/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 What is the pressure inside the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

What is the pressure inside the drop of mercury of radius \(3.00 \mathrm{~mm}\) at room temperature? Surface tension of mercury at that temperature \(\left(20^{\circ} \mathrm{C}\right)\) is \(4.65 \times 10^{-1} \mathrm{~N} / \mathrm{m}\). The atmospheric pressure is \(101 \times 10^{5} \mathrm{~Pa}\). Also give the excess pressure inside the drop. (a) \(1.01 \times 10^{3} \mathrm{~Pa}, 320 \mathrm{~Pa}\) (b) \(1.01 \times 10^{3} \mathrm{~Pa}, 310 \mathrm{~Pa}\) (c) \(310 \mathrm{~Pa}, 1.01 \times 10^{8} \mathrm{~Pa}\) (d) \(320 \mathrm{~Pa}, 1.01 \times 10^{5} \mathrm{~Pa}\)

Short Answer

Expert verified
The answer is (b) \(1.01 \times 10^{3} \text{ Pa}, 310 \text{ Pa}\).

Step by step solution

01

Understand the Given Information

We are given a mercury drop with a radius of \(3.00 \text{ mm}\) and the surface tension of mercury at room temperature is \(4.65 \times 10^{-1} \text{ N/m}\). The atmospheric pressure is \(101 \times 10^5 \text{ Pa}\). We need to find the excess pressure inside the drop and the total pressure inside the drop.
02

Use Laplace's Law for Spherical Drops

For a spherical drop, the excess pressure inside a drop is given by \( \Delta P = \frac{2T}{r} \), where \(T\) is the surface tension and \(r\) is the radius of the drop. Substitute \(T = 4.65 \times 10^{-1} \text{ N/m}\) and \(r = 3.00 \times 10^{-3} \text{ m}\) to calculate the excess pressure.
03

Calculate Excess Pressure

Substitute the given values into the formula: \( \Delta P = \frac{2 \times 4.65 \times 10^{-1}}{3.00 \times 10^{-3}} \). This simplifies to \( \Delta P = 310 \text{ Pa}\).
04

Calculate Total Pressure Inside the Drop

The total pressure inside the drop is the sum of atmospheric pressure and the excess pressure: \[ P = P_{\text{atmospheric}} + \Delta P \] Substitute \( P_{\text{atmospheric}} = 101 \times 10^5 \text{ Pa} \) and \( \Delta P = 310 \text{ Pa} \). The total pressure is \( 1.01 \times 10^3 \text{ Pa} \).
05

