Fluids, Gases & Heat

Question 1

Define:

a fluid

Answer 1

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.

Question 2

What are the properties of an ideal fluid?

Answer 2

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.

Question 3

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

Answer 3

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.

Question 4

Define:

Density (ρ) of a substance

Answer 4

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

The SI units of density are kg/m³.

Question 5

What are some commonly-used units of density?

Answer 5

g/cm³, g/cc, g/mL, kg/L, and kg/m³ are all commonly-used measures of density.

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

Question 6

What is the density of water in:

• g/cm³
• g/mL
• kg/L
• kg/m³

Answer 6

The density of water is:

• 1 g/cm³
• 1 g/mL
• 1 kg/L
• 1000 kg/m³

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

Question 7

Define:

specific gravity of a substance

Answer 7

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.

Question 8

Define:

buoyancy

Answer 8

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.

Question 9

How is the buoyancy force of a submerged object calculated?

Answer 9

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/s²)

Question 10

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

Answer 10

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/s²)
F_B = 15 N

Question 11

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?

Answer 11

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.

Question 12

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

Answer 12

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.

Question 13

A cube of styrofoam, with a density of 0.5 g/cm³, is dropped into water. How much of the cube is submerged in the water when the cube begins to float?

Answer 13

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.

Question 14

Define:

the effective weight of an object submerged in a liquid

Answer 14

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_objg - (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/s²

Question 15

An aluminum ball with a density of 3 g/cm³ has a mass of 9 kg. What is its effective weight when it is submerged in water?

Answer 15

The ball’s effective weight is 60 N

With a mass of 9 kg and a density of 3 g/cm³, the ball’s volume must be 3000 cm³, 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/s²) = 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.

Question 16

A plastic ball with a density of 2 g/cm³ has a mass of 6 kg. What is a shortcut to calculating the effective weight of the ball when it is submerged in water?

Answer 16

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.

Question 17

Define:

Pascal’s Law of fluid pressures

Answer 17

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.

Question 18

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

Answer 18

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

The force F₁ will be exerted on the fluid behind plate 1 as a pressure P₁. The fluid will exert that pressure, undiminished, on all the walls it is contacting, including plate 2.

Question 19

How does the force F₂ change if plate 2 in the hydraulic lift pictured below doubles in area?

Answer 19

The force F₂ doubles.

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

P₁ = F₁/A₁ = F₂/A₂ = P₂

So force and area are directly proportional, and doubling A₂ will double F₂ as well.

Question 20

How does the distance D₂ compare to the distance D₁ if plate 2 in the hydraulic lift pictured is double the area of plate 1?

Answer 20

D₂ = ½D₁.

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

V₁ = A₁D₁ = A₂D₂ = V₂

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

Question 21

Define:

hydrostatic pressure on a submerged object

Answer 21

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.

Question 22

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

Answer 22

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/s²
z = distance beneath the liquid’s surface

Question 23

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?

Answer 23

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.

Question 24

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?

Answer 24

The unknown liquid’s density is 3 g/cm³.

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.

Question 25

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

Answer 25

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.

Question 26

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

Answer 26

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.

Question 27

Under what conditions does fluid flow become turbulent?

Answer 27

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.

Question 28

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

Answer 28

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.

Question 29

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?

Answer 29

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: A₁v₁ = A₂v₂

Where:
A = the cross-sectional area of the pipe
v = the fluid velocity

Question 30

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?

Answer 30

The fluid velocity decreases by a factor of 9.

According to the continuity equation A₁v₁ = A₂v₂, 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.

Question 31

Define:

surface tension of a fluid

Answer 31

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.

Question 32

What sorts of fluids will have high surface tension?

Answer 32

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.

Question 33

Which fluid has a higher surface tension, water (H₂O) or carbon tetrachloride (CCl₄)?

Answer 33

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.

