Fluids, Gases & Heat Flashcards

Question

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

Answer 1

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. ## Footnote 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.

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

The fluid velocity doubles as the area reduces by half. ## Footnote 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

Answer 6

The fluid velocity decreases by a factor of 9. ## Footnote 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.

Answer 7

**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.

Answer 8

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

Answer 9

Water has a higher surface tension. ## Footnote 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.

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

As altitude increases, pressure decreases. ## Footnote 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.

Answer 14

The air pressure on top of the mountain is less. ## Footnote Bernoulli's Equation predicts that pressure decreases as altitude increases.

Answer 15

An object's coefficient of thermal expansion is a measure of how much the object's size changes with temperature change. ## Footnote Most objects expand when they heat, and shrink when they cool.

Answer 16

An object's length change depends on its coefficient of thermal expansion: ΔL/L =α_L ΔT ## Footnote 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

Answer 17

The gold bar expands by 0.1 mm. ## Footnote 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

Answer 18

**STP** is 0º C and 1 atm ## Footnote Physics exams occasionally refer to Standard Ambient Temperature and Pressure (SATP), which you should know is 25ºC and 1 atm.

Answer 19

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.

Answer 20

**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**

Answer 21

The system's volume will double also. ## Footnote Charles's Law indicates that at a constant pressure, the temperature and volume of a gas are directly proportional.

Answer 22

**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**

Answer 23

The system's volume will double. ## Footnote Boyle's Law indicates that at a constant temperature, the pressure and volume of a gas are inversely proportional.

Answer 24

**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 ## Footnote where: V = volume in L n = number of moles k = proportionality constant of the specific gas

Answer 25

The system's volume will decrease to 1/3 also. ## Footnote Avogadro's Law indicates that at a constant pressure and temperature, the number of moles and volume of a gas are directly proportional.

Answer 26

The **Ideal Gas Law** combines Charles', Boyle's, and Avogadro's Laws into one: PV = nRT ## Footnote 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

Answer 27

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

Answer 28

P_A = *x*_AP_total ## Footnote 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

Answer 29

P_total = P_A + P_B + P_C + ... ## Footnote **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.

Answer 30

U = KE_avg = (3/2)nRT ## Footnote 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

Answer 31

Gases deviate from ideal at: 1. Low temperatures 2. High pressures ( or very low volume)

Answer 32

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.

Answer 33

(P + a(n/V)²)(V - nb) = nRT ## Footnote 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.

Answer 34

Polar gases will deviate more from ideal. ## Footnote (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.

Answer 35

Larger molecule gases will deviate more from ideal. ## Footnote (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.

Answer 36

The **absolute temperature** is any temperature that is given in Kelvin (K). ## Footnote 0 K is the temperature at which all molecules are assumed to cease moving completely. To convert Kelvin to Celsius: K = C + 273.15

Answer 37

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.

Answer 38

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.

Answer 39

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.

Answer 40

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.

Answer 41

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

Answer 42

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. ## Footnote 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.*

Answer 43

ΔX = X₂ - X₁ ## Footnote 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.

Answer 44

**Enthalpy** is a measure of the heat contained in a system. ## Footnote Enthalpy's absolute value cannot be directly measured, so the change in enthalpy's value, ΔH, is measured instead.

Answer 45

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. ## Footnote (Exo = exit)

Answer 46

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. ## Footnote (Endo = into)

Answer 47

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

Answer 48

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

Answer 49

**Standard conditions** for a reaction require that: * Pressure = 1 atm * Temperature = 25º C = 298 K * Concentration = 1 M for all products and reactants

Answer 50

ΔH₁ = ΔH₂ + ΔH₃ ## Footnote **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.

Answer 51

-152 kJ ## Footnote 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.

Answer 52

∆Hº_rxn= Σ∆Hº_f(*products*)-Σ∆Hº_f(*reactants*) ## Footnote Sum the enthalpies of formation of the products, and subtract the sum of the enthalpies of formation of the reactants.

Answer 53

The reaction is exothermic. ## Footnote ∆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

Answer 54

**Bond enthalpy** is the energy absorbed or released when a particular chemical bond is broken. ## Footnote Most chemical bonds are stabilizing, so most bond-breaking reactions are endothermic, and most bond enthalpies are positive.

