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1 The properties of gases
1A The perfect gas
Answers to discussion questions
1A.2 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. Dalton’s law 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.
Solutions to exercises
1A.1(b) The perfect gas law [1A.5] 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. n = 25 g
- (^95) g mol−^1
= 0. 626 mol
so p = (0.^62 6 mol)^ ×^ (8.^31 ×^10
− (^2) dm (^3) bar K− (^1) mol− (^1) ) × (30 + 273) K 1 .5 dm^3
= 10.5bar So no, the sample would not exert a pressure of 2.0 bar.
1A.2(b) Boyle’s law [1A.4a] applies. pV = constant so p f V f = p i V i Solve for the initial pressure: (i) p i = p f V f V i
= (1.97 bar)^ ×^ (2.14 dm
(2. 14 + 1 .80) dm^3
= 1.07 bar (ii) The original pressure in Torr is p i = (1.07 bar) × 1.013 bar1 atm
× 760 Torr1 atm
= 803 Torr
1A.3(b) The relation between pressure and temperature at constant volume can be derived from the perfect gas law, pV = nRT [1A.5] so p ∝ T and (^) Tp i i
= (^) Tp f f The final pressure, then, ought to be p f = p i T f T i
= (125 kPa)^ ×^ (11^ +^ 273)K (23 + 273)K
= 120 kPa
1A.4(b) According to the perfect gas law [1.8], one can compute the amount of gas from pressure, temperature, and volume. pV = nRT so n = (^) RTpV = (1.00 atm)^ ×^ (1.^013 ×^10
(^5) Pa atm− (^1) ) × (4. 00 × 103 m (^3) ) (8.3145 J K−^1 mol−^1 ) × (20 + 273)K
= 1. 66 × 105 mol Once this is done, the mass of the gas can be computed from the amount and the molar mass: m = (1. 66 × 105 mol) × (16.04 (^) g mol −^1 ) = 2. 67 × 106 g = 2.67 × 103 kg
1A.5(b) The total pressure is the external pressure plus the hydrostatic pressure [1A.1], making the total pressure
p = p ex + ρ gh. Let p ex be the pressure at the top of the straw and p the pressure on the surface of the liquid (atmospheric pressure). Thus the pressure difference is
p − p ex = ρ gh = (1. 0 g cm −^3 ) × 1 kg
103 g
× 1 cm 10 −^2 m
3 × (9.81 m s−^2 ) × (0.15m)
= 1.5 × 103 Pa = 1.5 × 10 −^2 atm
1A.6(b) The pressure in the apparatus is given by p = p ex + ρ gh [1A.1] where p ex = 760 Torr = 1 atm = 1.013× 105 Pa,
and ρ gh = 13.55 g cm−^3 × 1 kg
103 g
× 1 cm 10 −^2 m
3 × 0.100 m × 9.806 m s−^2 = 1.33 × 104 Pa
p = 1.013 × 105 Pa + 1.33 × 104 Pa = 1.146 × 105 Pa = 115 kPa
1A.7(b) Rearrange the perfect gas equation [1A.5] to give R = pV nT
= pV m T All gases are perfect in the limit of zero pressure. Therefore the value of pV m/ T extrapolated to zero pressure will give the best value of R. The molar mass can be introduced through pV = nRT = (^) Mm RT
which upon rearrangement gives M = (^) Vm^ RTp = ρ RTp The best value of M is obtained from an extrapolation of ρ/ p versus p to zero pressure; the intercept is M / RT. Draw up the following table:
From Figure 1A.1(a), (^) R = lim p → 0^ pV m T
= 0.082 062 dm^3 atm K−^1 mol−^1
Figure 1A.
(a)
p /atm ( pV m/ T )/(dm^3 atm K–1^ mol–1) ( ρ/ p )/(g dm–3^ atm–1) 0.750 000 0.082 0014 1.428 59 0.500 000 0.082 0227 1.428 22 0.250 000 0.082 0414 1.427 90
so n = (1.^49 ×^10
(^3) Pa) × (250 m (^3) ) (8.3145 J K−^1 mol−^1 ) × (23 + 273) K
= 151 mol
and m = (151 mol) × (18. (^0) g mol− 1 ) = 2. 72 × 103 g = 2.72 kg
1A.10(b) (i) The volume occupied by each gas is the same, since each completely fills the container. Thus solving for V we have (assuming a perfect gas, eqn. 1A.5) V = n J^ RT p J We have the pressure of neon, so we focus on it n Ne = 0.225 g 20.18 g mol−^1
= 1.115 × 10 −^2 mol Thus
V = 1.11^5 ×^10
− (^2) mol × 8.3145 Pa m (^3) K− (^1) mol− (^1) × 300 K 8.87 × 103 Pa
= 3.14 × 10 −^3 m^3 = 3.14 dm^3 (ii) The total pressure is determined from the total amount of gas, n = n CH 4 + n Ar + n Ne.