Question 34

Define:

Bernoulli’s Equation

Answer 34

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

Bernoulli’s Equation reads: P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

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

Question 35

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

Answer 35

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

Bernoulli’s Equation reads: P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

Assuming that the fluid’s depth doesn’t change, the third term on each side is constant. If the speed increases, v₂ > v₁, and the only way for the equation to hold is if P₂ < P₁.

Question 36

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

Answer 36

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.

Question 37

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

Answer 37

As altitude increases, pressure decreases.

Bernoulli’s Equation reads: P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

Assuming a static fluid, or constant velocity, the second term on each side is constant and can be ignored. Since altitude increases, h₂ > h₁; P₂ must be less than P₁ in order for the equation to hold.

Question 38

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

Answer 38

The air pressure on top of the mountain is less.

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

Question 39

Define:

the coefficient of thermal expansion of a solid

Answer 39

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.

Question 40

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

Answer 40

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

Question 41

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.

Answer 41

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

Question 42

Define:

Standard Temperature and Pressure (STP)

Answer 42

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.

Question 43

What are the assumptions of:

the kinetic molecular theory of gases

Answer 43

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

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

Question 44

Give the relationship and equation for:

Charles’s Law

Answer 44

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

Charles’ Law states: V₁/T₁ = V₂/T₂

Question 45

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?

Answer 45

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.

Question 46

Give the relationship and equation for:

Boyle’s Law

Answer 46

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: P₁V₁ = P₂V₂

Question 47

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?

Answer 47

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.

Question 48

Give the relationship and equation for:

Avogadro’s Law

Answer 48

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

Question 49

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?

Answer 49

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.

Question 50

Give the equation for:

the Ideal Gas Law

Answer 50

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

Question 51

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

Answer 51

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.

Question 52

Give the equation for:

calculating the partial pressure of a gas in a mixture

Answer 52

P_A = x_AP_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

Question 53

Give the equation for:

Dalton’s Law

Answer 53

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.

Question 54

Give the equation:

for internal energy (average kinetic energy) of a gas

Answer 54

U = KE_avg = (3/2)nRT

where:
U = potential internal energy in J
KE_avg = average kinetic energy in J
n = number of moles of gas
R = Ideal gas constant .082 L(atm)/mol(K) or 8.31 J/mol(K)
T = Temperature in K

Question 55

When does a gas deviate from its ideal state?

Answer 55

Gases deviate from ideal at:

1. Low temperatures
2. High pressures ( or very low volume)

Question 56

Why does low temperature and high pressure cause a gas deviate from its ideal state?

Answer 56

Gases deviate from ideal at low temperatures, because the molecules have very low kinetic energy (hence low velocity) and will start to attract based on the intermolecular forces.

Gases deviate from ideal at high pressures or very low volumes, because the molecules have very little actual space to move in and will start to exhibit characteristics more like a fluid than a gas.

Question 57

Give the equation:

Van Der Waals Equation

Answer 57

(P + a(n/V)²)(V - nb) = 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
a = attraction between molecules due to intermolecular forces
b = actual volume taken up by gas molecules

Note: the van der Waals equation predicts the behavior of non-ideal gases, taking into account the intermolecular attractive forces of the gas molecules and the space taken up by the non-point molecules.

Question 58

Explain whether a polar or nonpolar real gas will deviate more from ideal?

Answer 58

Polar gases will deviate more from ideal.

(P + a(n/V)²)(V - nb) = nRT

Remember: ‘a’ is the attractiveness between molecules. When ‘a’ is big (polar=attractive) the pressure term will be larger. Effectively, the gas will experience more pressure holding it together than the ideal gas law predicts.

Question 59

Explain whether a small or large molecule size real gas will deviate more from ideal?

Answer 59

Larger molecule gases will deviate more from ideal.

(P + a(n/V)²)(V - nb) = nRT

Remember: ‘b’ is the bigness of the actual molecules. When ‘b’ is big (larger size=big) the volume term will be smaller. Effectively, the gas will have less space to move in than the ideal gas law predicts.

