Atkins' Physical Chemistry, 12th Edition,
Peter Atkins, Chapter 1-19
SOLUTION MANUAL
, ❚❛❜❧❡ ♦❢ ❝♦♥t❡♥ts
Pr❡❢❛❝❡ ✈✐✐
✶ ❚❤❡ ♣r♦♣❡rt✐❡s ♦❢ ❣❛s❡s ✶
✶❆ ❚❤❡ ♣❡r❢❡❝t ❣❛s ✶
✶❇ ❚❤❡ ❦✐♥❡t✐❝ ♠♦❞❡❧ ✶✵
✶❈ ❘❡❛❧ ❣❛s❡s ✶✾
✷ ■♥t❡r♥❛❧ ❡♥❡r❣② ✸✺
✷❆ ■♥t❡r♥❛❧ ❡♥❡r❣② ✸✺
✷❇ ❊♥t❤❛❧♣② ✹✵
✷❈ ❚❤❡r♠♦❝❤❡♠✐str② ✹✸
✷❉ ❙t❛t❡ ❢✉♥❝t✐♦♥s ❛♥❞ ❡①❛❝t ❞✐❢❢❡r❡♥t✐❛❧s ✹✾
✷❊ ❆❞✐❛❜❛t✐❝ ❝❤❛♥❣❡s ✺✺
✸ ❚❤❡ s❡❝♦♥❞ ❛♥❞ t❤✐r❞ ❧❛✇s ✻✸
✸❆ ❊♥tr♦♣② ✻✸
✸❇ ❊♥tr♦♣② ❝❤❛♥❣❡s ❛❝❝♦♠♣❛♥②✐♥❣ s♣❡❝✐❢✐❝ ♣r♦❝❡ss❡s ✻✽
✸❈ ❚❤❡ ♠❡❛s✉r❡♠❡♥t ♦❢ ❡♥tr♦♣② ✽✵
✸❉ ❈♦♥❝❡♥tr❛t✐♥❣♦♥t❤❡s②st❡♠ ✽✼
✸❊ ❈♦♠❜✐♥✐♥❣ t❤❡ ❋✐rst ❛♥❞ ❙❡❝♦♥❞ ▲❛✇s ✾✷
✹ P❤②s✐❝❛❧ tr❛♥s❢♦r♠❛t✐♦♥s ♦❢ ♣✉r❡ s✉❜st❛♥❝❡s ✾✾
✹❆ P❤❛s❡❞✐❛❣r❛♠s ♦❢ ♣✉r❡s✉❜st❛♥❝❡s ✾✾
✹❇ ❚❤❡r♠♦❞②♥❛♠✐❝❛s♣❡❝ts ♦❢ ♣❤❛s❡tr❛♥s✐t✐♦♥s ✶✵✶
✺ ❙✐♠♣❧❡♠✐①t✉r❡s ✶✶✺
✺❆ ❚❤❡ t❤❡r♠♦❞②♥❛♠✐❝ ❞❡s❝r✐♣t✐♦♥ ♦❢ ♠✐①t✉r❡s ✶✶✺
✺❇ ❚❤❡♣r♦♣❡rt✐❡s♦❢s♦❧✉t✐♦♥s ✶✷✻
✺❈ P❤❛s❡❞✐❛❣r❛♠s♦❢❜✐♥❛r②s②st❡♠s✿❧✐q✉✐❞s ✶✹✶
✺❉ P❤❛s❡ ❞✐❛❣r❛♠s ♦❢ ❜✐♥❛r② s②st❡♠s✿s♦❧✐❞s ✶✹✽
✺❊ P❤❛s❡❞✐❛❣r❛♠s ♦❢ t❡r♥❛r②s②st❡♠s ✶✺✹
✺❋ ❆❝t✐✈✐t✐❡s ✶✺✽
✻ ❈❤❡♠✐❝❛❧ ❡q✉✐❧✐❜r✐✉♠ ✶✼✶
✻❆ ❚❤❡ ❡q✉✐❧✐❜r✐✉♠ ❝♦♥st❛♥t ✶✼✶
,✐✈ ❚❆❇▲❊ ❖❋ ❈❖◆❚❊◆❚❙
✻❇ ❚❤❡ r❡s♣♦♥s❡ ♦❢ ❡q✉✐❧✐❜r✐❛ t♦ t❤❡ ❝♦♥❞✐t✐♦♥s ✶✼✾
✻❈ ❊❧❡❝tr♦❝❤❡♠✐❝❛❧ ❝❡❧❧s ✶✾✵
✻❉ ❊❧❡❝tr♦❞❡ ♣♦t❡♥t✐❛❧s ✶✾✼
✼ ◗✉❛♥t✉♠ t❤❡♦r② ✷✶✺
✼❆ ❚❤❡ ♦r✐❣✐♥s ♦❢ q✉❛♥t✉♠ ♠❡❝❤❛♥✐❝s ✷✶✺
✼❇ ❲❛✈❡❢✉♥❝t✐♦♥s ✷✷✸
✼❈ ❖♣❡r❛t♦rs ❛♥❞ ♦❜s❡r✈❛❜❧❡s ✷✷✼
✼❉ ❚r❛♥s❧❛t✐♦♥❛❧ ♠♦t✐♦♥ ✷✸✻
✼❊ ❱✐❜r❛t✐♦♥❛❧ ♠♦t✐♦♥ ✷✹✽
✼❋ ❘♦t❛t✐♦♥❛❧ ♠♦t✐♦♥ ✷✺✼
✽ ❆t♦♠✐❝ str✉❝t✉r❡ ❛♥❞ s♣❡❝tr❛ ✷✼✶
✽❆ ❍②❞r♦❣❡♥✐❝ ❆t♦♠s ✷✼✶
✽❇ ▼❛♥②✲❡❧❡❝tr♦♥ ❛t♦♠s ✷✼✼
✽❈ ❆t♦♠✐❝ s♣❡❝tr❛ ✷✼✾
✾ ▼♦❧❡❝✉❧❛r ❙tr✉❝t✉r❡ ✷✽✼
✾❆ ❱❛❧❡♥❝❡✲❜♦♥❞ t❤❡♦r② ✷✽✼
✾❇ ▼♦❧❡❝✉❧❛r ♦r❜✐t❛❧ t❤❡♦r②✿ t❤❡ ❤②❞r♦❣❡♥ ♠♦❧❡❝✉❧❡✲✐♦♥ ✷✾✷
✾❈ ▼♦❧❡❝✉❧❛r ♦r❜✐t❛❧ t❤❡♦r②✿ ❤♦♠♦♥✉❝❧❡❛r ❞✐❛t♦♠✐❝ ♠♦❧❡❝✉❧❡s ✷✾✽
✾❉ ▼♦❧❡❝✉❧❛r ♦r❜✐t❛❧ t❤❡♦r②✿ ❤❡t❡r♦♥✉❝❧❡❛r ❞✐❛t♦♠✐❝ ♠♦❧❡❝✉❧❡s ✸✵✷
✾❊ ▼♦❧❡❝✉❧❛r ♦r❜✐t❛❧ t❤❡♦r②✿ ♣♦❧②❛t♦♠✐❝ ♠♦❧❡❝✉❧❡s ✸✵✼
✶✵ ▼♦❧❡❝✉❧❛r s②♠♠❡tr② ✸✷✶
✶✵❆ ❙❤❛♣❡ ❛♥❞ s②♠♠❡tr② ✸✷✶
✶✵❇ ●r♦✉♣ t❤❡♦r② ✸✷✾
✶✵❈ ❆♣♣❧✐❝❛t✐♦♥s ♦❢ s②♠♠❡tr② ✸✸✼
✶✶ ▼♦❧❡❝✉❧❛r❙♣❡❝tr♦s❝♦♣② ✸✹✾
✶✶❆ ●❡♥❡r❛❧ ❢❡❛t✉r❡s ♦❢ ♠♦❧❡❝✉❧❛r s♣❡❝tr♦s❝♦♣② ✸✹✾
✶✶❇ ❘♦t❛t✐♦♥❛❧ s♣❡❝tr♦s❝♦♣② ✸✺✼
✶✶❈ ❱✐❜r❛t✐♦♥❛❧ s♣❡❝tr♦s❝♦♣② ♦❢ ❞✐❛t♦♠✐❝ ♠♦❧❡❝✉❧❡s ✸✼✶
✶✶❉ ❱✐❜r❛t✐♦♥❛❧ s♣❡❝tr♦s❝♦♣② ♦❢ ♣♦❧②❛t♦♠✐❝ ♠♦❧❡❝✉❧❡s ✸✽✹
✶✶❊ ❙②♠♠❡tr② ❛♥❛❧②s✐s ♦❢ ✈✐❜r❛t✐♦♥❛❧ s♣❡❝tr♦s❝♦♣② ✸✽✺
✶✶❋ ❊❧❡❝tr♦♥✐❝ s♣❡❝tr❛ ✸✽✽
✶✶● ❉❡❝❛② ♦❢ ❡①❝✐t❡❞ st❛t❡s ✸✾✽
, ❚❆❇▲❊ ❖❋ ❈❖◆❚❊◆❚❙ ✈
✶✷ ▼❛❣♥❡t✐❝ r❡s♦♥❛♥❝❡ ✹✵✾
