Resolução Atkins' Physical Chemistry Solution Manual (7th Ed)

Resolução Atkins' Physical Chemistry Solution Manual (7th Ed)

(Parte 1 de 3)

Part 1: Equilibrium Part 1: Equilibrium

1 The properties of gases

Solutions to exercises Discussion questions

E1.1(b) The partial pressure of a gas in a mixture of gases is the pressure the gas would exert if it occupied alone the same container as the mixture at the same temperature. It is a limiting law because it holds exactly only under conditions where the gases have no effect upon each other. This can only be true in the limit of zero pressure where the molecules of the gas are very far apart. Hence, Dalton’s law holds exactly only for a mixture of perfect gases; for real gases, the law is only an approximation.

E1.2(b) The critical constants represent the state of a system at which the distinction between the liquid and vapour phases disappears. We usually describe this situation by saying that above the critical temperature the liquid phase cannot be produced by the application of pressure alone. The liquid and vapour phases can no longer coexist, though fluids in the so-called supercritical region have both liquid and vapour characteristics. (See Box 6.1 for a more thorough discussion of the supercritical state.)

E1.3(b) The van der Waals equation is a cubic equation in the volume, V . Any cubic equation has certain properties, one of which is that there are some values of the coefficients of the variable where the number of real roots passes from three to one. In fact, any equation of state of odd degree higher than 1 can in principle account for critical behavior because for equations of odd degree in V there are necessarily some values of temperature and pressure for which the number of real roots of V passes from n(odd) to 1. That is, the multiple values of V converge from n to1a s T → Tc. This mathematical result is consistent with passing from a two phase region (more than one volume for a given T and p) to a one phase region (only one V for a given T and p and this corresponds to the observed experimental result as the critical point is reached.

Numerical exercises E1.4(b) Boyle’s law applies.

pV = constant so pfVf = piVi

pf = piVi Vf

E1.5(b) (a) The perfect gas law is pV = nRT implying that the pressure would be p = nRT V

All quantities on the right are given to us except n, which can be computed from the given mass of Ar.

(b) The van der Waals equation is

Vm −b

E1.6(b) (a) Boyle’s law applies.

pV = constant so pfVf = piVi

and pi = pfVf Vi

(b) The original pressure in bar is


E1.7(b) Charles’s law applies.

V ∝ T so Vi

= Vf Tf

and Tf = VfTi Vi

E1.8(b) The relation between pressure and temperature at constant volume can be derived from the perfect gas law pV = nRT so p ∝ T and pi

Ti = pf Tf

The final pressure, then, ought to be

pf = piTf Ti

E1.9(b) According to the perfect gas law, one can compute the amount of gas from pressure, temperature, and volume. Once this is done, the mass of the gas can be computed from the amount and the molar mass using pV = nRT

E1.10(b) All gases are perfect in the limit of zero pressure. Therefore the extrapolated value of pVm/T will give the best value of R.


The molar mass is obtained from pV = nRT = m M RT which upon rearrangement gives M = mV RTp = ρ RT

The best value of M is obtained from an extrapolation of ρ/p versus p to p = 0; the intercept is M/RT.

Draw up the following table

From Fig. 1.1(a), ( pVm

From Fig. 1.1(b), ( ρ

Figure 1.1(a)

1.4278 1.4280 1.4282

1.4284 1.4286

Figure 1.1(b)

The value obtained for R deviates from the accepted value by 0.005 per cent. The error results from the fact that only three data points are available and that a linear extrapolation was employed. The molar mass, however, agrees exactly with the accepted value, probably because of compensating plotting errors.

E1.1(b) The mass density ρ is related to the molar volume Vm by

Vm = M ρ where M is the molar mass. Putting this relation into the perfect gas law yields pVm = RT so pMρ

Rearranging this result gives an expression for M; once we know the molar mass, we can divide by the molar mass of phosphorus atoms to determine the number of atoms per gas molecule

The number of atoms per molecule is

suggesting a formula of P4 E1.12(b) Use the perfect gas equation to compute the amount; then convert to mass.

pV = nRT so n = pV RT

We need the partial pressure of water, which is 53 per cent of the equilibrium vapour pressure at the given temperature and standard pressure.

E1.13(b) (a) The volume occupied by each gas is the same, since each completely fills the container. Thus solving for V from eqn 14 we have (assuming a perfect gas)

V = nJRT


(b) The total pressure is determined from the total amount of gas, n = nCH4 + nAr + nNe.

E1.14(b) This is similar to Exercise 1.14(a) with the exception that the density is first calculated.

RTp [Exercise 1.1(a)]

E1.15(b) This exercise is similar to Exercise 1.15(a) in that it uses the definition of absolute zero as that temperature at which the volume of a sample of gas would become zero if the substance remained a gas at low temperatures. The solution uses the experimental fact that the volume is a linear function of the Celsius temperature.

which is close to the accepted value of −273◦C.