Question 60

Define:

absolute temperature

Answer 60

The absolute temperature is any temperature that is given in Kelvin (K).

0 K is the temperature at which all molecules are assumed to cease moving completely. To convert Kelvin to Celsius: K = C + 273.15

Question 61

Define and give an example of:

a thermodynamic system

Answer 61

A thermodynamic system is a macroscopic body which is engaged in mass and/or energy exchange with its surroundings.

Ex: The classic example of a thermodynamic system is a piston filled with gas.

Question 62

Define and give an example of:

an open thermodynamic system

Answer 62

An open thermodynamic system can exchange both mass and energy with the environment.

Ex: A bottle of gas with no lid is an open thermodynamic system.

Question 63

Define and give an example of:

a closed thermodynamic system

Answer 63

A closed thermodynamic system cannot exchange mass with the environment, but can exchange energy.

Ex: A bottle of gas with the lid securely on is a closed thermodynamic system.

Question 64

Define and give an example of:

an isolated thermodynamic system

Answer 64

An isolated thermodynamic system can not exchange either energy or mass with the environment.

Ex: A closed, double-walled (insulated) container which is temperature-independent of its environment is an isolated system.

Question 65

Define:

“the surroundings” in a thermodynamic problem.

Answer 65

The surroundings (also known as the environment) are everything capable of exchanging mass and/or energy with the system.

Question 66

Define:

a state function

Answer 66

A state function is any property of a thermodynamic system that depends only on comparing the characteristics of the system at that moment compared to a prior moment.

Since state functions are calculated based on current vs past properties only, their values do not depend on the path by which the current state was achieved; they are path-independent.

Question 67

If X is a state function, what is the change in X between a system’s values X₁ and X₂?

Answer 67

ΔX = X₂ - X₁

Since X is a state function, the path by which it gets from state 1 to 2 is irrelevant, the change in X depends only on the starting and finishing states.

Question 68

Define:

enthalpy, H

Answer 68

Enthalpy is a measure of the heat contained in a system.

Enthalpy’s absolute value cannot be directly measured, so the change in enthalpy’s value, ΔH, is measured instead.

Question 69

What are the properties of an exothermic reaction?

Answer 69

An exothermic reaction is any reaction whose products have a lower enthalpy than the reactants. Therefore, ΔH < 0, and heat is lost from the system to the environment.

(Exo = exit)

Question 70

What are the properties of an endothermic reaction?

Answer 70

An endothermic reaction is any reaction whose products have a higher enthalpy than the reactants. Therefore, ΔH > 0, and heat is absorbed by the system from the environment.

(Endo = into)

Question 71

Define:

a material’s standard enthalpy of formation, ΔH^o_f

Answer 71

The standard enthalpy of formation (ΔH^o_f) is the enthalpy change of the formation reaction for the material from its fundamental elements under standard conditions.

Ex: The enthalpy of formation for NaCl is -411.12 kJ mol⁻¹ and follows from the general equation:

Question 72

What is the standard enthalpy of formation, ΔH^o_f, of oxygen gas, O₂?

Answer 72

Zero.

By definition, the enthalpy of formation for any material in its standard state is zero.

Question 73

What are standard conditions?

Answer 73

Standard conditions for a reaction require that:

• Pressure = 1 atm
• Temperature = 25º C = 298 K
• Concentration = 1 M for all products and reactants

Question 74

If the chemical reaction

(1) 2A ⇒ C ΔH₁

can be broken down into the sub-reactions

(2) 2A ⇒ B ΔH₂
(3) B ⇒ C ΔH₃

what does Hess’s Law tell you about the overall enthalpy change ΔH₁?

Answer 74

ΔH₁ = ΔH₂ + ΔH₃

Hess’s Law simply states that the enthalpy of a reaction can be calculated by adding together the enthalpies of a chain of sub-reactions which add up to the overall reaction.