✶✷❆ ●❡♥❡r❛❧ ♣r✐♥❝✐♣❧❡s ✹✵✾
✶✷❇ ❋❡❛t✉r❡s ♦❢ ◆▼❘ s♣❡❝tr❛ ✹✶✷
✶✷❈ P✉❧s❡t❡❝❤♥✐q✉❡s ✐♥ ◆▼❘ ✹✷✶
✶✷❉ ❊❧❡❝tr♦♥ ♣❛r❛♠❛❣♥❡t✐❝ r❡s♦♥❛♥❝❡ ✹✷✾
✶✸ ❙t❛t✐st✐❝❛❧ t❤❡r♠♦❞②♥❛♠✐❝s ✹✸✺
✶✸❆ ❚❤❡ ❇♦❧t③♠❛♥♥ ❞✐str✐❜✉t✐♦♥ ✹✸✺
✶✸❇ P❛rt✐t✐♦♥ ❢✉♥❝t✐♦♥s ✹✸✾
✶✸❈ ▼♦❧❡❝✉❧❛r❡♥❡r❣✐❡s ✹✹✾
✶✸❉ ❚❤❡ ❝❛♥♦♥✐❝❛❧ ❡♥s❡♠❜❧❡ ✹✺✻
✶✸❊ ❚❤❡ ✐♥t❡r♥❛❧ ❡♥❡r❣② ❛♥❞ ❡♥tr♦♣② ✹✺✼
✶✸❋ ❉❡r✐✈❡❞ ❢✉♥❝t✐♦♥s ✹✼✸
✶✹ ▼♦❧❡❝✉❧❛r■♥t❡r❛❝t✐♦♥s ✹✽✺
✶✹❆ ❊❧❡❝tr✐❝ ♣r♦♣❡rt✐❡s ♦❢ ♠♦❧❡❝✉❧❡s ✹✽✺
✶✹❇ ■♥t❡r❛❝t✐♦♥s❜❡t✇❡❡♥♠♦❧❡❝✉❧❡s ✹✾✻
✶✹❈ ▲✐q✉✐❞s ✺✵✹
✶✹❉ ▼❛❝r♦♠♦❧❡❝✉❧❡s ✺✵✻
✶✹❊ ❙❡❧❢✲❛ss❡♠❜❧② ✺✶✼
✶✺ ❙♦❧✐❞s ✺✷✸
✶✺❆ ❈r②st❛❧ str✉❝t✉r❡ ✺✷✸
✶✺❇ ❉✐❢❢r❛❝t✐♦♥t❡❝❤♥✐q✉❡s ✺✷✻
✶✺❈ ❇♦♥❞✐♥❣ ✐♥s♦❧✐❞s ✺✸✹
✶✺❉❚❤❡♠❡❝❤❛♥✐❝❛❧♣r♦♣❡rt✐❡s♦❢s♦❧✐❞s ✺✸✾
✶✺❊ ❚❤❡❡❧❡❝tr✐❝❛❧♣r♦♣❡rt✐❡s♦❢s♦❧✐❞s ✺✹✶
✶✺❋ ❚❤❡ ♠❛❣♥❡t✐❝ ♣r♦♣❡rt✐❡s ♦❢s♦❧✐❞s ✺✹✸
✶✺● ❚❤❡♦♣t✐❝❛❧♣r♦♣❡rt✐❡s♦❢s♦❧✐❞s ✺✹✻
✶✻ ▼♦❧❡❝✉❧❡s ✐♥ ♠♦t✐♦♥ ✺✺✶
✶✻❆ ❚r❛♥s♣♦rt ♣r♦♣❡rt✐❡s ♦❢❛ ♣❡r❢❡❝t❣❛s ✺✺✶
✶✻❇ ▼♦t✐♦♥✐♥❧✐q✉✐❞s ✺✺✽
✶✻❈ ❉✐❢❢✉s✐♦♥ ✺✻✸
✶✼ ❈❤❡♠✐❝❛❧ ❦✐♥❡t✐❝s ✺✼✸
✶✼❆ ❚❤❡ r❛t❡s ♦❢ ❝❤❡♠✐❝❛❧ r❡❛❝t✐♦♥s ✺✼✸
,✈✐ ❚❆❇▲❊ ❖❋ ❈❖◆❚❊◆❚❙
✶✼❇ ■♥t❡❣r❛t❡❞ r❛t❡ ❧❛✇s ✺✼✾
✶✼❈ ❘❡❛❝t✐♦♥s ❛♣♣r♦❛❝❤✐♥❣ ❡q✉✐❧✐❜r✐✉♠ ✺✾✸
✶✼❉ ❚❤❡ ❆rr❤❡♥✐✉s ❡q✉❛t✐♦♥ ✺✾✼
✶✼❊ ❘❡❛❝t✐♦♥♠❡❝❤❛♥✐s♠s ✻✵✶
✶✼❋ ❊①❛♠♣❧❡s ♦❢ r❡❛❝t✐♦♥♠❡❝❤❛♥✐s♠s ✻✵✼
✶✼● P❤♦t♦❝❤❡♠✐str② ✻✶✹
✶✽ ❘❡❛❝t✐♦♥❞②♥❛♠✐❝s ✻✷✶
✶✽❆ ❈♦❧❧✐s✐♦♥ t❤❡♦r② ✻✷✶
✶✽❇ ❉✐❢❢✉s✐♦♥✲❝♦♥tr♦❧❧❡❞r❡❛❝t✐♦♥s ✻✷✻
✶✽❈❚r❛♥s✐t✐♦♥✲st❛t❡t❤❡♦r② ✻✷✾
✶✽❉ ❚❤❡ ❞②♥❛♠✐❝s ♦❢ ♠♦❧❡❝✉❧❛r ❝♦❧❧✐s✐♦♥s ✻✸✾
✶✽❊ ❊❧❡❝tr♦♥ tr❛♥s❢❡r ✐♥ ❤♦♠♦❣❡♥❡♦✉s s②st❡♠s ✻✹✵
✶✾ Pr♦❝❡ss❡s❛ts♦❧✐❞s✉r❢❛❝❡s ✻✹✼
✶✾❆ ❆♥ ✐♥tr♦❞✉❝t✐♦♥ t♦ s♦❧✐❞s✉r❢❛❝❡s ✻✹✼
✶✾❇ ❆❞s♦r♣t✐♦♥❛♥❞❞❡s♦r♣t✐♦♥ ✻✺✵
✶✾❈ ❍❡t❡r♦❣❡♥❡♦✉s ❝❛t❛❧②s✐s ✻✻✵
✶✾❉ Pr♦❝❡ss❡s ❛t ❡❧❡❝tr♦❞❡s ✻✻✷
, ✶ ❚❤❡♣r♦♣❡rt✐❡s♦❢❣❛s❡s
✶❆ ❚❤❡ ♣❡r❢❡❝t ❣❛s
❆♥s✇❡rs t♦ ❞✐s❝✉ss✐♦♥ q✉❡st✐♦♥s
DÔA.ò e partial pressure of gas J, pJ , in a mixture of gases is given by [ÔA.â–À], pJ =
xJ p, where p is the total pressure and xJ is the mole fraction of J.
If the gases are perfect, the partial pressure is also the pressure the
gas would exert if it occupied on its own the same container as the
mixture at the same temperature. is leads to Dalton’s law, which is
that the pressure of a mixture of gases is the sum of the pressures that
each one would exert if it occupied the container alone.
Dalton’s law is a limiting law because it holds exactly only in the limit that
there are no interactions between the molecules, which for real gases
