Introduction
In the liquid state or phase, a substance has two properties that distinguish it from the solid phase. First, a liquid changes shape so that it conforms to the inside boundaries of any container in which it is placed. Second, a liquid placed in an open container (such as a jar or bucket) flows to the bottom of the container and develops a defined, flat surface. At least this is the way a sample of liquid behaves in an environment where there is gravity.
Diffusion Of Liquids
Imagine a jar on board a space ship in which the environment is weightless (there is no acceleration force). Suppose that the jar is filled with liquid, and then another liquid that does not react chemically with the first liquid is introduced into the jar. Gradually, the two liquids blend together until the mixture is uniform throughout the jar. This blending process is called diffusion .
In a liquid, diffusion takes place rather slowly; some liquids undergo the process faster than others. Alcohol diffuses into water at room temperature much more quickly than heavy motor oil into light motor oil. Eventually, however, when any two liquids are mixed (as long as they don’t react chemically, as do an acid and a base), the mixture will become uniform throughout any container of finite size. This happens without the need for shaking the container because the molecules of a liquid are always in motion, and this motion literally causes them to push and jostle each other until they become uniformly mixed.
If the same experiment is conducted in a bucket on Earth where there is acceleration force produced by gravity, diffusion will occur, but “heavier” liquids will sink toward the bottom and “lighter” liquids will rise toward the surface. Alcohol, for example, will float on water. However, the “surface” between the alcohol and water will not be sharply defined, as is the surface between the water and the air. The motion of the molecules constantly tries to mix the two liquids. However, gravitation prevents the mixture from becoming uniform throughout the bucket unless the two liquids are of exactly the same density. We’ll talk about the meaning of density for liquids shortly.
Viscosity Of Liquids
Some liquids are “runnier” than others. You know there is a difference at room temperature between, say, water and thick molasses. If you fill a glass with water and another glass with an equal amount of molasses and then pour the contents of both glasses into the sink, the glass containing the water will empty much faster. The molasses is said to have higher viscosity than the water at room temperature. On an extremely hot day, the difference is less obvious than it is on a cold day, unless, of course, you have air conditioning that keeps the air in your house at the same temperature all the time.
Some liquids are far more viscous even than thick molasses. An example of a liquid with extremely high viscosity is hot tar as it is poured to make the surface of a new highway. Another example is warm petroleum jelly. These substances meet the criteria as defined above to qualify as liquids, but they are thick indeed. As the temperature goes down, these substances become less and less liquid-like and more solid-like. In fact, it’s impossible to draw an exact line between the liquid and the solid phases for either of these two substances. They aren’t like water; they don’t freeze into ice and change state in an obvious way. As hot tar cools, where do we draw the line? How can we say, “Now, this stuff is liquid,” and then 1 second later say, “Now, this stuff is solid,” and be sure of the exact point of transition?
Liquid Or Solid?
There is not always a defined answer to the question, “Is this substance a solid or a liquid?” It can depend on the observer’s point of reference. Some substances can be considered solid in the short-term time sense but liquid in the long-term sense. An example is the mantle of the Earth, the layer of rock between the crust and the core. In a long-term time sense, pieces of the crust, known as tectonic plates , float around on top of the mantle like scum on the surface of a hot vat of liquid. This is manifested as continental drift and is apparent when the Earth is evaluated over periods of millions of years. From one moment (as we perceive it) to the next, however, and even from hour to hour or from day to day, the crust seems rigidly fixed on the mantle. The mantle behaves like a solid in the short-term sense but like a liquid in the long-term sense.
Imagine that we could turn ourselves into creatures whose life spans were measured in trillions (units of 1012) of years so that 1 million years seemed to pass like a moment. Then, from our point of view, Earth’s mantle would behave like a liquid with low viscosity, just as water seems to us in our actual state of time awareness. If we could become creatures whose entire lives lasted only a tiny fraction of a second, then liquid water would seem to take eons to get out of a glass tipped on its side, and we would conelude that this substance was solid, or a else a liquid with extremely high viscosity.
The way we define the state of a substance can depend on the temperature, and it also can depend on the time frame over which the substance is observed.
Density Of Liquids
The density of a liquid is defined in three ways: mass density, weight density , and particle density . The difference between these quantities might seem theoretically subtle, but in practical situations, the difference becomes apparent.
