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Chapter 1: Fundamental Concepts of Thermodynamics

Problem numbers in italics indicate that the solution is included in the Student’s Solutions Manual. Questions on Concepts Q1.1) The location of the boundary between the system and the surroundings is a choice that must be made by the thermodynamicist. Consider a beaker of boiling water in an airtight room. Is the system open or closed if you place the boundary just outside the liquid water? Is the system open or closed if you place the boundary just inside the walls of the room? If the system boundaries are just outside of the liquid water, the system is open because water can escape from the top surface. The system is closed if the boundary is just inside the walls, because the room is airtight. Q1.2) Real walls are never totally adiabatic. Order the following walls in increasing order with respect to being diathermal: 1-cm-thick concrete, 1-cm-thick vacuum, 1-cm-thick copper, 1-cm-thick cork. 1-cm-thick vacuum < 1-cm-thick cork < 1-cm-thick concrete < 1-cm-thick copper Q1.3) Why is the possibility of exchange of matter or energy appropriate to the variable of interest a necessary condition for equilibrium between two systems? Equilibrium is a dynamic process in which the rates of two opposing processes are equal. However, if the rate in each direction is zero, no exchange is possible, and therefore the system can not reach equilibrium. Q1.4) At sufficiently high temperatures, the van der Waals equation has the form

m

RTPV b

≈−

. Note that the attractive part of the potential has no influence in this

expression. Justify this behavior using the potential energy diagram in Figure 1.6. At high temperatures, the energy of the molecule is large as indicated by the colored rectangular area in the figure below.

Chapter 1/Fundamental Concepts of Thermodynamics

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In this case, the well depth is a small fraction of the total energy. Therefore, the particle is unaffected by the attractive part of the potential. Q1.5) The parameter a in the van der Waals equation is greater for H2O than for He. What does this say about the form of the potential function in Figure 1.6 for the two gases? It says that the depth of the attractive potential is greater for H2O than for He. Q2.1) Electrical current is passed through a resistor immersed in a liquid in an adiabatic container. The temperature of the liquid is raised by 1ºC. The system consists solely of the liquid. Does heat or work flow across the boundary between the system and surroundings? Justify your answer. Although work is done on the resistor, this work is done in the surroundings. Heat flows across the boundary between the surroundings and the system because of the temperature difference between them. Q2.2) Explain how a mass of water in the surroundings can be used to determine q for a process. Calculate q if the temperature of 1.00 kg of water in the surroundings increases by 1.25ºC. Assume that the surroundings are at a constant pressure. If heat flows across the boundary between the system and the surroundings, it will lead to

a temperature change in the surroundings given by P

qTC

Δ = . For the case of interest,

-1 -1 31000 g 4.19 J g K 1.25 K 5.24 10 Jsurroundings Pq q mC T= − = − Δ = − × × = − × . Q2.3) Explain the relationship between the terms exact differential and state function.


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In order for a function f(x,y) to be a state function, it must be possible to write the total

differential df in the form y x

f fdf dx dyx y

⎛ ⎞∂ ∂⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠. If the form df as written exists, it is

an exact differential. Q2.4) Why is it incorrect to speak of the heat or work associated with a system? Heat and work are transients that exist only in the transition between equilibrium states. Therefore, a state at equilibrium is not associated with values of heat or work. Q2.5) Two ideal gas systems undergo reversible expansion starting from the same P and V. At the end of the expansion, the two systems have the same volume. The pressure in the system that has undergone adiabatic expansion is lower that in the system that has undergone isothermal expansion. Explain this result without using equations. In the system undergoing adiabatic expansion, all the work done must come through the lowering of ΔU, and therefore of the temperature. By contrast, some of the work done in the isothermal expansion can come at the expense of the heat that has flowed across the boundary between the system and surroundings. Q2.6) A cup of water at 278 K (the system) is placed in a microwave oven and the the oven is turned on for one minute, during which it begins to boil. Which of q, w, and ΔU are positive, negative or zero? The heat q is positive because heat flows across the system-surrounding boundary into the system. The work w is negative because the vaporizing water does work on the surroundings. ΔU is positive because the temperature increases and some of the liquid is vaporized. Q2.7) What is wrong with the following statement?: Because the well insulated house stored a lot of heat, the temperature didn't fall much when the furnace failed. Rewrite the sentence to convey the same information in a correct way. Heat can’t be stored because it exists only as a transient. A possible rephrasing follows. Because the house was well insulated, the walls were nearly adiabatic. Therefore, the temperature of the house did not fall as rapidly when in contact with the surroundings at a lower temperature as would have been the case if the walls were diathermal. Q2.8) What is wrong with the following statement?: Burns caused by steam at 100ºC can be more severe than those caused by water at 100ºC because steam contains more heat than water. Rewrite the sentence to convey the same information in a correct way. Heat is not a substance that can be stored. When steam is in contact with your skin, it condenses to the liquid phase. In doing so, energy is released that is absorbed by the