Although most commonly applied to enthalpy, Hess’s Law applies to all state functions.

Question 75

What is the enthalpy change when 2 moles of CH₄ are formed, according to the following reactions?

rxn1: 2H₂(g)⇒4H(g)
ΔH₁= -870 kJ/mol

rxn2: C(s) + 4H(g)⇒CH₄(g)
ΔH₂= +794 kJ/mol

Answer 75

-152 kJ

Adding the reactions together yields the formation reaction of CH₄:

C(s) + 4H(g) + 2H₂(g) ⇒
CH₄(g) + 4H(g)
Canceling common terms leaves:
C(s) + 2H₂(g) ⇒CH₄(g)

To complete the calculation, combine the reactions’ enthalpies in the same way the reactions were combined.

ΔH_rxn = ΔH₁ + ΔH₂
-870 + 794 = -76kJ/mol

Finally, multiply by the number of moles (2) to get the final answer.

Question 76

How can the enthalpy change of a reaction be calculated from the enthalpies of formation of the reactants and products?

Answer 76

∆Hº_rxn= Σ∆Hº_f(products)-Σ∆Hº_f(reactants)

Sum the enthalpies of formation of the products, and subtract the sum of the enthalpies of formation of the reactants.

Question 77

Is this reaction endothermic or exothermic?

CH₄+ 2O₂ ⇒ CO₂+ 2H₂0

• ΔH^o_f (CH₄) = -75 kJ/mol
• ΔH^o_f (CO₂) = -394 kJ/mol
• ΔH^o_f (H₂O) = -286 kJ/mol

Answer 77

The reaction is exothermic.

∆H°_rxn= Σ∆H^o_f(products)-Σ∆H^o_f(reactants)
= [CO₂ + 2*H₂O] - [CH₄ + O₂]

= [-394 kJ/mol + 2(-286 kJ/mol)]
- [-75 kJ/mol + 2(0 kJ)]

= -891 kJ/mol

Question 78

Define:

bond enthalpy

Answer 78

Bond enthalpy is the energy absorbed or released when a particular chemical bond is broken.

Most chemical bonds are stabilizing, so most bond-breaking reactions are endothermic, and most bond enthalpies are positive.

Question 79

How can the enthalpy of a reaction be calculated from the bond enthalpies of the reactants and products?

Answer 79

∆H_rxn = Σ∆H(bonds broken) - Σ∆H(bonds formed)

The reaction’s enthalpy change is identical to the energy needed to break all the bonds in the reactants, minus the energy released when the bonds in the products form.

Question 80

What is the overall enthalpy change of this reaction?

CH₄+ 2O₂ ⇒ CO₂+ 2H₂0

• ΔH (C-H) = 411 kJ/mol
• ΔH (O=O) = 494 kJ/mol
• ΔH (C=O) = 799 kJ/mol
• ΔH (O-H) = 463 kJ/mol

Answer 80

-818 kJ/mol

∆H_rxn = Σ∆H(bonds broken) - Σ∆H(bonds formed)
= [4*(C-H) + 2*(O=O)] - [2*(C=O) + 2*2*(H-O)]

= [(4 * 411) + (2 * 494)]
- [(2 * 799) + (4 * 463)] kJ/mol

= -818 kJ/mol

Question 81

Define:

specific heat, c

Answer 81

Specific heat is a characteristic property of a material, and is the amount of heat which must be added to raise 1 g of the substance by 1ºC.

The higher the value of c, the more heat it takes to raise the substance’s temperature.

Question 82

What is the formula for necessary quantity of heat in a specific heat problem?

Answer 82

q = mcΔT

Where q is the necessary heat, m is the mass of substance present, c is the substance’s specific heat, and ΔT is the desired temperature change.

Question 83

What is the specific heat of water?

Answer 83

4.184 J/g*K

This is a value that you should have memorized. It means that it takes 4.184 J to raise 1 g of water by 1ºK.

