# Fluids, Gases & Heat Flashcards

## From fluid flow to Gibbs' Free Energy, use these cards to master the topic of Fluids, Gases, and Heat as tested in most introductory undergrad physics courses and even on the AP Physics exam.

Define:

a **fluid**

A **fluid** is a phase of matter that is capable of yielding to pressure and changing its shape to fit a container.

On the AP exam, fluids come in two forms: liquids and gases which are in motion. Static gases are technically fluids, but they are non-ideal ones and are not tested as such.

What are the properties of an ideal fluid?

An ideal fluid:

- is incompressible
- is non-viscous (no friction)
- changes shape to fit its container
- flows without turbulence

On the AP Physics exam, all fluids are assumed to be ideal unless otherwise noted.

Water is flowing through a level pipe. What can be said about the nature of the flow?

Since water can be assumed to be an ideal fluid, the flow will be frictionless and non-turbulent. The speed of the flow will be constant at all points along the pipe, and the pressure on the walls of the pipe will be constant at all points.

Define:

**Density (ρ)** of a substance

A substance’s **density (ρ)** is the mass of a unit volume of that substance.

The SI units of density are kg/m^{3}.

What are some commonly-used units of density?

g/cm^{3}, g/cc, g/mL, kg/L, and kg/m^{3} are all commonly-used measures of density.

The first four are equivalent in magnitude, while the fifth differs by a factor of 1000.

What is the density of water in:

- g/cm
^{3} - g/mL
- kg/L
- kg/m
^{3}

The density of water is:

- 1 g/cm
^{3} - 1 g/mL
- 1 kg/L
- 1000 kg/m
^{3}

This is the one density you should memorize for the AP Physics exam.

Define:

**specific gravity** of a substance

A substance’s **specific gravity** is that substance’s density, divided by the density of water.

The magnitude of a substance’s specific gravity is identical to the substance’s density, as expressed in g/mL, but specific gravity is a unitless quantity.

Define:

**buoyancy**

**Buoyancy** is the tendency of an object to weigh less when partially or fully submerged in a liquid.

Buoyancy is caused by the liquid displaced by the object. The liquid pushes up on the object with a force equal to the weight of the liquid displaced.

How is the buoyancy force of a submerged object calculated?

The buoyancy force of a submerged object is simply the weight of the liquid displaced by the portion of the object that is submerged.

F_{B} = ρ_{liq }* V_{sub}* g

Where:

F_{B} = the buoyancy force pushing up on the object

ρ_{liq} = the density of the liquid

V_{sub} = the volume of the object that is submerged in the liquid

g = the gravitational acceleration (commonly 10 m/s^{2})

An object with a volume of 1.5 L is fully submerged in water. What is the buoyancy force on the object?

The buoyancy force on the object is 15 N.

Using the definition of buoyancy force:

F_{B} = ρ_{liq} * V_{sub} * g

F_{B} = (1 kg/L) * (1.5 L) * (10 m/s^{2})

F_{B} = 15 N

Equivalent objects are submerged in ethyl alcohol (ρ = 0.79 kg/L) and mercury (ρ = 5.43 kg/L). Which one experiences the larger buoyancy force?

The object submerged in mercury will experience the larger buoyancy force.

Remember, the buoyancy force is equal to ρ_{liq} * V_{sub} * g. Since the objects are identical, the submerged volume is equal in each case, and presumably gravity is constant, so the fluid with the larger density will have the higher buoyancy force.

If a solid object is dropped into a liquid, under what conditions will it float?

The object will float if its density is less than or equal to the density of the surrounding liquid. This is known as the **buoyancy rule**.

A cube of styrofoam, with a density of 0.5 g/cm^{3}, is dropped into water. How much of the cube is submerged in the water when the cube begins to float?

One half of the cube will be submerged in the water.

The proportion of an object that is under the surface of a liquid when the object floats is simply equal to ρ_{obj}/ρ_{liq}, where ρ_{obj} is the object’s density and ρ_{liq} is the liquid’s density.

In this case, the object is one-half the density of the liquid, and thus one-half of it will be submerged when it floats.

Note that if ρ_{obj} > ρ_{liq}, this fraction is greater than 1, and the object sinks to the bottom.

Define:

the **effective weight** of an object submerged in a liquid

The **effective weight** of an object submerged in a fluid is equal to the object’s weight on dry land minus the buoyancy force due to the liquid displaced by the object’s submersion.