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skin. Hot water does not release as much energy in the same situation, because no phase change occurs. Q2.9) Describe how reversible and irreversible expansions differ by discussing the degree to which equilibrium is maintained between the system and the surroundings. In a reversible expansion, the system and surroundings are always in equilibrium with one another. In an irreversible expansion, they are not in equilibrium with one another. Q2.10) A chemical reaction occurs in a constant volume enclosure separated from the surroundings by diathermal walls. Can you say whether the temperature of the surroundings increase, decrease, or remain the same in this process? Explain. No. The temperature will increase if the reaction is exothermic, decrease if the reaction is endothermic, and not change if no energy is evolved in the reaction. Q3.1) Why is CP,m a function of temperature for ethane, but not for argon? Argon has only translational degrees of freedom, which are fully excited at very low temperatures because ,Δ <<translationalE kT where translationalEΔ is the spacing between translational levels. The translational degrees of freedom for ethane are also fully excited at 298 K. This condition is not fulfilled for the spacing between vibrational levels for ethane, and CP,m increases with temperature as the vibrational degrees of freedom become excited. Q3.2) Why is qV = ΔU only for a constant volume process? Is this formula valid if work other than P–V work is possible? Because ΔU = q + w, qV = ΔU only if w is zero. Therefore, the formula is not valid if work other than P–V work is possible.

Q3.3) Refer to Figure 1.6 and explain why T

UV

∂⎛ ⎞⎜ ⎟∂⎝ ⎠

is generally small for a real gas.

The depth of the minimum in the potential is generally very small. Therefore, it takes very little energy to separate the atoms or molecules that make up the gas.

Q3.4) Explain without using equations why T

HP

∂⎛ ⎞⎜ ⎟∂⎝ ⎠

is generally small for a real gas.

The variation of H with P is a measure of how the energy of a gas changes with the spacing between the molecules of the gas. Because the depth of the minimum in the potential in Figure 1.6 is generally very small, it takes very little energy to separate the atoms or molecules that make up the gas.


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Q3.5) Why is it reasonable to write PdH C dT VdP≈ + for a liquid or solid sample?

This approximation is valid because T

UT VV

∂⎛ ⎞ <<⎜ ⎟∂⎝ ⎠for a liquid or solid.

Q3.6) Why is the equation ( ) ( ),

f f

i i

T T

P P mT T

H C T dT n C T dTΔ = =∫ ∫ valid for an ideal gas

even if P is not constant in the process? Is this equation also valid for a real gas? Why or why not? It is valid because for an ideal gas, H is a function of T only, and is not a function of V or P. The formula is not accurate for a real gas, because H is a weak function of V and P. Q3.7) Heat capacity CP,m is less than CV,m for H2O(l) between 4º and 5ºC. Explain this result. This unusual behavior occurs because the density of water decreases with temperature in this range. Therefore, work is done by the surroundings on the system as water is heated at constant P between 4ºC and 5ºC. Q3.8) What is the physical basis for the experimental result that U is a function of V at constant T for a real gas? Under what conditions will U decrease as V increases? U is a function of V for a real gas because of the interaction potential between gas molecules. U decreases as V increases if the density is such that the gas is dominated by the repulsive part of the potential. Q3.9) Why does the relation CP > CV always hold for a gas? Can CP < CV be valid for a liquid? This is the case because the volume of a real gas always increases with increasing T. Therefore, heating a real gas at constant P results in the system doing work on the surroundings. CP < CV is valid for a liquid if V decreases with T at constant P. Although unusual, this condition is satisfied for water between 4ºC and the freezing point. Q3.10) Can a gas be liquefied through an isoenthalpic expansion if 0?J Tμ − = No. The gas would not be cooled if 0.J Tμ − = Q4.1) Under what conditions are ΔH and ΔU for a reaction involving gases and/or liquids or solids identical?