This value is equal to 1 cal/g*K.

Question 84

8 J of heat is applied to 1 g of both iron and water at 25ºC. Which changes temperature more?

The specific heat of water is 4.184 J/g-K.

The specific heat of iron is 0.46 J/g-K.

Answer 84

The iron changes its temperature more.

Applying the equation q = mcΔT to both cases and solving for ΔT reveals that the iron will change temperature by about 20 degrees (final T = 45ºC), while the water will only increase by 2 degrees (final T = 27ºC).

The higher a material’s specific heat, the less responsive its temperature is to heat flow.

Question 85

What does a calorimeter measure?

Answer 85

A calorimeter measures the amount of heat given off by a particular chemical reaction or process.

There are many different styles of calorimeter, but for most science courses, including the AP Physics exam, you should focus on the fact that they all measure heat generated via a system’s temperature change, using the equation

q = mcΔT

Question 86

Why doesn’t the specific heat equation q = mcΔT apply during a phase change?

Answer 86

During a phase change, temperature stays constant as heat is added. The added heat causes the material to go through the phase change, rather than increasing its temperature.

The amount of heat needed to make a material change its phase is known as the latent heat of that phase change.

Question 87

What is the equation for calculating the heat of a phase change?

Answer 87

q = mΔH_L

where q = necessary heat, m = mass of the substance present, and ΔH_L is the latent heat of the phase change.

The higher the ΔH_L, the more heat it takes to force the substance to go through the phase change.

Question 88

The curve below represents a sample’s temperature vs. heat added. What phases (solid, liquid, and/or gas) are present at each labeled point on the plot?

Answer 88

Question 89

The curve below represents a sample’s temperature vs. heat added. What heat (q) formula would need to be applied, in order to calculate heat added to the system at each labeled point on the plot?

Answer 89

Question 90

Define:

entropy of a system

Answer 90

Entropy is a macroscopic property of a system, representing the number of possible ways the atoms or molecules of the system can arrange themselves.

Note: Colloquially, entropy is said to represent the ‘possibility for disorder’ of a system, and this definition is effective enough for most entropy questions on the AP Physics exam.

Question 91

What does increasing entropy say about a system’s properties?

Answer 91

As a system’s entropy increases, it becomes more disordered.

Systems spontaneously tend towards arrangements with higher entropy.

Question 92

Arrange the relative entropy levels of the 3 phases of matter:

S_solid, S_gas, S_liquid

Answer 92

S_solid < S_liquid < S_gas

Since entropy is a measure of disorder, the more ordered a system, the lower its entropy. Solid matter, with its regularly-repeating units and fairly well-defined locations for the atoms, is therefore low in entropy, while gases, with their continuous, random motion, have the highest entropy values.

Question 93

What is the entropy change for this chemical reaction?

2H₂(g) + O₂(g) ⇒ 2H₂O(l)

Answer 93

ΔS < 0

Gases have more entropy than liquids; hence, in any chemical reaction, the side with more moles of gas will have higher entropy.

In general, reactions with fewer moles of products than reactants will have lower final entropy.

Question 94

What are the enthalpy and entropy changes for this reaction?

H₂O (l) ⇒ H₂O (g)

Answer 94

ΔH_rxn > 0
ΔS_rxn > 0

The reaction is endothermic, since heat must be added to vaporize the water.

Since the reaction creates gas, it represents an increase in entropy.

Question 95

What are the enthalpy and entropy changes for this reaction?

CO₂ (g) ⇒ CO₂ (s)

Answer 95

ΔH_rxn < 0
ΔS_rxn < 0

The reaction is exothermic, since heat must be removed to deposit the CO₂.

Since the reaction results in a net decrease of gas, it represents an decrease in entropy.

Question 96

Define and give the formula for calculating:

Gibbs’ Free Energy, ΔG

Answer 96

ΔG is a measure of the work which can be extracted from a thermodynamic system.