n CH 4 = 0.320 g 16.04 g mol−^1
= 1.99 5 × 10 −^2 mol n Ar = 0.175 g 39.95 g mol−^1
= 4.38 × 10 −^3 mol
n = (^) (1.99 5 + 0.438 + 1.11 (^5) )× 10 −^2 mol = 3.55 × 10 −^2 mol
and p = nRT V
= 3.55^ ×^10
− (^2) mol × 8.3145 Pa m (^3) K− (^1) mol− (^1) × 300 K 3.14 × 10 −^3 m^3 = 2.82 × 104 Pa = 28.2 kPa
1A.11(b) This exercise uses the formula, M =^ ρ RT
p
, which was developed and used in Exercise 1A.8(b). First the density must first be calculated.
ρ = 33.^5 ×^10
− (^3) g 250 cm^3
× 10
(^3) cm 3 dm^3
= 0 .134 g dm−^3
M = (0.134 g dm
− (^3) ) × (62. (^36) dm (^3) torr K −^1 mol− 1 ) × (298 K) 152 torr
= 16.4 g mol−^1
1A.12(b) This exercise is similar to Exercise 1.12(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: V = V 0 + αθ where V 0 = 20.00 dm^3 and α = 0.0741 dm^3 °C–^. At absolute zero, V = 0 = V 0 + αθ
so θ(abs.zero) = −
V 0
α = −^
20.00 dm^3 0.0741 dm^3 ¡C−^1
= Ğ 270 °C
which is close to the accepted value of –273C.
1A.13(b) (i) Mole fractions are
x N =
n N n total^ [1A.9]^ =^
2 .5 mol (2. 5 + 1 .5) mol =^0.^63 Similarly, x H = 0. 37
According to the perfect gas law p tot V = n tot RT
so p tot = n tot^ RT V
= (4.0 mol)^ ×^ (0.08206 dm
(^3) atm mol− (^1) K− (^1) ) × (273.15 K) 22 .4 dm^3
= 4 .0 atm
(ii) The partial pressures are p N = x N p tot = (0.63) × (4.0 atm) = 2 .5 atm and p H = (0.37) × (4.0 atm) = 1 .5 atm
(iii) p^ =^ p H +^ p N [1A.10]^ =^ (2.^5 +^1 .5) atm^ =^4 .0 atm
Solutions to problems
1A.2 Solving for n from the perfect gas equation [1A.5] yields n = pVRT. From the definition of
molar mass n = Mm , hence ρ = mV = MpRT
Rearrangement yields the desired relation, namely
p RT
= ρ M.
Therefore, for ideal gases p ρ = RTM and M = pRT / ρ. For real gases, find the zero-pressure
limit of p
by plotting it against p. Draw up the following table.
p /(kPa) 12.223 25.20 36.97 60.37 85.23 101. ρ/(kg m–3) 0.225^ 0.456^ 0.664^ 1.062^ 1.468^ 1. p / ρ 103 m^2 s−^2
Bear in mind that 1 kPa = 10^3 kg m–1^ s–2. p
ρ is plotted in Figure 1A.2. A straight line fits the data rather well. The extrapolation to^ p^ = 0
yields an intercept of 54.0× 103 m^2 s–2^. Then M = RT
- 40 × 104 m^2 s−^2
= (8.3145 J K
− (^1) mol− (^1) ) × (298.15 K)
- 40 × 104 m^2 s−^2 = 0 .0459 kg mol−^1 = 45.9 g mol−^1 Figure 1A.
n V
= 1.12^ ×^10
− (^3) mol (1.00 dm^2 ) × (40 × 103 m) × (10 dm m−^1 )
= 2.8 × 10 −^9 mol dm−^3
and n V
= 4.46^ ×^10
− (^4) mol (1.00 dm^2 ) × (40 × 103 m) × (10 dm m−^1 )
= 1.1 × 10 −^9 mol dm−^3 respectively.