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ΔH = ΔU + Δ(PV). ΔH ≈ ΔU for reactions involving only liquids or solids because, to a good approximation, the volume does not change as a result of the reaction. If gases are iunvolved, Δ(PV)= Δ nRT. Therfore, ΔH ≈ ΔU if the number of moles of reactants and products that are gases are the same. Q4.2) If fHΔ o for the chemical compounds involved in a reaction are available at a given temperature, how can the reaction enthalpy be calculated at another temperature? The reaction enthalpy can be calculated to high accuracy at another temperature can be calculated only if the heat capacities of all reactants and products are known using

( )298.15298.15

T

T K PK

H H C T dT′ ′Δ = Δ + Δ∫o o . If the heat capacities of reactants and products

are similar, the reaction enthalpy will not vary greatly over a limited temperature range. Q4.3) Does the enthalpy of formation of compounds containing a certain element change if the enthalpy of formation of the element under standard state conditions is set equal to 100 kJ mol-1 rather than to zero? If it changes, how will it change for the compound AnBm if the formation enthalpy of element A is set equal to 100 kJ mol-1? Yes, because part of the enthalpy change of the reaction will be attributed to the element. For the reaction nA +mB → AnBm,

( ) ( ) ( )

( ) ( ) ( )n m

n m

A B A B

A B A Breaction f f f

f reaction f f

H H n H m H

H H n H m H

Δ = Δ − Δ − Δ

Δ = Δ + Δ + Δ

o o o o

o o o o

Therefore ( )n mA BfHΔ o for the compound will increase by 100n kJ mol-1. Q4.4) Is the enthalpy for breaking the first C-H bond in methane equal to the average C-H bond enthalpy in this molecule? Explain your answer. No. The average bond enthalpy is the average of the enthalpies of the four steps leading to complete dissociation. The enthalpy of each successive dissociation step will increase. Q4.5) Why is it valid to add the enthalpies of any sequence of reactions to obtain the enthalpy of the reaction that is the sum of the individual reactions? Because H is a state function, any path between the reactants and products, regardless of which intermediate products are involved, has the same value for ΔH. Q4.6) The reactants in the reaction 2NO(g) + O2(g) → 2NO2(g) are initially at 298 K. Why is the reaction enthalpy the same if the reaction is (a) constantly kept at 298 K or (b)