will be in the limit of zero pressure.
❙♦❧✉t✐♦♥s t♦ ❡①❡r❝✐s❡s
EÔA.Ô(b) From inside the front cover the conversion between pressure units is: Ô atm
≡
101.325 kPa ≡ Þâý Torr.
(i) A pressure of òò. kPa is converted to atm as follows
1 atm
22.5 kPa × = ý.òòò atm
101.325 kPa
(ii) A pressure of ÞÞý Torr is converted to Pa as follows
1 atm
101.325 kPa
= 103 kPa =1.03 × 10 Pa
770 Torr × 5
760 Torr × 1 atm
EÔA.ò(b) e perfect gas law [ÔA.¥–—], pV = nRT , is rearranged to give the
pressure, p = nRT ~V . e amount n is found by dividing the mass by the
molar mass of Ar, çÀ.À g mol−1.
n
³¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ·¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹µ −2
3 −1 −1
(25 g) ‰8.3145 × 10 dm bar K mol Ž × (303.15 K)
p = (39.95 g mol−1
1.5 dm3
)
= 10.4 bar
So no , the sample would not exert a pressure of ò.ý Ôý.¥ bar if it were
bar, but a perfect gas.
,✷ ✶❚❍❊P❘❖P❊❘❚■❊❙❖❋●❆❙❊❙
EÔA.ç(b) Because the temperature is constant (isothermal) Boyle’s law applies, pV =
const. erefore the product pV is the same for the initial and final states
pf Vf = pi Vi hence pi = pf Vf~Vi
e initial volume is Ô.—ý dm3 greater than the final volume so Vi = 2.14+1.80 =
3.94 dm3 .
Vf 2.14
pi = × pf = × (1.97 bar) = 1.07 bar
Vi dm3
3.94 dm
3
(i) e in initial pressure is Ô.ýÞ bar
(ii) Because Ô atm is equivalent to Ô.ýÔçò bar and also to Þâý Torr,
the initial pressure expressed in Torr is
1 atm 760 Torr
× × 1.07 bar = —ýç Torr
1.01325 bar 1 atm
EÔA.¥(b) If the gas is assumed to be perfect, the equation of state is [ÔA.¥–—], pV =
nRT . In this case the volume and amount (in moles) of the gas are
constant, so it follows that the pressure is proportional to the
temperature: p ∝ T . e ratio of the final and initial pressures is therefore
equal to the ratio of the temperatures: pf ~ pi = Tf ~Ti. Solving for the
final pressure pf (remember to use absolute temperatures) gives
Tf
pf = × pi
Ti
(11 + 273.15) K 125 kPa =
= ×( ) Ôòý kPa
(23 + 273.15) K
EÔA. (b) e perfect gas law pV = nRT is rearranged to give n = pV ~ RT.
pV
n = RT
(1.00 × 1.01325 × 105 Pa) × (4.00 × 103 m3 ) 1.66... × 105 mol
= =
(8.3145 J K−1 mol− ) × ([20 + 273.15] K)
1
where 1 J = 1 kg m2 s−2 and 1 Pa = 1 kg m−1 s−2 have been used.