1A.10 The perfect gas law [1A.5] can be rearranged to n = pVRT
The volume of the balloon is V = 43 π r^3 = 43 π × (3.0 m)^3 = 113 m^3
(a) n = (1.0 atm)^ ×^ (11^3 ×^10
(^3) dm (^3) ) (0.08206 dm^3 atm mol−^1 K−^1 ) × (298 K)
= 4.6 2 × 103 mol (b) The mass that the balloon can lift is the difference between the mass of displaced air and the mass of the balloon. We assume that the mass of the balloon is essentially that of the gas it encloses: m = m (H 2 ) = nM (H 2 ) = (4. 62 × 103 mol) × (2. (^02) g mol −^1 ) = 9. 33 × 103 g Mass of displaced air = (113 m^3 ) × (1. (^22) kg m −^3 ) = 1. 38 × 102 kg Therefore, the mass of the maximum payload is 13 8 kg − 9. 3 3 kg = 1.3 × 102 kg (c) For helium, m = nM (He) = (4. 62 × 103 mol) × (4.00 g mol−^1 ) = 18 kg The maximum payload is now 13 8 kg − 18 kg = 1.2 × 102 kg
1A.12 Avogadro’s principle states that equal volumes of gases contain equal amounts (moles) of the gases, so the volume mixing ratio is equal to the mole fraction. The definition of partial pressures is p J = x J p (^). The perfect gas law is pV = nRT so n J V
= p J RT
= x J^ p RT (a) n (CCl^3 F) V
= (261^ ×^10
− (^12) ) × (1.0 atm) (0.08206 dm^3 atm K−^1 mol−^1 ) × (10 + 273) K
= 1.1 × 10 −^11 mol dm-
and n (CCl^2 F^2 ) V
= (509^ ×^10
− (^12) ) × (1.0 atm) (0.08206 dm^3 atm K−^1 mol−^1 ) × (10 + 273) K
= 2.2 × 10 −^11 mol dm-
(b) n (CCl^3 F) V
= (261^ ×^10
− (^12) ) × (0.050 atm) (0.08206 dm^3 atm K−^1 mol−^1 ) × (200 K)
= 8.0 × 10 −^13 mol dm-
and n (CCl^2 F^2 ) V
= (509^ ×^10
− (^12) ) × (0.050 atm) (0.08206 dm^3 atm K−^1 mol−^1 ) × (200 K)
= 1.6 × 10 −^12 mol dm-
1B The kinetic model
Answers to discussion questions
1B.2 The formula for the mean free path [eqn 1B.13] is
λ = kT σ p In a container of constant volume, the mean free path is directly proportional to temperature and inversely proportional to pressure. The former dependence can be rationalized by noting that the faster the molecules travel, the farther on average they go between collisions. The latter also makes sense in that the lower the pressure, the less frequent are collisions,
and therefore the further the average distance between collisions. Perhaps more fundamental than either of these considerations are dependences on size. As pointed out in the text, the ratio T / p is directly proportional to volume for a perfect gas, so the average distance between collisions is directly proportional to the size of the container holding a set number of gas molecules. Finally, the mean free path is inversely proportional to the size of the molecules as given by the collision cross section (and therefore inversely proportional to the square of the molecules’ radius).
Solutions to exercises
1B.1(b) The mean speed is [1B.8]
v mean = 8 RT
π M
1/
The mean translational kinetic energy is E k = 12 mv^2 = 12 m v^2 = 12 mv rms^2 = m 2 ^3 RTM
^
[1B.3] = 3 kT 2 The ratios of species 1 to species 2 at the same temperature are v mean, v mean,
= M^2
M 1
1/ and
E k (^1) E k (^2)
(i)
v mean,H 2 v mean,Hg
1/ = 7.
(ii) The mean translation kinetic energy is independent of molecular mass and depends upon temperature alone! Consequently, because the mean translational kinetic energy for a gas is proportional to T , the ratio of mean translational kinetic energies for gases at the same temperature always equals 1.