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if the reaction temperature is not controlled and the heat flow to the surroundings is measured only after the temperature of the products is returned to 298 K? Q4.7) In calculating reactionHΔ o at 285.15 K, only the fHΔ o of the compounds that take part in the reactions listed in Tables 4.1 and 4.2 (Appendix A, Data Tables) are needed. Is this statement also true if you want to calculate reactionHΔ o at 500 K? No. At any temperature other than 298.15 K, the heat capacities of all elements and compounds that appear in the overall reaction enter into the calculation. Q4.8) What is the point of having an outer water bath in a bomb calorimeter (see Figure 4.3), especially if its temperature is always equal to that of the inner water bath? This water bath effectively isolates the calorimeter and the inner water bath from the rest of the universe. Because the temperatures of the water baths are the same, there is no heat follow between them. Because the container of the inner water bath has rigid walls, no work is done on the composite system consisting of the calorimeter and the inner water bath. Therefore, the alorimeter and inner water bath form an isolated composite system. Q4.9) What is the advantage of a differential scanning calorimeter over a bomb calorimeter in determining the enthalpy of fusion of a series of samples? All the samples can be measured in parallel, rather than sequentially. This reduces both the measurement time, and the possibility of errors. Q4.10) You wish to measure the heat of solution of NaCl in water. Would the calorimetric technique of choice be at constant pressure or constant volume? Why? Constant pressure calorimetry is the technique of choice because none of the reactants or products is gaseous, and it is therefore not necessary to contain the reaction. Constant pressure calorimetry is much easier to carry out than constant volume calorimetry. Q5.1) Classify the following processes as spontaneous or not spontaneous and explain your answer. a) The reversible isothermal expansion of an ideal gas. b) The vaporization of superheated water at 102ºC and 1 bar. c) The constant pressure melting of ice at its normal freezing point by the addition of an infinitesimal quantity of heat. d) The adiabatic expansion of a gas into a vacuum. a) is not spontaneous because the system and surroundings are in equilibrium. b) is spontaneous because the equilibrium phase under the stated conditions is a gas. c) is not spontaneous because the process is reversible. d) is spontaneous because at equilibrium, the density of a gas is uniform throughout its container. Q5.2) Why are ΔSfusion and ΔSvaporization always positive?


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This is the case because ΔHfusion and ΔHvaporization are always positive. In each of these transitions, attractive forces must be overcome. Q5.3) Why is the efficiency of a Carnot heat engine the upper bound to the efficiency of an internal combustion engine? This the case because the maximum work than can be done on the surroundings in the expansion of a gas is in a reversible process. Q5.4) The amplitude of a pendulum consisting of a mass on a along wire is initially adjusted to have a very small value. The amplitude is found to decrease slowly with time. Is this process reversible? Would the process be reversible if the amplitude did not decrease with time? No, because dissipative forces are acting on the system. If the amplitude did not decrease, no dispersive forces act on the system and the oscillation is reversible in the limit of very small amplitudes. If the amplitude did not decrease with time, no dissipative forces act on the system, and because the amplitude is very small, the process is reversible. Q5.5) A process involving an ideal gas is carried out in which the temperature changes at constant volume. For a fixed value ΔT, the mass of the gas is doubled. The process is repeated with the same initial mass and ΔT is doubled. For which of these processes is ΔS greater? Why? ΔS is greater if the mass is doubled, because ΔS increases linearly with the amount of material. By contrast, ΔS only increases as the logarithm of the temperature. This increase is much slower than a linear increase.

Q5.6) Under what conditions does the equality HST

ΔΔ = hold?

Because reversibledqST

/Δ = , the equality holds if reversibledq H/ = Δ . This is the case for a

reversible process at constant pressure. Q5.7) Under what conditions is ΔS < 0 for a spontaneous process? For a spontaneous process, ΔS + ΔS surroundings < 0. The inequality ΔS < 0 is satisfied only if ΔS surroundings > 0 and surroundingsS SΔ > Δ .

Q5.8) Is the equation ( )lnf f

i i

T VfV

V f iiT V

TCS dT dV C V VT T

β βκ κ

Δ = + = + −∫ ∫ valid for an

ideal gas?


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No. Because β and κ are not independent of V for an ideal gas, they can’t be taken out of the integral. Q5.9) Without using equations, explain why ΔS for a liquid or solid is dominated by the temperature dependence of S as both P and T change. It is useful to think of S = S(V,T). Because V changes very little with P for a liquid or solid, entropy changes are dominated by changes in T rather than P for liquids and solids. Q5.10) You are told that ΔS = 0 for a process in which the system is coupled to its surroundings. Can you conclude that the process is reversible? Justify your answer. No. The criterion for reversibility is ΔS + ΔS surroundings = 0. To decide if this criterion is satisfied, ΔS surroundings must be known. Q6.1) Under what conditions is 0dA ≤ a condition that defines the spontaneity of a process? This is the case at constant T and V if no nonexpansion work is possible. Q6.2) Under what conditions is 0dG ≤ a condition that defines the spontaneity of a process? This is the case at constant T and P if no nonexpansion work is possible. Q6.3) Which thermodynamic state function gives a measure of the maximum electric work that can be carried out in a fuel cell? ΔG because nonexpansiondG dw≤ / Q6.4) By invoking the pressure dependence of the chemical potential, show that if a valve separating a vessel of pure A from a vessel containing a mixture of A and B is opened, mixing will occur. Both A and B are ideal gases, and the initial pressure in both vessels is 1 bar. The chemical potential of A in the mixture than in the pure gas because