−1
e molar mass of CH4 is 12.01 + 4 × 1.0079 = 16.0416 g , so the mass of
mol −1
CH4 is (1.66... ×105 mol)×(16.0416 g ) = 2.67 × 106 g or 2.67 × 103 kg.
mol
EÔA.â(b) e vapour is assumed to be a perfect gas, so the gas law pV = nRT
applies. e task is to use this expression to relate the measured
mass density to the molar mass.
First, the amount n is expressed as the mass m divided by the molar
mass M to give pV = (m~ M)RT ; division of both sides by V gives p =
(m~ V )(RT~ M).
, ❙❖▲❯❚■❖◆❙ ▼❆◆❯❆▲ ❚❖ ❆❈❈❖▼P❆◆❨ ❆❚❑■◆❙✬ P❍❨❙■❈❆▲ ❈❍❊▼■❙❚❘❨ ✸
e quantity (m~ V ) is the mass density ρ, so p = ρRT ~ M, which
rearranges to M = ρRT ~ p; this is the required relationship between M
and the density.
−3 −1 −1
ρRT (0.6388 kg m ) × (8.3145 J K mol ) × ([100 + 273.15] K)
M= =
p 16.0 × 103 Pa
= 0.123... kg
mol−1
where 1 J = 1 kg m2 s−2 and 1 Pa = 1 kg m−1 s−2 have been used. e
molar mass of P is 30.97 g mol−1 , so the number of P atoms in the
−1 −1
molecules comprising
the vapour is (0.123... × 103 g mol )~(30.97 g mol ) = 4.00. e result is
expected to be an integer, so the formula is likely to be P4 .
EÔA.Þ(b) e vapour is assumed to be a perfect gas, so the gas law pV = nRT
applies; the task is to use this expression to relate the measured data
to the mass m. is is done by expressing the amount n as m~ M,
where M is the the molar mass. With this substitution it follows that m =
MPV ~ RT.
e partial pressure of water vapour is 0.53 times the saturated vapour pressure
M pV
m = RT
−1
(18.0158 g mol ) × (0.53 × 0.0281 × 105 Pa) × (250 m3 )
=
(8.3145 J K−1 mol− ) × ([23 + 273.15] K)
1
= 2.7 × 103 g = ò.Þ kg
EÔA.—(b) Once the total amount and the total pressure ptot are known, the
volume is found using the perfect gas law. e total amount in moles of
the mixture of gases is
ntot = nCH4 + nAr + nNe = 4m CH + mAr + mNe
MCH4 MAr MNe
0.320 g + 0.175 g 0.225 g
= −1 −1
+
(12.01 + 4 × 1.0079) g mol 39.95 g mol 20.18 g mol−1
= 3.54... × 10−2 mol
e mole fraction of neon is
nN 0.225 g 1
xNe = e = −1× 3.54... × 10−2 mol
= 0.314...
20.18 g mol
nto
t
Because pNe = xNe × ptot it follow that
pN 8.87 kPa ò—.ò kPa
ptot = e = =
0.314...
xN
e
e volume is calculated using the perfect gas equation with the known
total pressure and total amount
−2 −1 −1
nRT (3.54... × 10 mol) × (8.3145 J K mol ) × (300 K)
V= =
p 28.2 × 103 Pa
= 3.14 × 10−3 m3 ç.Ô¥ dm3
=
,✹ ✶❚❍❊P❘❖P❊❘❚■❊❙❖❋●❆❙❊❙
EÔA.À(b) e vapour is assumed to be a perfect gas, so the gas law pV = nRT applies.
e task is to use this expression to relate the measured pressure and
volume of a known mass of gas to the molar mass.
e amount n is expressed as the mass m divided by the molar
mass M to give pV = (m~ M)RT ; this rearranges to M = mRT~ pV
which is the required relationship. e pressure in Torr is converted to Pa
by noting that Þâý Torr is equivalent to Ô atm.
mRT
M = pV
−6 −1 −1
(33.5 × 10 kg) × (8.3145 J K mol ) × (298 K)
=
(152 Torr~760 Torr) × (1.01325 × 105 Pa) × (250 × 10−6 m3 )
= ý.ýÔ⥠kg
mol−1
e relationships 1 J = 1 kg m2 s−2 and 1 Pa = 1 kg m−1 s−2 have
been used; note the conversion of the volume to m3 .
EÔA.Ôý(b) e idea here is that the volume will go to zero at absolute zero. e data
given are the slope of the volume/temperature plot, together with one
fixed point, so the equation of the straight line can be found, and then
the required intercept.
e equation of the line is
3 ○ 3
(V ~dm ) = (0.0741) × (θ~ C) + (c~dm )
e fixed point given is that the volume at θ = 0 ○C is òý.ýý dm3 , so the constant
c is equal to this volume
3 ○
(V ~dm ) = (0.0741)
○
× (θ~ C) + 20.00
−20.00)~(0.0741) = −270, hence
is is solved for V = 0 to give (θ~ C) = (
θ = −270 ○ C . is is the estimate of absolute zero.
EÔA.ÔÔ(b) (i) e mole fractions
are 1.5 mol
nH2 3
8
5
8
xH2 = n + n = = xN2 = 1 − xH2 =
H2 N2 1.5 mol + 2.5 mol
(ii) e partial pressures are given by pi = xi ptot. e total pressure is
given by the perfect gas law: ptot = ntot RT~ V
−1 −1 ) × (273.15 K)
pH2 = xH2 ptot = 3 × (4.0 mol) × (8.3145 J K mol
8 22.4 × 10−3 m3
5
= 1.5 × 10 Pa
5 (4.0 mol) × (8.3145 J K−1 mol−1 ) × (273.15 K)
pN2 = xN2 ptot = ×
8 22.4 × 10−3 m3
5
= 2.5 × 10
Pa
Expressed in atmospheres these are Ô. atm and ò. atm, respectively.