1B.2(b) The root mean square speed [1B.3] is
v rms = ^3 MRT
1/
For CO 2 the molar mass is M = (12.011 + 2×15.9994)× 10 –3^ kg mol–1^ = 44.010× 10 –3^ kg mol–
so v rms = 3(8.3145 J K
− (^1) mol− (^1) )(20 + 273) K 44.01 × 10 −^3 kg mol−^1
1/ = 408 m s−^1
For He
v rms = 3(8.3145 J K
− (^1) mol− (^1) )(20 + 273) K 4.003 × 10 −^3 kg mol−^1
1/ = 1.35 × 103 m s−^1 = 1.35 km s−^1
1B.3(b) The Maxwell-Boltzmann distribution of speeds [1B.4] is
f ( v ) = 4 π 2 π M RT
3/ v^2 e−^ Mv^2 /2^ RT and the fraction of molecules that have a speed between v and v +d v is f ( v )d v. The fraction of molecules to have a speed in the range between v 1 and v 2 is, therefore, (^) v f ( v ) d v 1
v 2 ∫. If the range is relatively small, however, such that f ( v ) is nearly constant over that range, the integral may be approximated by f ( v )∆ v , where f ( v ) is evaluated anywhere within the range and ∆ v = v 2 – v 1. Thus, we have, with M = 44.010× 10 –3^ kg mol–1^ [Exericse 1B.2(b)],
2 1
3 1 3/ 2 1 2 1 1 ( )d ( ) 4 44.010^10 kg mol (402.5 m s ) 2 (8.3145 J K mol )(400 K)
v
v f v^ v^ f v^ v^ π^ π
− − − − −
≈ ∆ = ^ ×
∫
p = kT
= (1.^381 ×^10
− 23 J K− 1 )(293 K)
- 36 × (10−^9 m)^2 (10 × 0.34 × 10 −^9 m)
= 3.3 × 106 J m−^3 = 3.3 MPa Comment. This pressure works out to 33 bar (about 33 atm), conditions under which the assumption of perfect gas behavior and kinetic model applicability at least begins to come into question.
1B.7(b) The mean free path [1B.13] is
λ = kT σ p
= (1.381^ ×^10
− 23 J K− 1 )(217 K)
- 43 × (10−^9 m)^2 (12.1 × 103 Pa atm−^1 )
= 5.8 × 10 −^7 m
Solutions to problems
1B.2 The number of molecules that escape in unit time is the number per unit time that would have collided with a wall section of area A equal to the area of the small hole. This quantity is readily expressed in terms of Z W, the collision flux (collisions per unit time with a unit area), given in eqn 19A.6. That is, d N d t
= − Z W A = −^ Ap
(2 π mkT )1/
where p is the (constant) vapour pressure of the solid. The change in the number of molecules inside the cell in an interval ∆ t is therefore ∆ N = − Z W A t ∆ , and so the mass loss is
∆ w = m ∆ N = − Ap m
2 π kT
1/ ∆ t = − Ap M
2 π RT
1/ ∆ t
Therefore, the vapour pressure of the substance in the cell is w 2 RT^ 1/ 2 p A t M
= ^ −∆^ ×^ π
For the vapour pressure of germanium
p = 43 ×^10
− (^9) kg
π (0.50 × 10 −^3 m)(7200 s)
× 2 π^ (8.3145 J K
− (^1) mol− (^1) )(1273 K) 72.64 × 10 −^3 kg mol−^1
1/
= 7.3 × 10 −^3 Pa = 7.3 mPa
1B.4 We proceed as in Justification 1B.2 except that, instead of taking a product of three one- dimensional distributions in order to get the three-dimensional distribution, we make a product of two one-dimensional distributions.
