.5\) pascals when converted (as \(1\) atm = \(101325\) Pa).When surface tension acts on a liquid droplet, it compresses the liquid inwardly, increasing the internal pressure. This pressure difference ensures the droplet’s surface remains stable instead of dispersing.
The magnitude of this difference depends on factors such as surface tension and the size of the droplet as per the Young-Laplace equation.
  • Consider smaller droplets: Higher curvature leads to greater pressure differences.
  • Larger droplets: Reduced curvature results in smaller pressure differences.
water droplet physics
The physics behind water droplets involves a fascinating interplay of forces, such as surface tension and pressure differences. These forces dictate the shape and stability of droplets.
For water, surface tension is quite significant at \(72.8\) dynes/cm. This tension helps to maintain the spherical shape of small droplets by pulling water molecules inward, counterbalancing any potential spread due to external forces.A spherical droplet minimizes the surface area for a given volume, which energetically favors maintaining the shape. This is due to the cohesive forces between water molecules. Inside a droplet, the molecules experience balanced attractions, but on the surface, molecules are only attracted inwards.
Understanding these interactions contributes to various applications in fields such as meteorology and material science:
  • Predicting raindrop behavior in the atmosphere.
  • Understanding emulsions in culinary science.
  • Designing materials with specific wettability properties.
Exploring water droplet physics offers us insights into natural processes and modern technological applications.

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Most popular questions from this chapter

An electrical short cuts off all power to a submersible diving vehicle when it is 30 m below the surface of the ocean. The crew must push out a hatch of area 0.75 \(\mathrm{m}^{2}\) and weight 300 \(\mathrm{N}\) on the bottom to escape. If the pressure inside is 1.0 atm, what downward force must the crew exert on the hatch to open it?

A hot-air balloon has a volume of 2200 \(\mathrm{m}^{3} .\) The balloon fabric (the envelope) weighs 900 \(\mathrm{N} .\) The basket with gear and full propane tanks weighs 1700 \(\mathrm{N}\) . If the balloon can barely lift an additional 3200 \(\mathrm{N}\) of passengers, breakfast, and champagne when the outside air density is \(1.23 \mathrm{kg} / \mathrm{m}^{3},\) what is the average density of the heated gases in the envelope?

An open cylindrical tank of acid rests at the edge of a table 1.4 \(\mathrm{m}\) above the floor of the chemistry lab. If this tank springs a small hole in the side at its base, how far from the foot of the table will the acid hit the floor if the acid in the tank is 75 \(\mathrm{cm}\) deep?

(a) Calculate the buoyant force of air (density 1.20 \(\mathrm{kg} / \mathrm{m}^{3} )\) on a spherical party balloon that has a radius of 15.0 \(\mathrm{cm}\) . (b) If the rubber of the balloon itself has a mass of 2.00 \(\mathrm{g}\) and the balloon is filled with helium (density 0.166 \(\mathrm{kg} / \mathrm{m}^{3}\) ), calculate the net upward force (the "lift") that acts on it in air.

Advertisements for a certain small car claim that it floats in water. (a) If the car's mass is 900 \(\mathrm{kg}\) and its interior volume is \(3.0 \mathrm{m}^{3},\) what fraction of the car is immersed when it floats? You can ignore the volume of steel and other materials. (b) Water gradually leaks in and displaces the air in the car. What fraction of the interior volume is filled with water when the car sinks?

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