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Peter Atkins, Chapter 1-19
SOLUTION MANUAL
, ❚❛❜❧❡ ♦❢ ❝♦♥t❡♥ts
Pr❡❢❛❝❡ ✈✐✐
✶ ❚❤❡ ♣r♦♣❡rt✐❡s ♦❢ ❣❛s❡s ✶
✶❆ ❚❤❡ ♣❡r❢❡❝t ❣❛s ✶
✶❇ ❚❤❡ ❦✐♥❡t✐❝ ♠♦❞❡❧ ✶✵
✶❈ ❘❡❛❧ ❣❛s❡s ✶✾
✷ ■♥t❡r♥❛❧ ❡♥❡r❣② ✸✺
✷❆ ■♥t❡r♥❛❧ ❡♥❡r❣② ✸✺
✷❇ ❊♥t❤❛❧♣② ✹✵
✷❈ ❚❤❡r♠♦❝❤❡♠✐str② ✹✸
✷❉ ❙t❛t❡ ❢✉♥❝t✐♦♥s ❛♥❞ ❡①❛❝t ❞✐❢❢❡r❡♥t✐❛❧s ✹✾
✷❊ ❆❞✐❛❜❛t✐❝ ❝❤❛♥❣❡s ✺✺
✸ ❚❤❡ s❡❝♦♥❞ ❛♥❞ t❤✐r❞ ❧❛✇s ✻✸
✸❆ ❊♥tr♦♣② ✻✸
✸❇ ❊♥tr♦♣② ❝❤❛♥❣❡s ❛❝❝♦♠♣❛♥②✐♥❣ s♣❡❝✐❢✐❝ ♣r♦❝❡ss❡s ✻✽
✸❈ ❚❤❡ ♠❡❛s✉r❡♠❡♥t ♦❢ ❡♥tr♦♣② ✽✵
✸❉ ❈♦♥❝❡♥tr❛t✐♥❣♦♥t❤❡s②st❡♠ ✽✼
✸❊ ❈♦♠❜✐♥✐♥❣ t❤❡ ❋✐rst ❛♥❞ ❙❡❝♦♥❞ ▲❛✇s ✾✷
✹ P❤②s✐❝❛❧ tr❛♥s❢♦r♠❛t✐♦♥s ♦❢ ♣✉r❡ s✉❜st❛♥❝❡s ✾✾
✹❆ P❤❛s❡❞✐❛❣r❛♠s ♦❢ ♣✉r❡s✉❜st❛♥❝❡s ✾✾
✹❇ ❚❤❡r♠♦❞②♥❛♠✐❝❛s♣❡❝ts ♦❢ ♣❤❛s❡tr❛♥s✐t✐♦♥s ✶✵✶
✺ ❙✐♠♣❧❡♠✐①t✉r❡s ✶✶✺
✺❆ ❚❤❡ t❤❡r♠♦❞②♥❛♠✐❝ ❞❡s❝r✐♣t✐♦♥ ♦❢ ♠✐①t✉r❡s ✶✶✺
✺❇ ❚❤❡♣r♦♣❡rt✐❡s♦❢s♦❧✉t✐♦♥s ✶✷✻
✺❈ P❤❛s❡❞✐❛❣r❛♠s♦❢❜✐♥❛r②s②st❡♠s✿❧✐q✉✐❞s ✶✹✶
✺❉ P❤❛s❡ ❞✐❛❣r❛♠s ♦❢ ❜✐♥❛r② s②st❡♠s✿s♦❧✐❞s ✶✹✽
✺❊ P❤❛s❡❞✐❛❣r❛♠s ♦❢ t❡r♥❛r②s②st❡♠s ✶✺✹
✺❋ ❆❝t✐✈✐t✐❡s ✶✺✽
✻ ❈❤❡♠✐❝❛❧ ❡q✉✐❧✐❜r✐✉♠ ✶✼✶
✻❆ ❚❤❡ ❡q✉✐❧✐❜r✐✉♠ ❝♦♥st❛♥t ✶✼✶
,✐✈ ❚❆❇▲❊ ❖❋ ❈❖◆❚❊◆❚❙
✻❇ ❚❤❡ r❡s♣♦♥s❡ ♦❢ ❡q✉✐❧✐❜r✐❛ t♦ t❤❡ ❝♦♥❞✐t✐♦♥s ✶✼✾
✻❈ ❊❧❡❝tr♦❝❤❡♠✐❝❛❧ ❝❡❧❧s ✶✾✵
✻❉ ❊❧❡❝tr♦❞❡ ♣♦t❡♥t✐❛❧s ✶✾✼
✼ ◗✉❛♥t✉♠ t❤❡♦r② ✷✶✺
✼❆ ❚❤❡ ♦r✐❣✐♥s ♦❢ q✉❛♥t✉♠ ♠❡❝❤❛♥✐❝s ✷✶✺
✼❇ ❲❛✈❡❢✉♥❝t✐♦♥s ✷✷✸
✼❈ ❖♣❡r❛t♦rs ❛♥❞ ♦❜s❡r✈❛❜❧❡s ✷✷✼
✼❉ ❚r❛♥s❧❛t✐♦♥❛❧ ♠♦t✐♦♥ ✷✸✻
✼❊ ❱✐❜r❛t✐♦♥❛❧ ♠♦t✐♦♥ ✷✹✽
✼❋ ❘♦t❛t✐♦♥❛❧ ♠♦t✐♦♥ ✷✺✼
✽ ❆t♦♠✐❝ str✉❝t✉r❡ ❛♥❞ s♣❡❝tr❛ ✷✼✶
✽❆ ❍②❞r♦❣❡♥✐❝ ❆t♦♠s ✷✼✶
✽❇ ▼❛♥②✲❡❧❡❝tr♦♥ ❛t♦♠s ✷✼✼
✽❈ ❆t♦♠✐❝ s♣❡❝tr❛ ✷✼✾
✾ ▼♦❧❡❝✉❧❛r ❙tr✉❝t✉r❡ ✷✽✼
✾❆ ❱❛❧❡♥❝❡✲❜♦♥❞ t❤❡♦r② ✷✽✼
✾❇ ▼♦❧❡❝✉❧❛r ♦r❜✐t❛❧ t❤❡♦r②✿ t❤❡ ❤②❞r♦❣❡♥ ♠♦❧❡❝✉❧❡✲✐♦♥ ✷✾✷
✾❈ ▼♦❧❡❝✉❧❛r ♦r❜✐t❛❧ t❤❡♦r②✿ ❤♦♠♦♥✉❝❧❡❛r ❞✐❛t♦♠✐❝ ♠♦❧❡❝✉❧❡s ✷✾✽
✾❉ ▼♦❧❡❝✉❧❛r ♦r❜✐t❛❧ t❤❡♦r②✿ ❤❡t❡r♦♥✉❝❧❡❛r ❞✐❛t♦♠✐❝ ♠♦❧❡❝✉❧❡s ✸✵✷
✾❊ ▼♦❧❡❝✉❧❛r ♦r❜✐t❛❧ t❤❡♦r②✿ ♣♦❧②❛t♦♠✐❝ ♠♦❧❡❝✉❧❡s ✸✵✼
✶✵ ▼♦❧❡❝✉❧❛r s②♠♠❡tr② ✸✷✶
✶✵❆ ❙❤❛♣❡ ❛♥❞ s②♠♠❡tr② ✸✷✶
✶✵❇ ●r♦✉♣ t❤❡♦r② ✸✷✾
✶✵❈ ❆♣♣❧✐❝❛t✐♦♥s ♦❢ s②♠♠❡tr② ✸✸✼
✶✶ ▼♦❧❡❝✉❧❛r❙♣❡❝tr♦s❝♦♣② ✸✹✾
✶✶❆ ●❡♥❡r❛❧ ❢❡❛t✉r❡s ♦❢ ♠♦❧❡❝✉❧❛r s♣❡❝tr♦s❝♦♣② ✸✹✾
✶✶❇ ❘♦t❛t✐♦♥❛❧ s♣❡❝tr♦s❝♦♣② ✸✺✼
✶✶❈ ❱✐❜r❛t✐♦♥❛❧ s♣❡❝tr♦s❝♦♣② ♦❢ ❞✐❛t♦♠✐❝ ♠♦❧❡❝✉❧❡s ✸✼✶
✶✶❉ ❱✐❜r❛t✐♦♥❛❧ s♣❡❝tr♦s❝♦♣② ♦❢ ♣♦❧②❛t♦♠✐❝ ♠♦❧❡❝✉❧❡s ✸✽✹
✶✶❊ ❙②♠♠❡tr② ❛♥❛❧②s✐s ♦❢ ✈✐❜r❛t✐♦♥❛❧ s♣❡❝tr♦s❝♦♣② ✸✽✺
✶✶❋ ❊❧❡❝tr♦♥✐❝ s♣❡❝tr❛ ✸✽✽
✶✶● ❉❡❝❛② ♦❢ ❡①❝✐t❡❞ st❛t❡s ✸✾✽
, ❚❆❇▲❊ ❖❋ ❈❖◆❚❊◆❚❙ ✈
✶✷ ▼❛❣♥❡t✐❝ r❡s♦♥❛♥❝❡ ✹✵✾