f ( vx , vy )d vx d vy = f ( vx^2 ) f ( vy^2 )d vx dvy = 2 π m kT e−^ mv
(^2) / 2 kT d vx d vy
where v^2 = vx^2 + vy^2. The probability f ( v )d v that the molecules have a two-dimensional speed, v , in the range v to v + d v is the sum of the probabilities that it is in any of the area elements d vx d vy in the circular shell of radius v. The sum of the area elements is the area of the circular shell of radius v and thickness d v which is π( ν+d ν)^2 – πν^2 = 2 πνd ν. Therefore,
f ( v ) = kTm
^
v e−^ mv^2 /^2 kT^ = RTM
^
v e−^ Mv^2 /^2 RT^^ MR = mk ^
The mean speed is determined as v mean = 0 vf ( v ) d v
∞ ∫ =^
m kT
^
0 v^2 e−^ mv^2 /^2 kT^ d v
∞ ∫ Using integral G.3 from the Resource Section yields
v mean = kTm
×^ π
1/ 4
× ^2 kTm
3/ =^ ^ π 2 kTm
1/ =^ ^ π 2 RT M
1/
1B.6 The distribution [1B.4] is
f ( v ) = 4 π
3 / 2 M 2 π RT
^
v^2 e−^ Mv^2 /^2 RT^. The proportion of molecules with speeds less than v rms is
P = ∫ 0^ v rms f ( v ) d v = 4 π
M^3 /^2
2 π RT
^
∫ 0^ v rms v^2 e−^ Mv^2 /^2 RT^ d v
Defining a ≡ R / 2 RT ,
P = 4 π
a^3 /^2
^
∫ 0^ v rms v^2 e−^ av^2 d v = −^4 π
a^3 /^2
^
d d a
∫ 0^ v rmse−^ av^2 d v
Defining χ^2 ≡ av^2. Then, d v = a −1/ 2d χand
P = − 4 π
a^3 /^2
^
d d a
a^1 /^2
e−^ χ
2
0 d^ χ
v rms a 1/ 2
a^3 /^2
^
13 /^2
() a e
− χ^2 d χ
0
v rms a 1/ 2
11 /^2
() a
d d a e
− χ^2 d χ
0
v rms a 1/ 2
Then we use the error function [Integral G.6]:
0 e−^ χ^2 d^ χ
v rms a 1/ 2
∫ =^ π
( 1/2^ / 2)erf ( v
rms a
d d a e
− χ^2 d χ
0
v rms a 1/ 2
∫ =^
d v rms a^1 /^2 d a
× (e−^ av rms
2 ) = 12^ c a^1 /^2
^ e
− av rms^2
where we have used (^) dd z 0 f ( y ) d y
z
∫ =^ f^ ( z )
Substituting and cancelling we obtain P = erf ( v rms a^1 /^2 ) − ( 2 v rms a^1 /^2 / π^1 /^2 )e−^ av rms
2
Now v rms = ^3 RTM
1/ so v rms a^1 /^2 = ^3 MRT
1/ × 2 MRT
1/21/ = ^32
1/
and P = erf (^32)
^ −
6 1 /^2
^ e
Therefore, (a) 1 – P = 39% have a speed greater than the root mean square speed. (b) P = 61% of the molecules have a speed less than the root mean square speed. (c) For the proportions in terms of the mean speed v mean, replace v rms by
v mean = ( 8 kT / π m )
1/
1/ v rms so v mean a 1/2^ = 2/ π1/^.
Then P = erf ( v mean a^1 /^2 ) − ( 2 v mean a^1 /^2 / π^1 /^2 )× (e−^ av^2 mean^ )
= erf 2 / ( π^1 /^2 )− (4 / π)e−^4 /^ π^ = 0. 889 − 0. 356 = 0.
That is, 53% of the molecules have a speed less than the mean, and 47% have a speed greater than the mean.
1B.8 The average is obtained by substituting the distribution (eqn 1B.4) into eqn 1B.7:
vn^ = 0 vn^ f ( v ) d v
∞
∫ =^4 π^
M
2 π RT
3/ 0^ vn +^2 e−^ Mv^2 /^2 RT^ d v
∞
For even values of n , use Integral G.8:
vn^ = 4 π 2 π M RT
3/2 ( n + 1)!!
n + 4 2
2 RT
M
n + 2 2
^2 π RT
M
1/ = ( n + 1)!! RTM
n 2
where ( n +1)!! = 1 × 3 × 5 ... × ( n +1)
Thus vn^
1/ n = ( n + 1)!! RTM
1/ even n
1C Real gases
Answers to discussion questions
1C.2 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 supercritical fluids have both liquid and vapour characteristics.
1C.4 The van der Waals equation is a cubic equation in the volume, V. Every cubic equation has some values of the coefficients for which the number of real roots passes from three to one. In fact, any equation of state of odd degree n > 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 to 1. That is, the multiple values of V converge from n to 1 as the temperature approaches the critical temperature. 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.