W_{eff} = m_{obj}g - (ρ_{liq} * V_{obj} * g)

Where:

m_{obj} = the object’s mass

ρ_{liq} = the liquid’s density

V_{obj} = the object’s volume

g = 10 m/s^{2}

An aluminum ball with a density of 3 g/cm^{3} has a mass of 9 kg. What is its effective weight when it is submerged in water?

The ball’s effective weight is 60 N

With a mass of 9 kg and a density of 3 g/cm^{3}, the ball’s volume must be 3000 cm^{3}, or 3 L. Therefore, it displaces 3 L of water when it is submerged. The buoyancy force is simply the weight of 3 L of water:

F_{B} = ρ_{liq} * V_{sub} * g

= (1 kg/L) (3 L) (10 m/s^{2}) = 30 N

With a mass of 9 kg, the ball’s dry land weight is 90 N. Subtracting the buoyancy force leaves the final answer, 60 N.

A plastic ball with a density of 2 g/cm^{3} has a mass of 6 kg. What is a shortcut to calculating the effective weight of the ball when it is submerged in water?

The shortcut is that the proportion of the object’s weight that is cancelled due to buoyancy is exactly equal to the ratio of the liquid’s density to the object’s density.

In this case, the liquid, water, has a density one-half that of the object. Thus, the proportion of the object’s weight cancelled by buoyancy is one-half the object’s dry land weight.

With a mass of 6 kg, the object’s dry land weight is 60 N. One half (30 N) of that is cancelled by buoyancy, so the remaining effective weight is 60 - 30 = 30 N.

Define:

**Pascal’s Law** of fluid pressures

**Pascal’s Law** states that a fluid will carry pressure undiminished; that is, pressure exerted on any part of a fluid will be carried equally to all the walls of the container.

Pascal’s Law is most commonly tested on the AP Physics exam using hydraulic lifts.

What does Pascal’s Law predict about the pressure on plates 1 and 2 in the diagram below?

Pascal’s Law predicts that the pressures will be equal.

The force F_{1} will be exerted on the fluid behind plate 1 as a pressure P_{1}. The fluid will exert that pressure, undiminished, on all the walls it is contacting, including plate 2.

How does the force F_{2} change if plate 2 in the hydraulic lift pictured below doubles in area?

The force F_{2} doubles.

According to Pascal’s Law, the pressures on plates 1 and 2 must be equivalent. From the definition of pressure, therefore:

P_{1} = F_{1}/A_{1} = F_{2}/A_{2} = P_{2}

So force and area are directly proportional, and doubling A_{2} will double F_{2} as well.

How does the distance D_{2} compare to the distance D_{1} if plate 2 in the hydraulic lift pictured is double the area of plate 1?

D_{2} = ½D_{1}.

Since the liquid between the plates in incompressible, the volume displaced by plate 1 must be equal to the volume displaced by plate 2:

V_{1} = A_{1}D_{1} = A_{2}D_{2} = V_{2}

So distance and area are inversely proportional, and a plate of double the area will move half the distance.

Define:

**hydrostatic pressure** on a submerged object

**Hydrostatic pressure** is the pressure exerted on a submerged object by the liquid in which the object is submerged.

Hydrostatic pressure increases proportionately with distance beneath the surface of the liquid.

What is the pressure exerted on an object submerged a distance z below the surface of a liquid?

Hydrostatic pressure is calculated:

P_{H} = ρ_{liq} * g * z

Where:

P_{H} = the hydrostatic pressure

ρ_{liq} = the liquid’s density

g = 10 m/s^{2}

z = distance beneath the liquid’s surface

What is the difference between the hydrostatic pressure at a depth of 20 and 50 cm beneath the surface of water in a large, round-bottomed flask?

The pressure 50 cm beneath the surface is 2.5x the pressure 20 cm beneath the surface.

Since the hydrostatic pressure is P_{H} = ρ_{liq} * g * z, pressure is proportional to depth.

Note that the pressure is independent of vessel shape; the only variables are liquid density and depth beneath the surface.

Vessel 1 is filled with water, while vessel 2 is filled with an unknown liquid. The pressure 10 cm beneath the surface of the liquid in vessel 2 is 3 times the pressure the same depth beneath the surface of the water in vessel 1. What is the density of the unknown liquid?

The unknown liquid’s density is 3 g/cm^{3}.

Since the hydrostatic pressure is P_{H} = ρ_{liq} * g * z, the pressure is proportional with liquid density. If the pressure at an equal depth is higher, the liquid’s density must also be higher, by the same proportion.

What are the properties of ideal fluid as it flows through a pipe?

Ideal fluids undergo **laminar flow**, a form of motion in which the fluid is flowing at the same speed at all points in the pipe.