( ) ( ), ln AA A

PT P T RTP

μ μ= +o

o, and PA < Pº. Because mass flows from regions of high

chemical potential to regions of low chemical potential, pure A will flow into the mixture, and the contents of both vessels will be mixed. Q6.5) Under what condition is KP = Kx?


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This is the case for a reaction involving gases if Δn = 0. Q6.6) It is found that KP is independent of T for a particular chemical reaction. What does this tell you about the reaction? This is true if ΔHreaction = 0. Q6.7) The reaction A + B → C + D is at equilibrium for ξ = 0.1. What does this tell you about the variation of Gpure with ξ ? It tells you that fGΔ o for A + B is less than that for C + D. If the fGΔ o of reactants and

products were nearly the same, ξ would be near 0.5. If the fGΔ o for the products were less than those for the reactants, ξ would be near 1. Q6.8) The reaction A + B → C + D is at equilibrium for ξ = 0.5. What does this tell you about the variation of Gpure with ξ ? It tells you that fGΔ o for A + B is equal to that for C + D. If the fGΔ o for the products

were less than those for the reactants, ξ would be near 1. If the fGΔ o for the reactants were less than those for the products, ξ would be much smaller than 0.5. Q6.9) Why is it reasonable to set the chemical potential of a pure liquid or solid substance equal to its standard state chemical potential at that temperature independent of the pressure in considering chemical equilibrium? The pressure dependence of G for solids and liquids is very small, because the volume of liquids and solids changes very little unless very high pressures are applied. Therefore, it is a good approximation to set the chemical potential of a pure liquid or solid substance equal to its standard state chemical potential at the temperature of interest.

Q6.10) Is the equation T

U T PV

β κκ

∂ −⎛ ⎞ =⎜ ⎟∂⎝ ⎠ valid for liquids, solids, and gases?

Yes. No assumptions have been made in deriving this differential relationship. Q6.11) What is the relationship between the KP for the two reactions 3/2H2 + 1/2N2 → NH3 and 3H2 + N2 → 2NH3?


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KP for the second reaction is the square of KP for the first reaction as can be seen the

reactions quotients

3 3

2 2 2 2

2

3 2 1 2 3 and

NH NH

H N H N

P PP P

P P P PP P P P

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

o o

o o o o

.

Q7.1) Explain why the oscillations in the two-phase coexistence region using the Redlich-Kwong and van der Waals equations of state (see Figure 7.4) do not correspond to reality. The oscillations predict that as V increases, P will increase. No real gas exhibits this behavior. Q7.2) Explain the significance of the Boyle temperature. The Boyle temperature provides a way to classify the way in which z varies with P at low values of P for different gases. If T > TB, z increases with increasing P; if T < TB, z decreases with increasing P. Q7.3) The value of the Boyle temperature increases with the strength of the attractive interactions between molecules. Arrange the Boyle temperatures of the gases Ar, CH4, and C6H6 in increasing order. Ar < CH4 < C6H6 Q7.4) Will the fugacity coefficient of a gas above the Boyle temperature be less than one at low pressures?

No. The integral 0

1P z dPP− ′′∫ is always greater than zero for this case. Therefore, γ > 1 for

all P. Q7.5) Using the concept of the intermolecular potential, explain why two gases in corresponding states can be expected to have the same value for z. Two different gases will have different values for the depth of the intermolecular potential and for the distance at which the potential becomes positive. By normalizing P, T, and V to their critical values, the differences in the intermolecular potential are to a significant extent also normalized. Q7.6) By looking at the a and b values for the van der Waals equation of state, decide whether 1 mole of O2 or H2O has the higher pressure at the same value of T and V.