Answer 55

∆H_rxn = Σ∆H(*bonds broken*) - Σ∆H(*bonds formed*) ## Footnote 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.

Answer 56

-818 kJ/mol ## Footnote ∆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

Answer 57

**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. ## Footnote The higher the value of c, the more heat it takes to raise the substance's temperature.

Answer 58

q = mcΔT ## Footnote 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.

Answer 59

4.184 J/g\*K ## Footnote 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.

Answer 60

The iron changes its temperature more. ## Footnote 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.

Answer 61

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

Answer 62

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

Answer 63

q = mΔH_L ## Footnote 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.

Answer 64

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

Answer 65

As a system's entropy increases, it becomes more disordered. ## Footnote Systems spontaneously tend towards arrangements with higher entropy.

Answer 66

S_solid \< S_liquid \< S_gas ## Footnote 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.

Answer 67

ΔS \< 0 ## Footnote 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.

Answer 68

ΔH_rxn \> 0 ΔS_rxn \> 0 ## Footnote The reaction is endothermic, since heat must be added to vaporize the water. Since the reaction creates gas, it represents an increase in entropy.

Answer 69

ΔH_rxn \< 0 ΔS_rxn \< 0 ## Footnote 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.

Answer 70

**ΔG** is a measure of the work which can be extracted from a thermodynamic system. ΔG = ΔH - TΔS ## Footnote 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'.

Answer 71

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

Answer 72

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.

Answer 73

An **exergonic reaction** is any reaction for which ΔG \< 0; hence all exergonic reactions are spontaneous. ## Footnote 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)

Answer 74

An **endergonic reaction** is any reaction for which ΔG \> 0; hence all endergonic reactions are non-spontaneous. ## Footnote 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)

Answer 75

This reaction will always be non-spontaneous. ## Footnote 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.

Answer 76

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

Answer 77

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

Answer 78

This reaction will always be spontaneous. ## Footnote 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.

Answer 79

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

Answer 80

**Temperature** and **internal energy** are equivalent concepts. ## Footnote 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.

Answer 81

U = KE_avg = (3/2)nkT ## Footnote 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)

Answer 82

**Work** is the flow of energy between a system and its surroundings in the form of changing pressure and volume of the system. ## Footnote In thermodynamics, work is signified by the symbol w.

Answer 83

**Heat** is the flow of energy between a system and its surroundings in any form other than work. ## Footnote In thermodynamics, heat is signified by the symbol q.

Answer 84

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

Answer 85

During a **compression**, work is being done *by* the environment, and *on* the system. ## Footnote Since energy is flowing into the system due to the work being done, ΔE \> 0.

Answer 86

During an **expansion**, work is being done *by* the system, and *on* the environment. ## Footnote Since energy is flowing out of the system due to the work being done, ΔE \< 0.

Answer 87

When the environment is at a higher temperature than the system, heat flows from the environment into the system, and Δq \> 0. ## Footnote This direction of heat flow leads to energy being added to the system, and ΔE \> 0.

Answer 88

When the environment is at a lower temperature than the system, heat flows from the system to the environment, and Δq \< 0. ## Footnote This direction of heat flow leads to energy being removed from the system, and ΔE \< 0.

Answer 89

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

Answer 90

The system's temperature must increase during an adiabatic compression. ## Footnote 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.

Answer 91

The system's temperature must decrease during an adiabatic expansion. ## Footnote 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.

Answer 92

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

Answer 93

Heat flows out of the system during an isothermal compression. ## Footnote 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.

Answer 94

Heat flows into the system during an isothermal expansion. ## Footnote 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.

Answer 95

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

Answer 96

ΔS_X \> 0 ## Footnote 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.

Answer 97

T_K = T_C + 273 ## Footnote 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

Answer 98

T_F = (9/5)\*T_C + 32 ## Footnote 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

Answer 99

**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.

Answer 100

**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.

Answer 101

**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.

Answer 102

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

Answer 103

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

Answer 104

Radiation ## Footnote 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.

Answer 105

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

Answer 106

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

Answer 107

Work = Δ(PV) ## Footnote 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.

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.