This is opposed to turbulent flow, where the fluid can have eddies, vortices, and local disruption to the fluid’s speed and direction of flow.

How does the flow of a viscous fluid through a pipe differ from the flow of an ideal fluid?

Viscous fluid undergoes **Poiseuille flow**, where the fluid near the walls of the pipe flows much more slowly due to viscous interaction with the walls.

Above a certain fluid speed and pipe diameter, viscous fluids begin to flow turbulently as well.

Under what conditions does fluid flow become turbulent?

Turbulent flow occurs in non-ideal fluids when the cross-sectional area of the fluid flow becomes large, and when the fluid velocity increases.

For a given system, there are threshold values of cross-sectional area and velocity, but the specifics won’t be tested on the AP Physics exam.

How does the speed of a fluid flowing through a pipe change, as the diameter of the pipe decreases?

As pipe diameter decreases, fluid flowing through the pipe begins to flow faster.

This is akin to placing your thumb partially over the end of a water hose, which causes the water to flow through faster, creating a jet of water.

Water is flowing through a pipe, and the cross-sectional area of the pipe decreases by one-half. How does the fluid velocity change as the pipe area decreases?

The fluid velocity doubles as the area reduces by half.

The equation to describe the relationship between cross-sectional area and velocity is the **continuity equation**: A_{1}v_{1} = A_{2}v_{2}

Where:

A = the cross-sectional area of the pipe

v = the fluid velocity

Water is flowing through a round pipe, and the radius of the pipe increases by a factor of 3. How does the fluid velocity change as the pipe radius increases?

The fluid velocity decreases by a factor of 9.

According to the continuity equation A_{1}v_{1} = A_{2}v_{2}, there is an inverse proportionality between cross-sectional area and velocity.

However, cross-sectional area of a round pipe is proportional to the radius squared, so if the radius triples, the area goes up by a factor of 9, and therefore the velocity decreases by the same amount.

Define:

**surface tension** of a fluid

**Surface tension** is the ability of a fluid’s surface to resist an external force.

Surface tension is caused by attractive forces between the molecules of the fluid. It is responsible for the spherical shape of soap bubbles and the ability to skip a stone off the surface of a lake.

What sorts of fluids will have high surface tension?

Fluids with high intermolecular attractive forces will have large surface tension.

For instance, fluids whose molecules are bound by hydrogen bonding will have larger surface tension than nonpolar fluids, whose molecules are only attracted by dispersion forces.

Which fluid has a higher surface tension, water (H_{2}O) or carbon tetrachloride (CCl_{4})?

Water has a higher surface tension.

Water molecules are attracted by hydrogen bonding, while carbon tetrachloride molecules are nonpolar, and only attracted by dispersion forces. Hydrogen bonding forces are much stronger, and lead to surface tension nearly four times as strong.

Define:

**Bernoulli’s Equation**

**Bernoulli’s Equation** relates the pressure exerted by a fluid to its speed and depth.

Bernoulli’s Equation reads: P_{1} + ½ρv_{1}^{2} + ρgh_{1} = P_{2} + ½ρv_{2}^{2} + ρgh_{2}

Where:

P = the pressure exerted by the fluid

ρ = the density of the fluid

v = the speed of the fluid

h = the depth of the fluid

How does the pressure exerted by a fluid change as its speed increases, according to Bernoulli’s Equation?

As fluid speed increases, the pressure exerted by the fluid decreases.

Bernoulli’s Equation reads: P_{1} + ½ρv_{1}^{2} + ρgh_{1} = P_{2} + ½ρv_{2}^{2} + ρgh_{2}

Assuming that the fluid’s depth doesn’t change, the third term on each side is constant. If the speed increases, v_{2} > v_{1}, and the only way for the equation to hold is if P_{2} < P_{1}.

Why do you put boards on the outside of your windows during a hurricane?

During a hurricane, the wind speed increases significantly. According to Bernoulli’s equation, this results in a severe drop in pressure outside the house.

The pressure difference between the inside and outside leads to the danger of the windows being blown out; that is why boards are put on the outside.

How does the pressure exerted by a fluid change as altitude increases, according to Bernoulli’s Equation?

As altitude increases, pressure decreases.

Bernoulli’s Equation reads: P_{1} + ½ρv_{1}^{2} + ρgh_{1} = P_{2} + ½ρv_{2}^{2} + ρgh_{2}

Assuming a static fluid, or constant velocity, the second term on each side is constant and can be ignored. Since altitude increases, h_{2} > h_{1}; P_{2} must be less than P_{1} in order for the equation to hold.