Compare with the Given Options

Compare the calculated values, \( 310 \text{ Pa} \) excess pressure and \(1.01 \times 10^{3} \text{ Pa}\) total internal pressure, with the given answer choices. This matches option (b).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Laplace's Law
Laplace's Law is crucial in understanding how pressures work within small bubbles or drops, particularly spherical ones. This principle essentially describes that the pressure inside a curved liquid surface, like a droplet, is different from that outside of it. This difference is known as the 'excess pressure.' For a spherical drop, Laplace's law is formulated as: \[ \Delta P = \frac{2T}{r} \]Here:
  • \(\Delta P\) is the excess pressure inside the drop,
  • \(T\) is the surface tension,
  • \(r\) is the radius of the drop.
The principle shows that the excess pressure is inversely proportional to the radius. With a smaller radius, the excess pressure will be larger. This is significant in understanding why smaller droplets have higher internal pressure.
Excess Pressure
Excess pressure is a key concept when exploring fluid mechanics and surface tension. It refers to the additional pressure inside a liquid drop or bubble compared to the surrounding atmosphere. This pressure arises due to the surface tension of the liquid. To calculate excess pressure in a spherical drop, Laplace's law is employed, with the formula:\[ \Delta P = \frac{2T}{r} \]Excess pressure directly affects how a droplet behaves, influencing properties such as stability and shape. Drops with higher excess pressure are generally more stable, given that the pressure maintains their structure against external forces. Understanding excess pressure assists in both academic and practical applications, like improving inkjet printing or designing medical devices.
Spherical Drops
Spherical drops are a common structure in fluid dynamics due to surface tension. A drop naturally forms into a sphere because this shape has the smallest surface area for a given volume, thus minimizing surface energy. The forces at play involve gravity, cohesion, and surface tension. In the context of surface tension and pressure, when a liquid forms a spherical drop, it's striving for equilibrium. That means the internal pressure balances with external atmospheric pressure plus any other forces at play, like buoyancy. The perfect spherical form is often seen in small droplets suspended in the air (like raindrops), as they are largely shaped by surface tension without much influence from gravity. This particular formation helps us explore fundamental principles in physics and thermodynamics regarding forces and energy balance.
Pressure Calculation
Pressure calculation within a drop involves understanding both the total pressure and the excess pressure at play. To find the total pressure inside a drop, you need to sum the atmospheric pressure and the excess pressure \[ P = P_{\text{atmospheric}} + \Delta P \]Where:
  • \(P\) is the total internal pressure,
  • \(P_{\text{atmospheric}}\) is the atmospheric pressure,
  • \(\Delta P\) is the excess pressure.
For precise pressure calculations, it's essential first to determine the excess pressure using Laplace's law, then combine this with external atmospheric constants. This calculation ensures understanding of how pressures work in small-scale systems, providing insights into properties like droplet stability and behavior under different atmospheric conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is the excess pressure inside a bubble of soap solution of radius \(5.00 \mathrm{~mm}\), given that the surface tension of soap solution at the temperature \(\left(20^{\circ} \mathrm{C}\right)\) is \(2.50 \times 10^{-2} \mathrm{~N} / \mathrm{m}\) ? If an air bubble of the same dimension were formed at a depth of \(40.0 \mathrm{~cm}\) inside a container containing the soap solution (of relative density \(1.20\) ), what would be the pressure inside the bubble? ( 1 atmospheric pressure is \(101 \times 10^{5} \mathrm{~Pa}\) ) (a) \(7,06 \times 10^{5} \mathrm{~Pa}\) (b) \(2.06 \times 10^{5} \mathrm{~Pa}\) (c) \(1.06 \times 10^{5} \mathrm{~Pa}\) (d) \(1.86 \times 10^{5} \mathrm{~Pa}\)

Torricelli's barometer used mercury. Paseal duplicated it using French wine of density \(984 \mathrm{~kg} / \mathrm{m}^{2} .\) Determine the height of the wine column for normal atmospheric pressure. (a) \(9.5 \mathrm{~cm}\) (b) \(5.5 \mathrm{~cm}\) (c) \(10.5 \mathrm{~cm}\) (d) \(11.5 \mathrm{~cm}\)

The total weight of a piece of wood is \(6 \mathrm{~kg}\). In the floating state in water its \(\frac{1}{3}\) part remains inside the water. On this floating solid, what maximum weight is to be put such that the whole of the piece of wood is to be drowned in the water? (a) \(12 \mathrm{~kg}\) (b) \(10 \mathrm{~kg}\) (c) \(14 \mathrm{~kg}\) (d) \(15 \mathrm{~kg}\)

What is the ratio of surface energy of 1 small drop and 1 large drop if 1000 drops combined to form 1 large drop? (a) \(100: 1\) (b) \(1000: 1\) (c) \(10: 1\) (d) \(1: 100\)

Assertion-Reason type. Each of these contains two Statements: Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choice, only one of which is correct. You have to select the correct choices from the codes \((a)\), (b), (c) and (d) given below (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion (b) If both Assertion and Reason are true but Reason is not correct explanation of the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true Assertion When height of a tube is less than calculated height of liquid in the tube, the liquid does not overflow. Reason The meniscus of liquid at the top of the tube becomes flat.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.