E1.16(b) (a) p = nRT V

= 270atm (2 significant figures)

p = nRT

E1.17(b) The critical constants of a van der Waals gas are pc = a

E1.18(b) The compression factor is

Z = pVmRT = Vm


(b) The molar volume is


(b) The van der Waals equation is a cubic equation in Vm. The most direct way of obtaining the molar volume would be to solve the cubic analytically. However, this approach is cumbersome, so we proceed as in Example 1.6. The van der Waals equation is rearranged to the cubic form

Vm − abp

THE PROPERTIES OF GASES 9 The coefficients in the equation are evaluated as

Calculators and computer software for the solution of polynomials are readily available. In this case we find

(b) Using p = RT

V2m and substituting into the expression for Z above we get

Z = Vm

Vm −b − a VmRT

Comment. Both values of Z are very close to the perfect gas value of 1.0, indicating that water vapour is essentially perfect at 1.0 bar pressure.

E1.21(b) The molar volume is obtained by solving Z = pVmRT [1.20b], for Vm, which yields

(b) An approximate value of B can be obtained from eqn 1.2 by truncation of the series expansion after the second term, B/Vm, in the series. Then,


xN = nN ntotal

Similarly, xH = 0.37 (c) According to the perfect gas law ptotalV = ntotalRT so ptotal = ntotalRT V

(b) The partial pressures are

By interpreting b as the excluded volume of a mole of spherical molecules, we can obtain an estimate of molecular size. The centres of spherical particles are excluded from a sphere whose radius is the diameter of those spherical particles (i.e., twice their radius); that volume times the Avogadro constant is the molar excluded volume b b = NA

The critical pressure is


But this problem is overdetermined. We have another piece of information

According to the constants we have already determined, Tc should be

However, the reported Tc is 305.4K, suggesting our computed a/b is about 25 per cent lower than it should be.

E1.24(b) (a) The Boyle temperature is the temperature at which lim Vm→∞ d(1/Vm) vanishes. According to the van der Waals equation

RT = Vm

Vm −b − a VmRT so dZ

dVm dVm


In the limit of large molar volume, we have

lim Vm→∞

= 0s o aRT = b

(b) By interpreting b as the excluded volume of a mole of spherical molecules, we can obtain an estimate of molecular size. The centres of spherical particles are excluded from a sphere whose radius is the diameter of those spherical particles (i.e. twice their radius); the Avogadro constant times the volume is the molar excluded volume b

b = NA

E1.25(b) States that have the same reduced pressure, temperature, and volume are said to correspond. The reduced pressure and temperature for N2 at 1.0atm and 25◦C are

pr = p pc

The corresponding states are (a) For H2S

(Critical constants of H2S obtained from Handbook of Chemistry and Physics.) (b) For CO2

(c) For Ar

E1.26(b) The van der Waals equation is

Vm −b which can be solved for b

Solutions to problems Solutions to numerical problems

P1.2 Identifying pex in the equation p = pex + ρgh [1.4] as the pressure at the top of the straw and p as the atmospheric pressure on the liquid, the pressure difference is

P1.4 pV = nRT [1.12] implies that, with n constant, pfVf

Tf = piVi Ti

Solving for pf, the pressure at its maximum altitude, yields pf = Vi Vf

Ti × pi


Ti × pi

P1.6 The value of absolute zero can be expressed in terms of α by using the requirement that the volume of a perfect gas becomes zero at the absolute zero of temperature. Hence

Then θ(abs. zero) =− 1 α

All gases become perfect in the limit of zero pressure, so the best value of α and, hence, θ(abs. zero) is obtained by extrapolating α to zero pressure. This is done in Fig. 1.2. Using the extrapolated value, α = 3.6637 × 10−3◦C−1,o r

which is close to the accepted value of −273.15◦C.

P1.7 The mass of displaced gas is ρV , where V is the volume of the bulb and ρ is the density of the gas. The balance condition for the two gases is m(bulb) = ρV (bulb), m(bulb) = ρ′V(bulb)

the balance condition is pM = p′M′

This relation is valid in the limit of zero pressure (for a gas behaving perfectly).

In a proper series of experiments one should reduce the pressure (e.g. by adjusting the balanced weight). Experiment 2 is closer to zero pressure than experiment 1; it may be safe to conclude that

P1.9 We assume that no H2 remains after the reaction has gone to completion. The balanced equation is

N2 +3H2 → 2NH3 We can draw up the following table

Specifically 0.33mol 0 1.33mol 1.66mol Mole fraction 0.20 0 0.80 1.0

(b) From p = RT

Then, with a and b from Table 1.6

Substitution of 12.3L mol−1 into the denominator of the first expression again results in Vm = 12.3L mol−1, so the cycle of approximation may be terminated.