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A is significantly larger for H2O. Therefore, the attractive forces between H2O molecules are greater than between O2 molecules. Consequently, O2 will have a higher pressure. Q7.7) Consider the comparison made between accurate results and those based on calculations using the van der Waals and Redlich-Kwong equations of state in Figures 7.1 and 7.5. Is it clear that one of these equations of state is better than the other under all conditions? The Redlich-Kwong gives more accurate results for almost all of the values of pressure shown in these figures. However, it is not better under all conditions. Q7.8) Why is the standard state of fugacity, f º, equal to the standard state of pressure, Pº? If this were not the case the fugacity would not become equal to the pressure in the limit of low pressures. Q7.9) For a given set of conditions, the fugacity of a gas is greater than the pressure. What does this tell you about the interaction between the molecules of the gas? If the fugacity is greater than the pressure, the repulsive pat of the potential dominates the interaction between the molecules. Q7.10) A system containing argon gas is at pressure P1 and temperature T1. How would you go about estimating the fugacity coefficient of the gas? Use the critical constants of the gas to determine the reduced pressure and temperature. With these results, estimate the fugacity coefficient using figure 7.11. Q8.1) At a given temperature, a liquid can coexist with its gas at a single value of the pressure. However, you can sense the presence of H2O(g) above the surface of a lake by the humidity, and it is still there if the barometric pressure rises or falls at constant temperature. How is this possible? The statement that at a given temperature, a liquid can coexist with its gas at a single value of the pressure holds for a system with only one substance. For the case described, the system consists of water and air. The change in barometric pressure is equivalent to an external pressure exerted on a liquid. As discussed in Section 8.5, this will change the vapor pressure only slightly. Q8.2) Why is it reasonable to show the μ versus T segments for the three phases as straight lines as is done in <LINK>Figure 8.1? More realistic curves would have some curvature. Is the curvature upward or downward on a μ versus T plot?


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P

STμ∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠

. Because S increases with T, the realistic curves will curve downward.

Q8.3) Figure 8.5 is not drawn to scale. What would be the relative lengths on the qP axis of the liquid + solid, liquid, and liquid + gas segments for water if the drawing were to scale and the system consisted of H2O? For the liquid + solid segment, the length of the segment is ΔHfusion, for the liquid segment, the length is ,

liquidP mC TΔ , and for the liquid + gas segment, the length is

ΔHvaporization. Numerically, for water, the relative lengths are 6008 : 7550 : 40656. Q8.4) Why is sublimation fusion vaporizationH H HΔ = Δ + Δ ? Because H is a state function, ΔH for the process solid → liquid → gas must be the same as for the process solid → gas if the initial and final states are the same. Q8.5) A triple point refers to a point in a P-T phase diagram for which three phases are in equilibrium. Do all triple points correspond to gas–liquid–solid equilibrium? No. If there are several solid phases, there can be a triple point corresponding to equilibrium between 3 solid phases. Q8.6) Why are the triple point temperature and the normal freezing point very close in temperature for most substances? This is the case because the freezing point changes only slightly with temperature. Q8.7) As the pressure is increased at –45ºC, ice I is converted to ice II. Which of these phases has the lower density? The higher density phase, in this case ice II, is more stable as the pressure is increased. Therefore ice I is the less dense phase. Q8.8) What is the physical origin of the pressure difference across a curved liquid–gas interface? The origin is the cohesive forces in the liquid. Across an interface, the resultant force vector is not zero in magnitude. If the interface is curved, the resultant force tends to minimize the surface area, leading to a pressure difference across the interface. Q8.9) Why does the triple point in a P-T diagram become a triple line in a P-V diagram? This is the case because a gas and a liquid are in equilibrium for the range of values of volume from that of the pure liquid to that of the pure gas at a given temperature.


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Q8.10) Give a molecular level explanation why the surface tension of Hg(l) is not zero. Hg atoms have a strong attractive interaction. Therefore, those atoms at the surface of a droplet that have fewer nearest neighbors than an atom in the liquid experience a net force perpendicular to the surface that will lead to a pressure increase inside the droplet.

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