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Chem340 Solution to problem set2
1. J or Pa*m3
(note that both dV and V have unit of volume and can cancel each other)
2. J/Pa = m3
3. J/m3 = Pa
4. Molar kinetic energy K.E= 1/2Mv2
Kinetic energy of a K atom = Mv2 / 2NA
= 39.0983 ×10-3kg mol-1 * (629 m/s)2 / (2*6.022×1023 mol−1) = 1.28 ×10-20 J
5. Molar kinetic energy K.E. = 1/2Mv2 = 1/2 RT for one dimensional system,
temperature T = Mv2 / R
= 39.0983 ×10-3kg mol-1 × (629 m/s)2 / (8.31447 JK-1mol-1)
= 1.86 ×103 K
Gas Laws
5. Problem 1.11 in Engle P. Chem Life Science (photosynthesis)
The mass of CO2 for 1 kg of C:
m(C ) * M (CO 2 ) (1.0 kg ) * 12 × 10-3 kg mol-1
m(CO 2 ) =
=
= 3.7 kg
M (C )
44 × 10- 3 kg mol-1
(
)
Then calculate the number of moles necessary to provide one 1 kg of CO2:
n (CO 2 ) =
(3.7 kg )
m(CO 2 )
=
= 84 mol
M(CO 2 ) 44 × 10-3 kg mol-1
(
)
With the composition of air, a certain volume of air containing one kg of CO2 also
contains:
n (O 2 ) =
n (N 2 ) =
m(O 2 )
M(O 2 )
m(N 2 )
M (N 2 )
(3.7 kg ) × ⎛⎜
0.2 ⎞
⎟
⎝ 0.00046 ⎠ = 5.0 × 104 mol
=
(32 × 10-3 kg mol-1 )
(3.7 kg ) × ⎛⎜
0.8 ⎞
⎟
⎝ 0.00046 ⎠ = 2.3 × 105 mol
=
(28 × 10-3 kg mol-1 )
The volume of air can then be obtained by considering the total number of moles:
n R T {n (CO 2 ) + n (O 2 ) + n (N 2 )} R T
=
Vair = tot
p
p
=
{(84 mol) + (5.0 × 10
= 6.8 × 103 m3
4
) (
)} (
)
mol + 2.3 × 105 mol × 8.314472 J K -1 mol-1 × (298 K )
(101325 Pa )
7. Problem 1.13 in Engle P. Chem Life Science (Hemoglobin problem)
With pV = nRT the number of moles of O2 in one liter of fully oxygenated blood is:
n (O 2 ) =
(
)
(101325 Pa ) × 0.00020 m3
pV
=
= 8.9 × 10-3 mol
R T 8.314472 Pa m 3 K −1 mol−1 × (273 K )
(
)
Converting to molecules:
(
) (
)
molecules O 2 = 8.9 ×10-3 mol × 6.02214 × 10 23 molecules mol-1 = 5.4 × 10 21 molecules Fi
nally, four binding sites per Hemoglobin molecule have to be considered, so that the
number of hemoglobin molecules required is:
hemoglobin molecules required = 5.4 × 10 21 molecules / 4 = 1.4 × 10 21 molecules
8. Problem 1.17 in Engle P. Chem Life Science (Carbon monoxide problem)
Converting the partial pressure of O2 in the atmosphere to ppm using xi =
x (O 2 ) =
pi
:
p
0.20 atm
= 0.20 = 20% = 2.0 × 105 ppm
1 atm
Therefore, the fatal O2/CO ratio is:
x (O 2 ) 2.0 × 105
=
= 2.5 × 10 2
x (CO )
800
9. Problem 1.26 in Engle P. Chem Life Science book. (photosynthesis problem)
First we need to calculate the number of moles of fixed CO2:
nCO2
(101191.68 Pa ) × ( 0.001 m 3 )
pV
=
=
= 0.0446 mol
RT (8.314472 J K−1 mol −1 ) × ( 273 K )
The number of moles of 3-phosphoglycerate formed is given by:
n3− phos = 2 nCO2 = 0.0892 mol
And finally, the mass of 3-phosphoglycerate formed is:
m = nM = 0.0892 mol × 182.97 g mol-1 = 16.3 g
10. Problem 1.8 in Atkins P. Chem (9th) (compressibility problem)
Removed to problem set3.
11. Problem 1.20 in Atkins P. Chem (9th) (van der Waals)
Method 1:
Method 2: p = RT/Vm + (a + bT)/Vm2
Multiply Vm2 on both sides, pVm2= RT Vm + a + bT
Derivative with respective to T on both sides, keep P constant,
⎛ ∂V ⎞
⎛ ∂V ⎞
2p Vm ⎜
⎟ =[R Vm+RT ⎜
⎟ ] +0 + b
⎝ ∂T ⎠ p
⎝ ∂T ⎠ p
Ö
[note: d(fg)=fdg+ gdf;
df df dg
=
×
]
dt dg dt
⎛ ∂V ⎞
⎛ ∂V ⎞
(2p Vm - RT ) ⎜
⎟ = R Vm + b ==> ⎜
⎟ = (R Vm + b) / (2p Vm - RT )
⎝ ∂T ⎠ p
⎝ ∂T ⎠ p
12. Problem 1.22 in Atkins P. Chem (9th) (compressibility)