Solutions to exercises
1C.1(b) The van der Waals equation [1C.5a] is
p = (^) VnRT − nb − an
2 V^2 From Table 1C.3 for H 2 S, a = 4.484 dm^6 atm mol–1^ and b = 0.0434 dm^3 mol–1. (i) p = (1.0 mol)^ ×^ (0.08206 dm
(^3) atm mol− (^1) K− (^1) ) × (273.15 K) 22 .414 dm^3 − (1.0 mol) × (4. 34 × 10 −^2 dm^3 mol−^1 ) − (4.484 dm
(^6) atm mol− (^2) ) × (1.0 mol) 2 (22.414 dm^3 )^2
= 0.99 atm
(ii) p = (1.0 mol)^ ×^ (0.08206 dm
(^3) atm mol− (^1) K− (^1) ) × (500 K) 0 .150 dm^3 − (1.0 mol) × (4. 34 × 10 −^2 dm^3 mol−^1 )
− (4.484 dm
(^3) atm mol− (^1) ) × (1.0 mol) 2 (0.150 dm^3 )^2
= 190 atm (2 sig. figures)
1C.2(b) The conversions needed are as follows: 1 atm = 1.013× 105 Pa, 1 Pa = 1 kg m–1^ s–2, 1 dm^6 = (10–1^ m)^6 = 10–6^ m^6 , 1 dm^3 = 10–3^ m^3. Therefore, a = 1.32 atm dm^6 mol−^2 × 1.013^ ×^10
(^5) kg m− (^1) s− 2 1 atm
× 10
− (^6) m 6 dm^6 = 1.34 × 10 −^1 kg m^5 s−^2 mol−^2
and b = 0.0426 dm^3 mol−^1 × 10
− (^3) m 3 dm^3
= 4.26 × 10 −^5 m^3 mol−^1
1C.3(b) The compression factor Z is [1C.1]
Z =
V m V m^ °^
pV m RT (i) Because (^) V m = V m^ ο^ + 0. 12 V m^ ο^ = (1.12) V m^ ο^ , we have (^) Z = 1. 12 (ii) The molar volume is
V m = (1.12) V m^ ο^ = (1.12) × RTp
= (1.12) × (0.08206 dm
(^3) atm mol− (^1) K− (^1) ) × (350 K) 12 atm
= 2.7 dm^3 mol−^1
Since V m > V mo^ repulsive forces dominate.
1C.4(b) (i) According to the perfect gas law
V mo^ = RTp = (8.3145 J K
− (^1) mol− (^1) ) × (298.15 K) (200 bar) × (10^5 Pa bar −^1 )
× 1 dm 10 −^1 m
3 = 0.124 dm^3 mol−^1
(ii) The van der Waals equation [1C.5b] is a cubic equation in V m. Cubic equations can be solved analytically. However, this approach is cumbersome, so we proceed as in Example 1C.1. The van der Waals equation is rearranged to the cubic form V m^3 − b + RT p
^
V m^2 + a p
^
V m − ab p
or x^3 − b + RT p
^
x^2 + a p
^
x − ab p
= 0 with x = V m/(dm^3 mol–1). It will be convenient to have the pressure in atm: 200 bar × (^) 1.013 bar1 atm = 197.4 atm The coefficients in the equation are b + RTp = (3.183 × 10 −^2 dm^3 mol−^1 ) + (0.08206 dm
(^3) atm mol− (^1) K− (^1) ) × (298.15 K) 197.4 atm = (3.183 × 10 −^2 + 0.1239) dm^3 mol−^1 = 0.1558 dm^3 mol−^1 a p
= 1.360 dm
(^6) atm mol− 2 197.4 atm
= 6.89 × 10 −^3 dm^6 mol−^2
ab p
= (1.360 dm
(^6) atm mol− (^2) ) × (3.183 × 10 − (^2) dm (^3) mol− (^1) ) 197.4 atm
= 2.19 3 × 10 −^4 dm^9 mol−^3
Thus, the equation to be solved is x^3 − 0.155 8 x^2 + (6.89 × 10 −^3 ) x − (2.19 3 × 10 −^4 ) = 0. Calculators and computer software for the solution of polynomials are readily available. In this case we find x = 0.112 and V m = 0.112 dm^3 mol–^. The perfect-gas value is about 15 percent greater than the van der Waals result.