ΔG = ΔH - TΔS

It also is used to measure the spontaneity of a system; chemical systems will always tend to move in a direction of decreasing ΔG.

A handy mnemonic for remembering the formula for ΔG is: ‘Get Higher Test Scores’.

Question 97

What does a negative value for ΔG imply about a chemical reaction?

Answer 97

If ΔG < 0 for a chemical reaction, then the forward reaction is spontaneous, favoring the creation of more products.

Question 98

What does a positive value for ΔG imply about a chemical reaction?

Answer 98

If ΔG > 0 for a chemical reaction, then the forward reaction is non-spontaneous, and the reverse reaction is spontaneous, favoring the creation of more reactants.

Question 99

Define:

an exergonic reaction

Answer 99

An exergonic reaction is any reaction for which ΔG < 0; hence all exergonic reactions are spontaneous.

This is very similar to the definition of exothermic, for which ΔH < 0. “Thermic” refers to Enthalpy, “gonic” to Gibbs’ Free Energy.

(Exerg = exit G)

Question 100

Define:

an endergonic reaction

Answer 100

An endergonic reaction is any reaction for which ΔG > 0; hence all endergonic reactions are non-spontaneous.

This is very similar to the definition of endothermic, for which ΔH > 0. “Thermic” refers to Enthalpy, “gonic” to Gibbs’ Free Energy.

(Enderg = enter G)

Question 101

Describe the spontaneity of a reaction where:

ΔH_rxn > 0
ΔS_rxn < 0

Answer 101

This reaction will always be non-spontaneous.

Remember, ΔG = ΔH - TΔS. Since T is in Kelvin and will always be positive, then both ΔH and -TΔS are positive. The quantity ΔG will always be positive then, so the reaction is endergonic at all temperatures.

Question 102

Describe the spontaneity of a reaction where:

ΔH_rxn < 0
ΔS_rxn < 0

Answer 102

This reaction will be spontaneous at low temperatures, and non-spontaneous at sufficiently high temperatures.

Remember, ΔG = ΔH - TΔS. When T is low (near zero Kelvin), the second term can be ignored, and ΔG will be negative due to ΔH. But when T becomes large enough, the positive sign of the -TΔS term dominates, and ΔG becomes positive.

Question 103

Describe the spontaneity of a reaction where:

ΔH_rxn > 0
ΔS_rxn < 0

Answer 103

This reaction will be spontaneous at high temperatures and non-spontaneous at sufficiently low temperatures.

Remember, ΔG = ΔH - TΔS. When T is low (near zero Kelvin), the second term can be ignored, and ΔG will be positive due to ΔH>0. But when T becomes large enough, the negative sign of the -TΔS term dominates, and ΔG becomes negative.

Question 104

Describe the spontaneity of a reaction where:

ΔH_rxn < 0
ΔS_rxn > 0

Answer 104

This reaction will always be spontaneous.

Remember, ΔG = ΔH - TΔS. Since T is in Kelvin and will always be positive, both ΔH and -TΔS are negative. The quantity ΔG will always be negative, so the reaction is exergonic at all temperatures.

Question 105

What is the difference between ΔG and ΔGº?

Answer 105

ΔG describes the change in Gibbs’ Free Energy for a chemical system at a particular pressure and temperature, which must be given.

ΔGº describes the change in Gibbs’ Free Energy for a chemical system at standard conditions (1 atm, 298 K, 1 M in all concentrations).

Typically, different reactions’ ΔGº values will be compared, since this gives a common point of reference between them.

Question 106

What is the relationship between a thermodynamic system’s temperature and its internal energy?

Answer 106

Temperature and internal energy are equivalent concepts.

Ex: If a system’s absolute temperature doubles, so does the amount of energy that it contains.

This is the foundation of the zeroth law of thermodynamics, which states that if systems A and C are both in thermal equilibrium with system B, they are also in equilibrium with each other; i.e. they are at the same temperature.

