# 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.

1
Q

Define:

a fluid

A

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.

2
Q

What are the properties of an ideal fluid?

A

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.

3
Q

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

A

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.

4
Q

Define:

Density (ρ) of a substance

A

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

The SI units of density are kg/m3.

5
Q

What are some commonly-used units of density?

A

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.

6
Q

What is the density of water in:

• g/cm3
• g/mL
• kg/L
• kg/m3
A

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.

7
Q

Define:

specific gravity of a substance

A

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.

8
Q

Define:

buoyancy

A

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.

9
Q

How is the buoyancy force of a submerged object calculated?

A

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

FB = ρliq * Vsub* g

Where:
FB = the buoyancy force pushing up on the object
ρliq = the density of the liquid
Vsub = the volume of the object that is submerged in the liquid
g = the gravitational acceleration (commonly 10 m/s2)

10
Q

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

A

The buoyancy force on the object is 15 N.

Using the definition of buoyancy force:
FB = ρliq * Vsub * g
FB = (1 kg/L) * (1.5 L) * (10 m/s2)
FB = 15 N

11
Q

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?

A

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

Remember, the buoyancy force is equal to ρliq * Vsub * 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.

12
Q

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

A

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.

13
Q

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?

A

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 ρobjliq, where ρobj is the object’s density and ρliq 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 ρobj > ρliq, this fraction is greater than 1, and the object sinks to the bottom.

14
Q

Define:

the effective weight of an object submerged in a liquid

A

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.

Weff = mobjg - (ρliq * Vobj * g)

Where:
mobj = the object’s mass
ρliq = the liquid’s density
Vobj = the object’s volume
g = 10 m/s2

15
Q

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?

A

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:

FB = ρliq * Vsub * 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.

16
Q

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?

A

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.

17
Q

Define:

Pascal’s Law of fluid pressures

A

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.

18
Q

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

A

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.

19
Q

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

A

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.

20
Q

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?

A

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.

21
Q

Define:

hydrostatic pressure on a submerged object

A

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.

22
Q

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

A

Hydrostatic pressure is calculated:

PH = ρliq * g * z

Where:
PH = the hydrostatic pressure
ρliq = the liquid’s density
g = 10 m/s2
z = distance beneath the liquid’s surface

23
Q

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?

A

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

Since the hydrostatic pressure is PH = ρliq * 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.

24
Q

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?

A

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

Since the hydrostatic pressure is PH = ρliq * 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.

25
Q

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

A

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.

26
Q

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

A

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.

27
Q

Under what conditions does fluid flow become turbulent?

A

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.

28
Q

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

A

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.

29
Q

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?

A

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

30
Q

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?

A

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.

31
Q

Define:

surface tension of a fluid

A

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.

32
Q

What sorts of fluids will have high surface tension?

A

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.

33
Q

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

A

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.

34
Q

Define:

Bernoulli’s Equation

A

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

35
Q

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

A

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.

36
Q

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

A

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.

37
Q

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

A

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.

38
Q

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

A

The air pressure on top of the mountain is less.

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

39
Q

Define:

the coefficient of thermal expansion of a solid

A

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.

40
Q

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

A

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

Where:
L = the object’s original length
ΔL = the change in the object’s length
αL = the object’s coefficient of thermal expansion
ΔT = the change in temperature

41
Q

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.

A

The gold bar expands by 0.1 mm.

The equation for thermal expansion is ΔL/L =αL ΔT

ΔT in this case is 70ºC, so αL ΔT is about 1x10-3.
ΔL = 1x10-3 x L = 10-4 m

42
Q

Define:

Standard Temperature and Pressure (STP)

A

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.

43
Q

What are the assumptions of:

the kinetic molecular theory of gases

A

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.
44
Q

Give the relationship and equation for:

Charles’s Law

A

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

45
Q

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?

A

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.

46
Q

Give the relationship and equation for:

Boyle’s Law

A

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

47
Q

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?

A

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.

48
Q

Give the relationship and equation for:

A

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

49
Q

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?

A

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.

50
Q

Give the equation for:

the Ideal Gas Law

A

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

51
Q

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

A

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.

52
Q

Give the equation for:

calculating the partial pressure of a gas in a mixture

A

PA = xAPtotal

where:
PA = pressure from gas A
xA = mole fraction of A (moles of A divided by total moles of gas)
Ptotal = total pressure of the sytem from all moles of gas

53
Q

Give the equation for:

Dalton’s Law

A

Ptotal = PA + PB + PC + …

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.

54
Q

Give the equation:

for internal energy (average kinetic energy) of a gas

A

U = KEavg = (3/2)nRT

where:
U = potential internal energy in J
KEavg = 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

55
Q

When does a gas deviate from its ideal state?

A

Gases deviate from ideal at:

1. Low temperatures
2. High pressures ( or very low volume)
56
Q

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

A

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.

57
Q

Give the equation:

Van Der Waals Equation

A

(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.

58
Q

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

A

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.

59
Q

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

A

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.

60
Q

Define:

absolute temperature

A

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

61
Q

Define and give an example of:

a thermodynamic system

A

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.

62
Q

Define and give an example of:

an open thermodynamic system

A

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.

63
Q

Define and give an example of:

a closed thermodynamic system

A

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.