1C.5(b) The molar volume is obtained by solving Z = pV m RT
[1C.2], for V m , which yields
V m = ZRT p
= (0.86)^ ×^ (0.08206 dm
(^3) atm mol− (^1) K− (^1) ) × (300 K) 20 atm
= 1. 0 6 dm^3 mol−^1
(i) Then, V = nV m = (8. 2 × 10 −^3 mol) × (1. 0 6 dm^3 mol−^1 ) = 8. 7 × 10 −^3 dm^3 = 8 .7 cm^3 (ii) An approximate value of B can be obtained from eqn 1C.3b by truncation of the series expansion after the second term, B / V m, in the series. Then, B = V m
pV m RT −^1
= V m × ( Z − 1)
= (1. 0 6 dm^3 mol−^1 ) × (0. 86 − 1) = −0.15 dm^3 mol−^1
1C.6(b) Equations 1C.6are solved for b and a , respectively, and yield b = V c/3 and a = 27 b^2 p c = 3 V c^2 p c. Substituting the critical constants b = 148 cm
(^3) mol− 1 3 =^ 49.3 cm
(^3) mol− (^1) = 0.0493 dm (^3) mol− 1
At the Boyle temperature B = 0 = b − (^) RTa B
so T B = (^) bRa =
27 T c 8 (i) From Table 1C.3, a = 4.484 dm^6 atm mol–2^ and b = 0.0434 dm^3 mol–1. Therefore, T B = (4.484 dm
(^6) atm mol− (^2) ) (0.08206 L atm mol−^1 K−^1 ) × (0.0434 dm^3 mol−^1 )
= 1259 K
(ii) As in Exercise 1C.6(b),
b = N A^4 π^ (2 r )
3 3
^
so r = (^12)
1 / 3 3 b
4 π N A
r = (^12)
1 / 3 3(0.0434 dm^3 mol−^1 ) 4 π(6.022 × 1023 mol−^1 )
^
= 1. 29 × 10 −^9 dm = 1. 29 × 10 −^10 m = 0 .129 nm
1C.8(b) States that have the same reduced pressure, temperature, and volume [1C.8] are said to correspond. The reduced pressure and temperature for N 2 at 1.0 atm and 25°C are [Table 1C.2] p r = p p c
= 1 .0 atm 33 .54 atm
= 0. 030 and T r = T T c
= (25^ +^ 273) K
126 .3K
The corresponding states are (i) For H 2 S (critical constants obtained from NIST Chemistry WebBook ) T = 2.36(373.3 K) = 881 K p = 0.030(89.7 atm) = 2.67 atm (ii) For CO 2 T = 2.36(304.2 K) = 718 K p = 0.030(72.9 atm) = 2.2 atm (iii) For Ar T = 2.36(150.7 K) = 356 K p = 0.030(48.0 atm) = 1.4 atm
1C.9(b) The van der Waals equation [1C.5b] is
p = (^) VRT m −^ b^
− a V m^2 which can be solved for b b = V m − RT p + a V m^2
= 4. 00 × 10 −^4 m^3 mol−^1 − (8.3145 J K
− (^1) mol− (^1) ) × (288 K)
- 0 × 106 Pa + 0 .76 m
(^6) Pa mol− 2 (4. 00 × 10 −^4 m^3 mol−^1 )^2
= 1.3 × 10 −^4 m^3 mol−^1 The compression factor is Z =
pV m RT [1C.2]^ =^
(4. 0 × 106 Pa) × (4. 00 × 10 −^4 m^3 mol−^1 ) (8.3145 J K−^1 mol−^1 ) × (288 K)
Solutions to problems
1C.2 From the definition of Z [1C.1] and the virial equation [1C.3b], Z may be expressed in virial form as
Z = 1 + B V^1
m
+ C
2 1 V m
^
+ L
Since V m = RT p
(by assumption of approximate perfect gas behavior), 1 V m
= p RT
; hence upon
substitution, and dropping terms beyond the second power of 1 V m
Z = 1 + B p RT
2
= 1 + (− 21. 7 × 10 −^3 dm^3 mol−^1 ) × 100 atm (0.08206 dm^3 atm mol−^1 K−^1 ) × (273K)
+(1. 200 × 10 −^3 dm^6 mol−^2 ) ×
2 100 atm (0.08206 dm^3 atm mol−^1 K−^1 ) × (273K)
V m = (0.927) RT p
= (0.927) (0.08206 dm
(^3) atm mol− (^1) K− (^1) )(273 K) 100 atm
= 0.208 dm^3 Question. What is the value of Z obtained from the next approximation using the value of V m just calculated? Which value of Z is likely to be more accurate?
1C.4 Since B ′( T B) = 0 at the Boyle temperature [Topic 1.3b]: B ′ ( T B ) = a + b e−^ c / T B
2 = 0
Solving for T B: ( )
1/ 2^ 1/ 2 (^22) B (^1) 1
(1131K ) 5.0 10 K
ln (^) ln ( 0 1993bar^ ) (0 2002 bar )
T c a b
− −
^
− ^ −
= = = ×
−^ ^ − −.
^.