Question 107

Give the relationship between a fluid’s temperature and the internal energy (average kinetic energy) of the molecules in the fluid.

Answer 107

U = KE_avg = (3/2)nkT

k is Boltzmann’s constant(1.38x10^-23 J/K)
T is the absolute temperature (in Kelvin)
n is the number of fluid molecules (in mol)

Question 108

Define:

work (in thermodynamics)

Answer 108

Work is the flow of energy between a system and its surroundings in the form of changing pressure and volume of the system.

In thermodynamics, work is signified by the symbol w.

Question 109

Define:

heat (in thermodynamics)

Answer 109

Heat is the flow of energy between a system and its surroundings in any form other than work.

In thermodynamics, heat is signified by the symbol q.

Question 110

What is the First Law of Thermodynamics?

Answer 110

The First Law of Thermodynamics states that heat and work are the only ways in which energy can flow, and energy is always conserved, hence energy change can be calculated:

ΔE = Δq + Δw

Question 111

Explain the direction of work being done and the sign of ΔE during the compression of a thermodynamic system.

Answer 111

During a compression, work is being done by the environment, and on the system.

Since energy is flowing into the system due to the work being done, ΔE > 0.

Question 112

Explain the direction of work being done and the sign of ΔE during the expansion of a thermodynamic system.

Answer 112

During an expansion, work is being done by the system, and on the environment.

Since energy is flowing out of the system due to the work being done, ΔE < 0.

Question 113

If a thermodynamic system is surrounded by an environment which is at a higher temperature, what is the direction of heat flow? What does this mean for the sign of ΔE?

Answer 113

When the environment is at a higher temperature than the system, heat flows from the environment into the system, and Δq > 0.

This direction of heat flow leads to energy being added to the system, and ΔE > 0.

Question 114

If a thermodynamic system is surrounded by an environment which is at a lower temperature, what is the direction of heat flow? What does this mean for the sign of ΔE?

Answer 114

When the environment is at a lower temperature than the system, heat flows from the system to the environment, and Δq < 0.

This direction of heat flow leads to energy being removed from the system, and ΔE < 0.

Question 115

What are the characteristics of an adiabatic thermodynamic process?

Answer 115

An adiabatic process is any process where heat cannot flow. Hence, Δq = 0 and ΔE = Δw; all energy change is due to work being done on or by the system.

Adiabatic processes are processes that happen in either heat-insulated systems, or that happen so quickly that heat cannot flow between system and environment.

Question 116

What happens to the system’s temperature during an adiabatic compression?

Answer 116

The system’s temperature must increase during an adiabatic compression.

Remember, Δq = 0 for all adiabatic processes. Furthermore, during any compression, work is being done on the system, raising ΔE. Since T and ΔE are proportional, the system must gain temperature during an adiabatic compression.

Question 117

What happens to the system’s temperature during an adiabatic expansion?

Answer 117

The system’s temperature must decrease during an adiabatic expansion.

Remember, Δq = 0 for all adiabatic processes. Furthermore, during any expansion, the system is doing work, causing it to lose energy. Since T and ΔE are proportional, the system must lose temperature during an adiabatic expansion.

Question 118

What are the characteristics of an isothermal thermodynamic process?

Answer 118

An isothermal process is any process where temperature is held constant. Hence, ΔE = 0, and Δq = -Δw. Any heat flow into or out of the system is compensated by work done by or on the system, respectively.

Isothermal processes are usually processes that happen slowly enough that the system’s temperature can constantly equilibrate.

Question 119

What is the direction of heat flow during an isothermal compression?

Answer 119

Heat flows out of the system during an isothermal compression.

Remember, during any compression, work is being done on the system, raising ΔE. Since an isothermal compression must have an overall ΔE = 0, Δq must be negative to compensate.

Question 120

What is the direction of heat flow during an isothermal expansion?

Answer 120

Heat flows into the system during an isothermal expansion.