Mass density is defined in terms of the number of kilograms per meter cubed (kg/m 3 ) in a sample of liquid. Weight density is defined in newtons per meter cubed (N/m 3 ) and is equal to the mass density multiplied by the acceleration in meters per second squared (m/s 2 ) to which the sample is subjected. Particle density is defined as the number of moles of atoms per meter cubed (mol/m 3 ), where 1 mol ≈ 6.02 × 10 23 .
Let d m be the mass density of a liquid sample (in kilograms per meter cubed), let d w be the weight density (in newtons per meter cubed), and let d p be the particle density (in moles per meter cubed). Let m represent the mass of the sample (in kilograms), let V represent the volume of the sample (in meters cubed), and let N represent the number of moles of atoms in the sample. Let a be the acceleration (in meters per second squared) to which the sample is subjected. Then the following equations hold:
d m = m/V
d w = ma/V
d p = N/V
Alternative definitions for mass density, weight density, and particle density use the liter , which is equal to a thousand centimeters cubed (1000 cm 3 ) or one-thousandth of a meter cubed (0.001 m 3 ), as the standard unit of volume. Once in awhile you’ll see the centimeter cubed (cm 3 ), also known as the milliliter because it is equal to 0.001 liter, used as the standard unit of volume.
These are simplified definitions because they assume that the density of the liquid is uniform throughout the sample.
Density of Liquids Practice Problems
Problem 1
A sample of liquid measures 0.275 m 3 . Its mass is 300 kg. What is its mass density in kilograms per meter cubed?
Solution 1
This is straightforward because the input quantities are already given in SI. There is no need for us to convert from grams to kilograms, from milliliters to meters cubed, or anything like that. We can simply divide the mass by the volume:
d m = m/V
= 300 kg/0.275 m 3
= 1090 kg/m 3
We’re entitled to go to three significant figures here because our input numbers are both given to three significant figures.
Problem 2
Given that the acceleration of gravity at the Earth’s surface is 9.81 m/s 2 , what is the weight density of the sample of liquid described in Problem 1?
Solution 2
All we need to do in this case is multiply our mass density answer by 9.81 m/s 2 . This gives us
d w = 1090 kg/m 3 × 9.81 m/s 2
= 10,700 N/m 3 = 1.07 × 10 4 N/m 3
Note the difference here between the nonitalicized uppercase N, which represents newtons, and the italicized uppercase N , which represents the number of moles of atoms in a sample.
Measuring Liquid Volume
The volume of a liquid sample is usually measured by means of a test tube or flask marked off in milliliters or liters. However, there’s another way to measure the volume of a liquid sample, provided we know its chemical composition and the weight density of the substance in question. This is to weigh the sample of liquid and then divide the weight by the weight density. We must, of course, pay careful attention to the units. In particular, the weight must be expressed in newtons, which is equal to the mass in kilograms times the acceleration of gravity (9.81 m/s 2 ).
Let’s do a mathematical exercise to show why we can measure volume in this way. Let d w be the known weight density of a huge sample of liquid too large for its volume to be measured using a flask or test tube. Suppose that this substance has a weight of w , in newtons. If V is the volume in meters cubed, we know from the preceding formula that
d w = w/V
because w = ma , where a is the acceleration of gravity. If we divide both sides of this equation by w , we get
d w / w = 1/ V
Then we can invert both sides of this equation and exchange the left-hand and the right-hand sides to obtain
V = w/d w
All this is based on the assumption that V , w , and d w are all nonzero quantities. This is always true in the real world; all materials occupy at least some volume, have at least some weight because of gravitation, and have some density because there is some “stuff in a finite amount of physical space.
Pressure In Liquids
Have you read or been told that liquid water can’t be compressed? In a simplistic sense, this is true, but it doesn’t mean liquid water never exerts pressure. Liquids can and do exert pressure, as anyone who has been in a flood or a hurricane or a submarine will tell you. You can experience “water pressure” for yourself by diving down several feet in a swimming pool and noting the sensation the water produces as it presses against your eardrums.