64
Q

Define and give an example of:

an isolated thermodynamic system

A

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.

65
Q

Define:

“the surroundings” in a thermodynamic problem.

A

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

66
Q

Define:

a state function

A

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.

67
Q

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

A

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.

68
Q

Define:

enthalpy, H

A

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.

69
Q

What are the properties of an exothermic reaction?

A

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)

70
Q

What are the properties of an endothermic reaction?

A

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)

71
Q

Define:

a material’s standard enthalpy of formation, ΔHof

A

The standard enthalpy of formation (ΔHof) 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:

72
Q

What is the standard enthalpy of formation, ΔHof, of oxygen gas, O2?

A

Zero.

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

73
Q

What are standard conditions?

A

Standard conditions for a reaction require that:

• Pressure = 1 atm
• Temperature = 25º C = 298 K
74
Q

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?

A

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.

75
Q

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

A

-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.

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

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

76
Q

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

A

∆Hºrxn= Σ∆Hºf(products)-Σ∆Hºf(reactants)

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

77
Q

Is this reaction endothermic or exothermic?

CH4+ 2O2 ⇒ CO2+ 2H20

• ΔHof (CH4) = -75 kJ/mol
• ΔHof (CO2) = -394 kJ/mol
• ΔHof (H2O) = -286 kJ/mol
A

The reaction is exothermic.

∆H°rxn= Σ∆Hof(products)-Σ∆Hof(reactants)
= [CO2 + 2*H2O] - [CH4 + O2]

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

= -891 kJ/mol

78
Q

Define:

bond enthalpy

A

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.

79
Q

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

A

∆Hrxn = 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.

80
Q

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
A

-818 kJ/mol

∆Hrxn = 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

81
Q

Define:

specific heat, c

A

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.

82
Q

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

A

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.

83
Q

What is the specific heat of water?

A

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.

84
Q

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.

A

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.

85
Q

What does a calorimeter measure?

A

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

86
Q

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

A

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.

87
Q

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

A

q = mΔHL

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

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

88
Q

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?

A
89
Q

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?

A
90
Q

Define:

entropy of a system

A

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.

91
Q

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

A

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

Systems spontaneously tend towards arrangements with higher entropy.

92
Q

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

Ssolid, Sgas, Sliquid

A

Ssolid < Sliquid < Sgas

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.

93
Q

What is the entropy change for this chemical reaction?

2H2(g) + O2(g) ⇒ 2H2O(l)

A

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.

94
Q

What are the enthalpy and entropy changes for this reaction?

H2O (l) ⇒ H2O (g)

A

ΔHrxn > 0
ΔSrxn > 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.

95
Q

What are the enthalpy and entropy changes for this reaction?

CO2 (g) ⇒ CO2 (s)

A

ΔHrxn < 0
ΔSrxn < 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.

96
Q

Define and give the formula for calculating:

Gibbs’ Free Energy, ΔG

A

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’.

97
Q

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

A

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

98
Q

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

A

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.

99
Q

Define:

an exergonic reaction

A

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)

100
Q

Define:

an endergonic reaction

A

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)

101
Q

Describe the spontaneity of a reaction where:

ΔHrxn > 0
ΔSrxn < 0

A

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.

102
Q

Describe the spontaneity of a reaction where:

ΔHrxn < 0
ΔSrxn < 0

A

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.

103
Q

Describe the spontaneity of a reaction where:

ΔHrxn > 0
ΔSrxn < 0

A

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.

104
Q

Describe the spontaneity of a reaction where:

ΔHrxn < 0
ΔSrxn > 0

A

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.

105
Q

What is the difference between ΔG and ΔGº?

A

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.

106
Q

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

A

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.

107
Q

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

A

U = KEavg = (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)

108
Q

Define:

work (in thermodynamics)

A

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.

109
Q

Define:

heat (in thermodynamics)

A

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.

110
Q

What is the First Law of Thermodynamics?

A

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

111
Q

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

A

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.

112
Q

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

A

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.

113
Q

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?

A

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.

114
Q

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?

A

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.

115
Q

What are the characteristics of an adiabatic thermodynamic process?

A

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.

116
Q

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

A

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.

117
Q

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

A

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.

118
Q

What are the characteristics of an isothermal thermodynamic process?

A

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.

119
Q

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

A

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.

120
Q

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

A

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.

121
Q

Define:

the Second Law of Thermodynamics

A

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

ΔSuniverse > 0

ΔSsystem+ ΔSsurroundings > 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.

122
Q

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

A

ΔSX > 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.

123
Q

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

A

TK = TC + 273

TK = Kelvin Temperature
TC = Celsius Temperature

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

124
Q

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

A

TF = (9/5)*TC + 32

TF = Fahrenheit Temperature
TC = Celsius Temperature

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

125
Q

Define:

conduction

A

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.

126
Q

Define:

convection

A

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.

127
Q

Define:

A

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.

128
Q

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?

A

Conduction

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

129
Q

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

A

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.

130
Q

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

A

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.

131
Q

Define:

a material’s heat of vaporization, ΔHvap.

A

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

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

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

132
Q

Define:

a material’s heat of fusion, ΔHfus.

A

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

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

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

133
Q

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

A

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.