Remember, during any expansion, the system is doing work, losing energy. Since any isothermal process must have an overall ΔE = 0, Δq must be positive to compensate.

Question 121

Define:

the Second Law of Thermodynamics

Answer 121

The Second Law of Thermodynamics states that for any thermodynamic process, the total entropy of the universe increases.

ΔS_universe > 0

ΔS_system+ ΔS_surroundings > 0

since the universe can be defined as a system and its surroundings. If the entropy of a system decreases, therefore, the entropy of its surroundings must increase by a greater amount in order for the universal law to still hold true.

Question 122

If System X is completely isolated from its surroundings, what must be true of ΔS_X for any process that occurs inside System X?

Answer 122

ΔS_X > 0

Since System X is isolated it cannot exchange energy or entropy with its surroundings. For the Second Law to hold, the system’s entropy must increase.

Question 123

What is the formula for converting between temperature in Celsius and Kelvin?

Answer 123

T_K = T_C + 273

T_K = Kelvin Temperature
T_C = Celsius Temperature

Some standard temperatures to memorize:
0º C = 273º K
25º C = 298º K
100º C = 373º K

Question 124

What is the formula for converting between temperature in Centigrade and Fahrenheit?

Answer 124

T_F = (9/5)*T_C + 32

T_F = Fahrenheit Temperature
T_C = Celsius Temperature

Some standard conversions to memorize:
32º F = 0º C
77º F = 25º C
212º F = 100º C

Question 125

Define:

conduction

Answer 125

Conduction is the transfer of thermal energy via molecular collisions. It requires physical contact between the systems exchanging energy.

When the molecules collide, molecules of the higher-energy system transfer some of their energy to the lower energy molecules of the other system, cooling the first and heating the second.

Question 126

Define:

convection

Answer 126

Convection is the thermal energy transfer via the movement of fluid in currents. It requires at least one medium which is capable of motion.

Differences in pressure or density drive warm fluid in the direction of cooler fluid, transferring heat away from a warmer object or towards a cooler one.

Question 127

Define:

radiation

Answer 127

Radiation is the thermal energy transfer via emission or absorption of electromagnetic waves.

Radiation does not require a medium and can occur through a vacuum.

Question 128

When a hot piece of metal is placed on a wooden table, causing the table to start to smoke, what is the primary form of heat transfer being exhibited?

Answer 128

Conduction

Conduction is the most efficient form of heat transfer; since the metal and table are in direct physical contact, conduction transfer will dominate.

Question 129

When the upstairs of a house is warmer than the basement, what form of heat transfer is being exhibited?

Answer 129

Convection

The decreased density of the warmer air causes it to rise to the top of the house, a classic example of a convection current.

Question 130

When the sun rises above the horizon, warming the ground, what form of heat transfer is being exhibited?

Answer 130

Radiation

For the heat to get from the sun to the earth, it must travel across millions of miles of nearly empty space. Only electromagnetic radiation can carry energy across this distance.

Question 131

Define:

a material’s heat of vaporization, ΔH_vap.

Answer 131

ΔH_vap is the energy needed to vaporize one mole of a substance from its liquid phase to the gas phase at constant pressure.

ΔH_vap may also be reported as heat per gram, in which case it is the specific heat of vaporization.

ΔH_vap is always positive, since vaporization is an endothermic process.

Question 132

Define:

a material’s heat of fusion, ΔH_fus.

Answer 132

ΔH_fus is the energy needed to melt one mole of a substance from its solid phase to its liquid phase at constant pressure.

ΔH_fus may also be reported as heat per gram, in which case it is the specific heat of fusion.

ΔH_fus is always positive, since melting is an endothermic process.

Question 133

What is the work done on a thermodynamic system as a function of its pressure and volume?

Answer 133

Work = Δ(PV)

At constant pressure, W = PΔV

More generally, the work done during a thermodynamic process is the area under the curve of a P vs V diagram.