✶✷❆ ●❡♥❡r❛❧ ♣r✐♥❝✐♣❧❡s ✹✵✾
✶✷❇ ❋❡❛t✉r❡s ♦❢ ◆▼❘ s♣❡❝tr❛ ✹✶✷
✶✷❈ P✉❧s❡t❡❝❤♥✐q✉❡s ✐♥ ◆▼❘ ✹✷✶
✶✷❉ ❊❧❡❝tr♦♥ ♣❛r❛♠❛❣♥❡t✐❝ r❡s♦♥❛♥❝❡ ✹✷✾
✶✸ ❙t❛t✐st✐❝❛❧ t❤❡r♠♦❞②♥❛♠✐❝s ✹✸✺
✶✸❆ ❚❤❡ ❇♦❧t③♠❛♥♥ ❞✐str✐❜✉t✐♦♥ ✹✸✺
✶✸❇ P❛rt✐t✐♦♥ ❢✉♥❝t✐♦♥s ✹✸✾
✶✸❈ ▼♦❧❡❝✉❧❛r❡♥❡r❣✐❡s ✹✹✾
✶✸❉ ❚❤❡ ❝❛♥♦♥✐❝❛❧ ❡♥s❡♠❜❧❡ ✹✺✻
✶✸❊ ❚❤❡ ✐♥t❡r♥❛❧ ❡♥❡r❣② ❛♥❞ ❡♥tr♦♣② ✹✺✼
✶✸❋ ❉❡r✐✈❡❞ ❢✉♥❝t✐♦♥s ✹✼✸
✶✹ ▼♦❧❡❝✉❧❛r■♥t❡r❛❝t✐♦♥s ✹✽✺
✶✹❆ ❊❧❡❝tr✐❝ ♣r♦♣❡rt✐❡s ♦❢ ♠♦❧❡❝✉❧❡s ✹✽✺
✶✹❇ ■♥t❡r❛❝t✐♦♥s❜❡t✇❡❡♥♠♦❧❡❝✉❧❡s ✹✾✻
✶✹❈ ▲✐q✉✐❞s ✺✵✹
✶✹❉ ▼❛❝r♦♠♦❧❡❝✉❧❡s ✺✵✻
✶✹❊ ❙❡❧❢✲❛ss❡♠❜❧② ✺✶✼
✶✺ ❙♦❧✐❞s ✺✷✸
✶✺❆ ❈r②st❛❧ str✉❝t✉r❡ ✺✷✸
✶✺❇ ❉✐❢❢r❛❝t✐♦♥t❡❝❤♥✐q✉❡s ✺✷✻
✶✺❈ ❇♦♥❞✐♥❣ ✐♥s♦❧✐❞s ✺✸✹
✶✺❉❚❤❡♠❡❝❤❛♥✐❝❛❧♣r♦♣❡rt✐❡s♦❢s♦❧✐❞s ✺✸✾
✶✺❊ ❚❤❡❡❧❡❝tr✐❝❛❧♣r♦♣❡rt✐❡s♦❢s♦❧✐❞s ✺✹✶
✶✺❋ ❚❤❡ ♠❛❣♥❡t✐❝ ♣r♦♣❡rt✐❡s ♦❢s♦❧✐❞s ✺✹✸
✶✺● ❚❤❡♦♣t✐❝❛❧♣r♦♣❡rt✐❡s♦❢s♦❧✐❞s ✺✹✻
✶✻ ▼♦❧❡❝✉❧❡s ✐♥ ♠♦t✐♦♥ ✺✺✶
✶✻❆ ❚r❛♥s♣♦rt ♣r♦♣❡rt✐❡s ♦❢❛ ♣❡r❢❡❝t❣❛s ✺✺✶
✶✻❇ ▼♦t✐♦♥✐♥❧✐q✉✐❞s ✺✺✽
✶✻❈ ❉✐❢❢✉s✐♦♥ ✺✻✸
✶✼ ❈❤❡♠✐❝❛❧ ❦✐♥❡t✐❝s ✺✼✸
✶✼❆ ❚❤❡ r❛t❡s ♦❢ ❝❤❡♠✐❝❛❧ r❡❛❝t✐♦♥s ✺✼✸
,✈✐ ❚❆❇▲❊ ❖❋ ❈❖◆❚❊◆❚❙
✶✼❇ ■♥t❡❣r❛t❡❞ r❛t❡ ❧❛✇s ✺✼✾
✶✼❈ ❘❡❛❝t✐♦♥s ❛♣♣r♦❛❝❤✐♥❣ ❡q✉✐❧✐❜r✐✉♠ ✺✾✸
✶✼❉ ❚❤❡ ❆rr❤❡♥✐✉s ❡q✉❛t✐♦♥ ✺✾✼
✶✼❊ ❘❡❛❝t✐♦♥♠❡❝❤❛♥✐s♠s ✻✵✶
✶✼❋ ❊①❛♠♣❧❡s ♦❢ r❡❛❝t✐♦♥♠❡❝❤❛♥✐s♠s ✻✵✼
✶✼● P❤♦t♦❝❤❡♠✐str② ✻✶✹
✶✽ ❘❡❛❝t✐♦♥❞②♥❛♠✐❝s ✻✷✶
✶✽❆ ❈♦❧❧✐s✐♦♥ t❤❡♦r② ✻✷✶
✶✽❇ ❉✐❢❢✉s✐♦♥✲❝♦♥tr♦❧❧❡❞r❡❛❝t✐♦♥s ✻✷✻
✶✽❈❚r❛♥s✐t✐♦♥✲st❛t❡t❤❡♦r② ✻✷✾
✶✽❉ ❚❤❡ ❞②♥❛♠✐❝s ♦❢ ♠♦❧❡❝✉❧❛r ❝♦❧❧✐s✐♦♥s ✻✸✾
✶✽❊ ❊❧❡❝tr♦♥ tr❛♥s❢❡r ✐♥ ❤♦♠♦❣❡♥❡♦✉s s②st❡♠s ✻✹✵
✶✾ Pr♦❝❡ss❡s❛ts♦❧✐❞s✉r❢❛❝❡s ✻✹✼
✶✾❆ ❆♥ ✐♥tr♦❞✉❝t✐♦♥ t♦ s♦❧✐❞s✉r❢❛❝❡s ✻✹✼
✶✾❇ ❆❞s♦r♣t✐♦♥❛♥❞❞❡s♦r♣t✐♦♥ ✻✺✵
✶✾❈ ❍❡t❡r♦❣❡♥❡♦✉s ❝❛t❛❧②s✐s ✻✻✵
✶✾❉ Pr♦❝❡ss❡s ❛t ❡❧❡❝tr♦❞❡s ✻✻✷
, ✶ ❚❤❡♣r♦♣❡rt✐❡s♦❢❣❛s❡s
✶❆ ❚❤❡ ♣❡r❢❡❝t ❣❛s
❆♥s✇❡rs t♦ ❞✐s❝✉ss✐♦♥ q✉❡st✐♦♥s
DÔA.ò e partial pressure of gas J, pJ , in a mixture of gases is given by [ÔA.â–À], pJ =
xJ p, where p is the total pressure and xJ is the mole fraction of J.
If the gases are perfect, the partial pressure is also the pressure the
gas would exert if it occupied on its own the same container as the
mixture at the same temperature. is leads to Dalton’s law, which is
that the pressure of a mixture of gases is the sum of the pressures that
each one would exert if it occupied the container alone.
Dalton’s law is a limiting law because it holds exactly only in the limit that
there are no interactions between the molecules, which for real gases