In a fluid, the pressure, which is defined in terms of force per unit area, is directly proportional to the depth. Pressure is also directly proportional to the weight density of the liquid. Let d w be the weight density of a liquid (in newtons per meter cubed), and let s be the depth below the surface (in meters). Then the pressure P (in newtons per meter squared) is given by
P = d w s
If we are given the mass density d m (in kilograms per meter cubed) rather than the weight density, the formula becomes
P = 9.8 l d m s
Pressure in Liquids Practice Problem
Problem
Liquid water generally has a mass density of 1000 kg/m 3. How much force is exerted on the outer surface of a cube measuring 10.000 cm on an edge that is submerged 1.00 m below the surface of a body of water?
Solution
First, figure out the total surface area of the cube. It measures 10.000 cm, or 0.10000 m, on an edge, so the surface area of one face is 0.10000 m × 0.10000 m = 0.010000 m 2 . There are six faces on a cube, so the total surface area of the object is 0.010000 m 2 × 6.0000 = 0.060000 m 2 . (Don’t be irritated by the “extra” zeroes here. They are important. They indicate that the length of the edge of the cube has been specified to five significant figures.)
Next, figure out the weight density of water (in newtons per meter cubed). This is 9.81 times the mass density, or 9,810 N/m 3 . This is best stated as 9.81 × 10 3 N/m 3 because we are given the acceleration of gravity to only three significant figures, and scientific notation makes this fact clear. From this point on let’s revert to power-of-10 notation so that we don’t fall into the trap of accidentally claiming more accuracy than that to which we’re entitled.
The cube is at a depth of 1.00 m, so the water pressure at that depth is 9.81 × 10 3 N/m 3 × 1.00 m = 9.81 × 10 3 N/m 2 . The force F (in newtons) on the cube is therefore equal to this number multiplied by the surface area of the cube:
F = 9.81 × 10 3 N/m 2 × 6.00000 × 10 −2 m 2
= 58.9 X 10 1 N = 589 N
Pascal’s Law For Incompressible Liquids
Imagine a watertight, rigid container. Suppose that there are two pipes of unequal diameters running upward out of this container. Imagine that you fill the container with an incompressible liquid such as water so that the container is completely full and the water rises partway up into the pipes. Suppose that you place pistons in the pipes so that they make perfect water seals, and then you leave the pistons to rest on the water surface (Fig. 10-4).
Fig. 10-4 . Pascal’s law for confined, incompressible liquids. The forces are directly proportional to the areas of the pistons.
Because the pipes have unequal diameters, the surface areas of the pistons are different. One of the pistons has area A 1 (in meters squared), and the other has area A 2 . Suppose that you push downward on piston number 1 (the one whose area is A1) with a force F 1 (in newtons). How much upward force F 2 is produced at piston number 2 (the one whose area is A 2 )? Pascal’s law provides the answer: The forces are directly proportional to the areas of the piston faces in terms of their contact with the liquid. In the example shown by Fig. 10-4, piston number 2 is smaller than piston number 1, so the force F 2 is proportionately less than the force F 1 . Mathematically, the following equations both hold:
F 1 / F 2 = A 1 / A 2
A 1 / F 2 = A 2 / F 1
When using either of these equations, we must be consistent with units throughout the calculations. In addition, the top equation is meaningful only as long as the force exerted is nonzero.
Pascal's Law Practice Problem
Problem
Suppose that the areas of the pistons shown in Fig. 10-4 are A 1 = 12.00 cm 2 and A 2 = 15.00 cm 2 . (This does not seem to agree with the illustration, where piston number 2 looks smaller than piston number 1, but forget about that while we solve this problem.) If you press down on piston number 1 with a force of 10.00 N, how much upward force will result at piston number 2?
Solution
At first, you might think that we have to convert the areas of the pistons to meters squared in order to solve this problem. In this case, however, it is sufficient to find the ratio of the areas of the pistons because both areas are given to us in the same units:
A 1 / A 2 = 12.00 cm 2 /15.00 cm 2
= 0.8000
Thus we know that F 1 / F 2 = 0.8000. We are given F 1 = 10.00 N, so it is easy to solve for F 2 :
10.00/ F 2 = 0.8000
1/ F 2 = 0.08000
F 2 = 1/0.08000 = 12.50 N
We are entitled to four significant figures throughout this calculation because all the input data were provided to this degree of precision.
Practice problems of these concepts at: Basic States Of Matter Practice Test
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