Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

Get the answer to your homework problem.

Try Numerade free for 7 days

Like

Report

At $20^{\circ} \mathrm{C}$ , the surface tension of water is 72.8 dynes/cm andthat of carbon tetrachloride $\left(\mathrm{CCl}_{4}\right)$ is 26.8 dynes/cm. If thegauge 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?

Physics 101 Mechanics

Chapter 13

Fluid Mechanics

Temperature and Heat

Rutgers, The State University of New Jersey

University of Washington

Hope College

University of Winnipeg

Lectures

03:45

In physics, a fluid is a s…

09:49

A fluid is a substance tha…

02:19

An air bubble of volume 20…

01:30

At $20^{\circ} \mathrm{C},…

02:39

Suppose that 3.11 mol of c…

02:44

The density of water is $1…

01:48

What radius must a water d…

05:04

A rigid vessel containing …

00:36

The density of water at $4…

01:35

The average coefficient of…

00:20

01:53

The density of acetylene g…

03:05

A $0.030-\mathrm{m}^{3}$ c…

00:47

To determine the volume of…

02:40

$10,000 \mathrm{cm}^{3}$ o…

00:23

The density of $\mathrm{CC…

03:31

How much is the change in …

01:38

A 400 -mL glass beaker at …

04:47

(II) It is observed that 5…

05:07

02:37

A piece of high-density St…

04:32

Density of liquid water is…

in this problem, we're told we have two drops of liquid, a drop of water, H 20 with a surface tension of 72.8 Dynes per centimeter and a drop of carbon tetrachloride, which is CCL four with a provided surface tension in which is 26.8 Dynes per centimeter. We're told that the pressure inside both of these drops is the same. And what we are looking for is we're looking for the ratio of the volume of the water drop to the volume of the carbon Tetrachloride drop. Now, let's make this a little bit easier. I'm going to convert these surface engines into s I units into standard units There, in Diane's per centimetre, I'm gonna put them into Newtons per meter. To do that, I need to know that one dying per centimeter is equal to tend to the negative three noons per meter. So what that means is that four hour water droplet the surface tension is equal to 72.8 times 10 to the negative three noons per meter and that four hour carbon tetrachloride, our surface tension is equal to 26.8 times 10 to the negative three noons per meter is going to make that a little bit easier to work with now. What we also need to know is we need to know that for a droplet of the material, it's with one surface. The gauge pressure in that droplet is equal to two times the surface tension divided by the radius of that droplet. And we know that the pressure in both of these droplets is the same. So what I can do is I can set up this equation, are gauge pressure equation with the variables for water and for carbon tetrachloride and set them equal to each other. So, for example, for water, the gauge pressure would be equal to two times the surface tension of water divided by the radius of the water droplet. And for the carbon tetrachloride, the pressure would be equal to two times the surface tension of carbon tetrachloride divided by the radius of the carbon tetrachloride bubble. Because those pressures the same I consent than equal to each other in order to work her way toward ratio. What I'm going to do is I'm going to cross multiply these two values to get them all in one nice line. So I'm gonna do that on our next peach. After I cross multiply that expression that I just had written. What I am left with is I have the radius of my water bubble multiplied by the surface tension of my carbon tetrachloride bubble which is equal to the radius of my carbon tetrachloride bubble multiplied by the surface tension of my water bubble. What I'm going to do is I'm going to substitute in my known values thes air, the known surface tensions that we found in S I units on the previous page. Now I have this nice little ratio set up. What I'm going to do is I'm going to use those coefficients that I've now kind of figured out for our, um ratio between our radius of water bubble in our radius of carbon tetrachloride and I'm going to turn those into volumes. In order to do that, I need to know that the volume of a sphere is equal to 4/3 pi r cubed. I'm going to use these coefficients, as are our values to come up with what their volumes would be has a ratio for these fears. So if I were to calculate a volume using that radius coefficient, I would end up on the left with a volume of 0.645 And our volume on the right would be 0.129 to 9. So that is our ratio between our volume up of our water using that radius value for water and are volume for our carbon tetrachloride. If you actually divide the value and left by the value on the right, get it down to one ratio. You end up with 0.0 for 989 for our final ratio on all of our values. Originally had three sig figs. So what we can do after we get that final answer is if we would like we can round it to 366 that would be 0.499 rounded up

View More Answers From This Book

Find Another Textbook

In physics, a fluid is a substance that continually deforms (flows) under an…

A fluid is a substance that continually deforms (flows) under an applied she…

02:11

$\bullet$ A particle has a charge of $-3.00 \mathrm{nC}$ . (a) Find the magn…

01:32

(1) What must your car's average speed be in order to travel 235 $\math…

01:00

(II) How much tension must a cable withstand if it is usedto accelerate …

03:26

If you began walking along one of Earth's lines of longitude and walked…

02:26

$\bullet$ Particle annihilation. In proton-antiproton annihilation,a pro…

01:52

$\bullet$ Two particles in a high-energy accelerator experiment areappro…

03:11

$\bullet$ When a toy car is rapidly scooted across the floor, it storese…

06:19

Figure 18.47 shows a system of four capacitors, where the potential differen…

05:45

Tritium $\left(_{1}^{3} \mathrm{H}\right)$ is an unstable isotope of hydroge…

03:55

A capacitor consists of two parallel plates, each with an area of 16.0 $\mat…