will be in the limit of zero pressure.
❙♦❧✉t✐♦♥s t♦ ❡①❡r❝✐s❡s
EÔA.Ô(b) From inside the front cover the conversion between pressure units is: Ô atm
≡
101.325 kPa ≡ Þâý Torr.
(i) A pressure of òò. kPa is converted to atm as follows
1 atm
22.5 kPa × = ý.òòò atm
101.325 kPa
(ii) A pressure of ÞÞý Torr is converted to Pa as follows
1 atm
101.325 kPa
= 103 kPa =1.03 × 10 Pa
770 Torr × 5
760 Torr × 1 atm
EÔA.ò(b) e perfect gas law [ÔA.¥–—], pV = nRT , is rearranged to give the
pressure, p = nRT ~V . e amount n is found by dividing the mass by the
molar mass of Ar, çÀ.À g mol−1.
n
³¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ·¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹µ −2
3 −1 −1
(25 g) ‰8.3145 × 10 dm bar K mol Ž × (303.15 K)
p = (39.95 g mol−1
1.5 dm3
)
= 10.4 bar
So no , the sample would not exert a pressure of ò.ý Ôý.¥ bar if it were
bar, but a perfect gas.
,✷ ✶❚❍❊P❘❖P❊❘❚■❊❙❖❋●❆❙❊❙
EÔA.ç(b) Because the temperature is constant (isothermal) Boyle’s law applies, pV =
const. erefore the product pV is the same for the initial and final states
pf Vf = pi Vi hence pi = pf Vf~Vi
e initial volume is Ô.—ý dm3 greater than the final volume so Vi = 2.14+1.80 =
3.94 dm3 .
Vf 2.14
pi = × pf = × (1.97 bar) = 1.07 bar
Vi dm3
3.94 dm
3
(i) e in initial pressure is Ô.ýÞ bar
(ii) Because Ô atm is equivalent to Ô.ýÔçò bar and also to Þâý Torr,
the initial pressure expressed in Torr is
1 atm 760 Torr
× × 1.07 bar = —ýç Torr
1.01325 bar 1 atm
EÔA.¥(b) If the gas is assumed to be perfect, the equation of state is [ÔA.¥–—], pV =
nRT . In this case the volume and amount (in moles) of the gas are
constant, so it follows that the pressure is proportional to the
temperature: p ∝ T . e ratio of the final and initial pressures is therefore
equal to the ratio of the temperatures: pf ~ pi = Tf ~Ti. Solving for the
final pressure pf (remember to use absolute temperatures) gives
Tf
pf = × pi
Ti
(11 + 273.15) K 125 kPa =
= ×( ) Ôòý kPa
(23 + 273.15) K
EÔA. (b) e perfect gas law pV = nRT is rearranged to give n = pV ~ RT.
pV
n = RT
(1.00 × 1.01325 × 105 Pa) × (4.00 × 103 m3 ) 1.66... × 105 mol
= =
(8.3145 J K−1 mol− ) × ([20 + 273.15] K)
1
where 1 J = 1 kg m2 s−2 and 1 Pa = 1 kg m−1 s−2 have been used.