How does the air pressure on top of a mountain compare to that at the base of the mountain.

The air pressure on top of the mountain is less.

Bernoulli’s Equation predicts that pressure decreases as altitude increases.

Define:

the **coefficient of thermal expansion** of a solid

An object’s coefficient of thermal expansion is a measure of how much the object’s size changes with temperature change.

Most objects expand when they heat, and shrink when they cool.

How much does the length of an object change when its temperature changes?

An object’s length change depends on its coefficient of thermal expansion: ΔL/L =α_{L} ΔT

Where:

L = the object’s original length

ΔL = the change in the object’s length

α_{L} = the object’s coefficient of thermal expansion

ΔT = the change in temperature

By how much does a 10 cm gold bar’s length change when it is heated from room temperature to 95 ºC?

The coefficient of thermal expansion of gold is 14x10^{-6}/ºC.

The gold bar expands by 0.1 mm.

The equation for thermal expansion is ΔL/L =α_{L} ΔT

ΔT in this case is 70ºC, so α_{L} ΔT is about 1x10^{-3}.

ΔL = 1x10^{-3} x L = 10^{-4} m

Define:

**Standard Temperature and Pressure (STP)**

**STP** is 0º C and 1 atm

Physics exams occasionally refer to Standard Ambient Temperature and Pressure (SATP), which you should know is 25ºC and 1 atm.

What are the assumptions of:

the **kinetic molecular theory of gases**

The **kinetic molecular theory** assumes an idealized version of gas, which makes calculating relationships easier. There are four assumptions:

- A gas molecule has no volume (point molecule).
- The collisions between gas molecules are completely elastic.
- There are no dissipative forces due to collisions.
- The average kinetic energy of gas molecules is directly proportional to the temperature of the gas.

Give the relationship and equation for:

**Charles’s Law**

**Charles’ Law** states that the volume of a gas is *directly proportional* to temperature while at a constant pressure.

Charles’ Law states: **V _{1}/T_{1} = V_{2}/T_{2}**

If the pressure of an ideal gas system is held constant but the temperature is doubled, what does Charles’s Law predict will happen to the volume?

The system’s volume will double also.

Charles’s Law indicates that at a constant pressure, the temperature and volume of a gas are directly proportional.

Give the relationship and equation for:

**Boyle’s Law**

**Boyle’s Law** states that the volume of a gas is *inversely proportional* to its pressure while at a constant temperature.

Boyle’s Law states: **P _{1}V_{1} = P_{2}V_{2}**

If the temperature of an ideal gas system is held constant but the pressure is reduced by 1/2, what does Boyle’s Law predict will happen to the volume?

The system’s volume will double.

Boyle’s Law indicates that at a constant temperature, the pressure and volume of a gas are inversely proportional.

Give the relationship and equation for:

**Avogadro’s Law**

**Avogadro’s Law** states that the volume of a gas is *directly proportional* to the number of moles at a constant temperature and pressure.

V / n = k

where:

V = volume in L

n = number of moles

k = proportionality constant of the specific gas

If the pressure and temperature of an ideal gas system are held constant but the number of moles is reduced to 1/3 of the original value, what does Avogadro’s Law predict will happen to the system’s volume?

The system’s volume will decrease to 1/3 also.

Avogadro’s Law indicates that at a constant pressure and temperature, the number of moles and volume of a gas are directly proportional.

Give the equation for:

the **Ideal Gas Law**

The **Ideal Gas Law** combines Charles’, Boyle’s, and Avogadro’s Laws into one:

PV = nRT

where:

P = Pressure in atm or Pa

V = Volume in L

n = Number of moles

R = Ideal gas constant .082 L(atm)/mol(K) or 8.31 J/mol(K)

T = Temperature in K

What is the volume of 1 mole of gas molecules at STP?

At STP, one mole of gas has a volume of 22.4 L

This is called the standard molar volume, and is true for a mole of any ideal gas at STP.

This value can be calculated using the Ideal Gas Law, but is worth memorizing.

Give the equation for:

calculating the partial pressure of a gas in a mixture

P_{A} = *x*_{A}P_{total}

where:

P_{A} = pressure from gas A

x_{A} = mole fraction of A (moles of A divided by total moles of gas)

P_{total} = total pressure of the sytem from all moles of gas

Give the equation for:

**Dalton’s Law**

P_{total} = P_{A} + P_{B} + P_{C} + …

**Dalton’s Law** states that the total pressure of a system of ideal gases can be thought of as the sum of all of the partial pressures that each gas exerts.