^
1C.6 From Table 1C.4 T c = 2 3
× 2 a 3 bR
1/ , p c = 1 12
× 2 aR 3 b^3
1/
( )
a bR
/ may be solved for from the expression for p c and yields 12 bp c R
^
Thus T c = 2 3
× 12 p c^ b R
× p c V c R
= ^83
× (40 atm)^ ×^ (160^ ×^10
− (^3) dm (^3) mol− (^1) ) 0.08206 dm^3 atm mol−^1 K−^1
= 21 0K
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 = N A^4 π(2 r )
3 3
^
so r = 1 2
1 / 3 3 b 4 π N A
^
[Exercise 1C.6(b)] = 1 2
V^1 /^3
c 4 π N A
r = 1 2
1 / 3 160 cm^3 mol−^1 4 π(6.022 × 1023 mol−^1 )
^
= 1. 38 × 10 −^8 cm = 0.138 nm
1C.8 Substitute the van der Waals equation [1C.5b] into the definition of the compression factor [1C.2]
summarized in Table 1C.2 where it is seen that the Dieterici equation prediction is often better.
1C.12^ pV m^^1 B p C p^2 RT
= + ′ + ′ + [1C.3a]
m (^1) m (^) m 2 pV (^) B C RT^ =^ +^ V +^ V +^ [1C.3b] Thus (^22) m (^) m
B p C p B^ C ′ + ′ + = (^) V + (^) V + Multiply through by V m, replace pV m by RT {1+( B / V m) + ...}, and equate coefficients of powers of 1/ V m: 2 2 m m
B RT BB RT^ C R T^ B C
V V
′ + ′^ + ′ + = + +
Hence, B ′ RT = B , implying that B ′= B RT
Also BB ′ RT + C ′ R^2 T^2 = C = B^2 + C ′ R^2 T^2 , implying that C ′ = C^ −^ B
2 R^2 T^2
1C.14 Write V m = f ( T , p ); then d V m = p
∂ V m ∂ T
^
d T + T
∂ V m ∂ p
^
d p
Restricting the variations of T and p to those which leave V m constant, that is d V m = 0, we obtain
p
∂ V m ∂ T
^
T
∂ V m ∂ p
^
×
V m
∂ p ∂ T
^
T
− 1 ∂ p ∂ V m
^
×
V m
∂ p ∂ T
From the equation of state
T
∂ p ∂ V m
^
= − RT
V m^2
− 2( a^ +^ bT^ ) V m^3
V m RT + 2( a + bT ) V m^3
and V m
∂ p ∂ T
^ =^
R
V m^ +^
b V m^2
RV m + b V m^2 Substituting
P
∂ V m ∂ T
^
= V m
3 V m RT + 2( a + bT )
^
RV m + b V m^2
RV m
(^2) + bV m V m RT + 2( a + bT ) From the equation of state, a + bT = pV m^2 – RTV m
Then P
∂ V m ∂ T
^
RV m^2 + bV m V m RT + 2 pV m^2 − 2 RTV m
RV m + b m 2 pV m − RT
1C.16 (^) Z = V m V mo^
[1C.1], where V m° = the molar volume of a perfect gas
From the given equation of state V m = b + RT p
= b + V mo For V m =10 b , we have 10 b = b + V m°, so V m° = 9 b.
Then Z = (^109) bb = 109 = 1.
1C.18 The virial equation is
m (^2) m m
pV RT 1 B^ C V V
= ^ + + +
[1C.3b]
or m 2 m m
pV (^) 1 B C RT V V
(a) If we assume that the series may be truncated after the B term, then a plot of pV m RT
vs 1 V m will have B as its slope and 1 as its y -intercept. Transforming the data gives p /MPa V m/(dm^3 mol–1) (1/ V m)/(mol dm–3) pV m/ RT 0.4000 6.2208 0.1608 0. 0.5000 4.9736 0.2011 0. 0.6000 4.1423 0.2414 0. 0.8000 3.1031 0.3223 0. 1.000 2.4795 0.4033 0. 1.500 1.6483 0.6067 0. 2.000 1.2328 0.8112 0. 2.500 0.98357 1.017 0. 3.000 0.81746 1.223 0. 4.000 0.60998 1.639 0.
Figure 1C.1(a)
The data are plotted in Figure 1C.1(a). The data fit a straight line reasonably well, and the y - intercept is very close to 1. The regression yields B = –1.324× 10 –2^ dm^3 mol–1.
(b) A quadratic function fits the data somewhat better (Figure 1C.1(b)) with a slightly better correlation coefficient and a y -intercept closer to 1. This fit implies that truncation of the virial series after the term with C is more accurate than after just the B term. The regression then yields