−1
e molar mass of CH4 is 12.01 + 4 × 1.0079 = 16.0416 g , so the mass of
mol −1
CH4 is (1.66... ×105 mol)×(16.0416 g ) = 2.67 × 106 g or 2.67 × 103 kg.
mol
EÔA.â(b) e vapour is assumed to be a perfect gas, so the gas law pV = nRT
applies. e task is to use this expression to relate the measured
mass density to the molar mass.
First, the amount n is expressed as the mass m divided by the molar
mass M to give pV = (m~ M)RT ; division of both sides by V gives p =
(m~ V )(RT~ M).
, ❙❖▲❯❚■❖◆❙ ▼❆◆❯❆▲ ❚❖ ❆❈❈❖▼P❆◆❨ ❆❚❑■◆❙✬ P❍❨❙■❈❆▲ ❈❍❊▼■❙❚❘❨ ✸
e quantity (m~ V ) is the mass density ρ, so p = ρRT ~ M, which
rearranges to M = ρRT ~ p; this is the required relationship between M
and the density.
−3 −1 −1
ρRT (0.6388 kg m ) × (8.3145 J K mol ) × ([100 + 273.15] K)
M= =
p 16.0 × 103 Pa
= 0.123... kg
mol−1
where 1 J = 1 kg m2 s−2 and 1 Pa = 1 kg m−1 s−2 have been used. e
molar mass of P is 30.97 g mol−1 , so the number of P atoms in the
−1 −1
molecules comprising
the vapour is (0.123... × 103 g mol )~(30.97 g mol ) = 4.00. e result is
expected to be an integer, so the formula is likely to be P4 .
EÔA.Þ(b) e vapour is assumed to be a perfect gas, so the gas law pV = nRT
applies; the task is to use this expression to relate the measured data
to the mass m. is is done by expressing the amount n as m~ M,
where M is the the molar mass. With this substitution it follows that m =
MPV ~ RT.
e partial pressure of water vapour is 0.53 times the saturated vapour pressure
M pV
m = RT
−1
(18.0158 g mol ) × (0.53 × 0.0281 × 105 Pa) × (250 m3 )
=
(8.3145 J K−1 mol− ) × ([23 + 273.15] K)
1
= 2.7 × 103 g = ò.Þ kg
EÔA.—(b) Once the total amount and the total pressure ptot are known, the
volume is found using the perfect gas law. e total amount in moles of
the mixture of gases is
ntot = nCH4 + nAr + nNe = 4m CH + mAr + mNe
MCH4 MAr MNe
0.320 g + 0.175 g 0.225 g
= −1 −1
+
(12.01 + 4 × 1.0079) g mol 39.95 g mol 20.18 g mol−1
= 3.54... × 10−2 mol
e mole fraction of neon is
nN 0.225 g 1
xNe = e = −1× 3.54... × 10−2 mol
= 0.314...
20.18 g mol
nto
t
Because pNe = xNe × ptot it follow that
pN 8.87 kPa ò—.ò kPa
ptot = e = =
0.314...
xN
e
e volume is calculated using the perfect gas equation with the known
total pressure and total amount
−2 −1 −1
nRT (3.54... × 10 mol) × (8.3145 J K mol ) × (300 K)
V= =
p 28.2 × 103 Pa
= 3.14 × 10−3 m3 ç.Ô¥ dm3
=
,✹ ✶❚❍❊P❘❖P❊❘❚■❊❙❖❋●❆❙❊❙
EÔA.À(b) e vapour is assumed to be a perfect gas, so the gas law pV = nRT applies.
e task is to use this expression to relate the measured pressure and
volume of a known mass of gas to the molar mass.
e amount n is expressed as the mass m divided by the molar
mass M to give pV = (m~ M)RT ; this rearranges to M = mRT~ pV
which is the required relationship. e pressure in Torr is converted to Pa
by noting that Þâý Torr is equivalent to Ô atm.
mRT
M = pV
−6 −1 −1
(33.5 × 10 kg) × (8.3145 J K mol ) × (298 K)
=
(152 Torr~760 Torr) × (1.01325 × 105 Pa) × (250 × 10−6 m3 )
= ý.ýÔ⥠kg
mol−1
e relationships 1 J = 1 kg m2 s−2 and 1 Pa = 1 kg m−1 s−2 have
been used; note the conversion of the volume to m3 .
EÔA.Ôý(b) e idea here is that the volume will go to zero at absolute zero. e data
given are the slope of the volume/temperature plot, together with one
fixed point, so the equation of the straight line can be found, and then
the required intercept.
e equation of the line is
3 ○ 3
(V ~dm ) = (0.0741) × (θ~ C) + (c~dm )
e fixed point given is that the volume at θ = 0 ○C is òý.ýý dm3 , so the constant
c is equal to this volume
3 ○
(V ~dm ) = (0.0741)
○
× (θ~ C) + 20.00
−20.00)~(0.0741) = −270, hence
is is solved for V = 0 to give (θ~ C) = (
θ = −270 ○ C . is is the estimate of absolute zero.
EÔA.ÔÔ(b) (i) e mole fractions
are 1.5 mol
nH2 3
8
5
8
xH2 = n + n = = xN2 = 1 − xH2 =
H2 N2 1.5 mol + 2.5 mol
(ii) e partial pressures are given by pi = xi ptot. e total pressure is
given by the perfect gas law: ptot = ntot RT~ V
−1 −1 ) × (273.15 K)
pH2 = xH2 ptot = 3 × (4.0 mol) × (8.3145 J K mol
8 22.4 × 10−3 m3
5
= 1.5 × 10 Pa
5 (4.0 mol) × (8.3145 J K−1 mol−1 ) × (273.15 K)
pN2 = xN2 ptot = ×
8 22.4 × 10−3 m3
5
= 2.5 × 10
Pa
Expressed in atmospheres these are Ô. atm and ò. atm, respectively.
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