(b) Perfect Gas Equation: Vm(p,T ) = RT p

VirialEquation(eqn1.21tofirstorder): Vm(p,T ) = RT

The perfect gas law predicts a molar volume that is 9% too large at 298K and 4% too large at 373K. The negative value of the second virial coefficient at both temperatures indicates the dominance of very weak intermolecular attractive forces over repulsive forces.



)1/2 may be solved for from the expression for pc and yields vmol = b NA

[Table 1.6]


Vm −b e−a/RTVm = nRT

V −nb e−na/RTV

Solutions to theoretical problems

P1.18 This expansion has already been given in the solutions to Exercise 1.24(a) and Problem 1.17; the result is

p = RT Vm

Compare this expansion with p = RT Vm

Vm + C and hence find B = b − a RT and C = b2

P1.2 For a real gas we may use the virial expansion in terms of p [1.21]

which rearranges to pρ = RTM


Therefore, the limiting slope of a plot of p ρ against p is B′RT

M . From Fig. 1.2 in the Student’s

Solutions Manual, the limiting slope is

Restricting the variations of T and p to those which leave Vm constant, that is dVm = 0, we obtain

∂T ) Vm


∂T ) Vm

Vm( ∂p

∂Vm) T

From the equation of state( ∂p

∂Vm) T

( ∂p

∂T ) Vm


Vm + b

From the equation of state (a + bT )


R + b Vm

R + b Vm

P1.25 Z = Vm

Vom , where Vom = the molar volume of a perfect gas

From the given equation of state

Vm = b+ RTp = b +Vom then Z = b +Vo


Vo m

P1.27 The two masses represent the same volume of gas under identical conditions, and therefore, the same number of molecules (Avogadro’s principle) and moles, n. Thus, the masses can be expressed as xArMAr and xAr =


Comment. This value for the mole fraction of argon in air is close to the modern value.

P1.29 Z = pVmRT


But Vr = V Vc

= RTc pcVc



To derive the alternative form, solve eqn 1 for V ′ r , substitute the result into eqn 2, and simplify into polynomial form.

V′r = ZTr pr

Z = ZTr/pr ZTr



The real root is Z = 0.611 and this prediction is independent of the specific gas. Figure 1.27 indicates that the experimental result for the listed gases is closer to 0.5.

Solutions to applications P1.31 Refer to Fig. 1.3.


Air (environment)

The buoyant force on the cylinder is

Fbuoy = Fbottom − Ftop = A(pbottom − ptop) according to the barometric formula.

ptop = pbottome−Mgh/RT where M is the molar mass of the environment (air). Since h is small, the exponential can be expanded


The buoyant force becomes

= Ah

( pbottomM

g = nMg [ n = pbottomV n is the number of moles of the environment (air) displaced by the balloon, and nM = m, the mass of the displaced environment. Thus Fbuoy = mg. The net force is the difference between the buoyant force and the weight of the balloon. Thus

Fnet = mg − mballoong = (m − mballoon)g This is Archimedes’ principle.

2 The First Law: the concepts

Solutions to exercises Discussion questions

E2.1(b) Work is a transfer of energy that results in orderly motion of the atoms and molecules in a system; heat is a transfer of energy that results in disorderly motion. See Molecular Interpretation 2.1 for a more detailed discussion.

E2.2(b) Rewrite the two expressions as follows: (1) adiabatic p ∝ 1/V γ (2) isothermal p ∝ 1/V

The physical reason for the difference is that, in the isothermal expansion, energy flows into the system as heat and maintains the temperature despite the fact that energy is lost as work, whereas in the adiabatic case, where no heat flows into the system, the temperature must fall as the system does work. Therefore, the pressure must fall faster in the adiabatic process than in the isothermal case. Mathematically this corresponds to γ> 1.

E2.3(b) Standard reaction enthalpies can be calculated from a knowledge of the standard enthalpies of formation of all the substances (reactants and products) participating in the reaction. This is an exact method which involves no approximations. The only disadvantage is that standard enthalpies of formation are not known for all substances.

Approximate values can be obtained from mean bond enthalpies. See almost any general chemistry text, for example, Chemical Principles, by Atkins and Jones, Section 6.21, for an illustration of the method of calculation. This method is often quite inaccurate, though, because the average values of the bond enthalpies used may not be close to the actual values in the compounds of interest.

Another somewhat more reliable approximate method is based on thermochemical groups which mimic more closely the bonding situations in the compounds of interest. See Example 2.6 for an illustration of this kind of calculation. Though better, this method suffers from the same kind of defects as the average bond enthalpy approach, since the group values used are also averages.

Computer aided molecular modeling is now the method of choice for estimating standard reaction enthalpies, especially for large molecules with complex three-dimensional structures, but accurate numerical values are still difficult to obtain.

Numerical exercises E2.4(b) Work done against a uniform gravitational field is w = mgh

E2.6(b) Work done by a system expanding against a constant external pressure is

(Parte 1 de 3)