Hi genetic_liability!!
a) the partial pressure in atm:
Recall the Dalton's Law of Partial Pressures:
The total pressure of a mixture of gases equals the sum of the
pressures that each would exert if it were present alone.
So if we call partial pressure of a gas to the pressure exerted by a
particular component of a mixture of k gases, then the Dalton's Law of
Partial Pressures is:
Pt = P1 + P2 + P3 + ... + Pk
Where
Pt is the total pressure of a sample which contains a mixture of gases.
P1, P2, P3, etc. are the partial pressures of each gas in the mixture.
For each gas in the mixture, its partial pressure is:
Pi = Ni*R*Ti / Vi
Where Ni is the number of moles of each gas in the mixture.
Since T and V are the same for all gases:
Pi = Ni * (R*T/V)
Therefore, Pt will be:
Pt = (N1 + N2 + ... + Nk)*(R*T/V) = Nt*R*T/V
Where Nt is the total number of moles of gas molecules in the mixture.
Assume for this problem we know that Pt = 1 atm (25ēC and may be you
forget the sea level or coastal place condition).
Assume also that the mixture is made of gases G1, G2, ... , Gk where
G1 are the oxygen radicals. Then:
Pt = Nt*(R*T/V) = 1 atm
The steady state concentration of oxygen radicals is measured to be
0.080 parts per million volume, this means:
N1 = (0.080 * 10^-6) * Nt
Then:
P1 = N1*R*T/V = (0.080 * 10^-6) * Nt*R*T/V =
= (0.080 * 10^-6) * Pt =
= (0.080 * 10^-6) * 1 atm =
= 8 *10^(-8) atm
The partial pressure of oxygen radicals is 0.00000008 atm.
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b) molarity scale
Molarity is defined as the number moles of solute divided by the
volume of the solution in liter units.
For mixtures, a molarity scale is represented by the ratio of the
number moles of gas component divided by the volume occuped by the
mixture in liter units.
We now that the volume occupied by the mixture of gases is:
V = Nt*R*T/Pt =
= 1 [mol] * 0.08206 [liter*atm/mol*ēK] * 298.15 [ēK] / 1 [atm] =
= 24.466 liters
The number of moles of oxygen radicals molecules is:
N1 = ppm(v) * Nt =
= 0.08 * 10^(-6) * 1 [mol] =
= 0.00000008 moles
A molarity scale:
Mi = Ni / V .
Then:
M1 = 0.00000008 moles / 24.466 liters =
= 3.27 *10^(-9) moles/liters
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c) molecules per cm^-3
We know M1 and we can do the conversion as follows:
M1 = 3.27 *10^(-9) moles/liters
Then we have that:
3.27 *10^(-9)moles/liters * 1 liter/1000 cm^3 = 3.27 *10^(-12)moles/cm^3 ;
Since each mol of molecules has 6.022 *10^23 molecules/mol we have
that the concentration of oxygen radical C1 is:
C1 = 3.27 *10^(-12) moles/cm^3 =
= 3.27 *10^(-12) moles/cm^3 * 6.022 *10^23 molecules/mol =
= 19.7 *10^11 molecules/cm^3 =
= 1.97 *10^12 molecules/cm^3
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I hope that this helps you. If you find something unclear and/or need
a clerification, please do not hesitate to request for it using the
clarification feature. I will gladly respond your requests for further
assistance on this.
Best regards.
livioflores-ga |