Take nitrogen to be a van der Waals gas with a=1.352 dm6atm mol−2andb=0.0387 dm3mol−1, and calculate ∆Hmwhen the pressure on the gas isdecreased from 500 atm to 1.00 atm at 300 K. For a van der Waals gas, μ={(2a/RT)−b}/Cp,m. Assume Cp,m=7–2R
Answers
Answer:
Preamble:
For any collection of N molecules, the number NE having an energy E or greater is given by
the expression
NE = Ne-E/kT
where k is the Boltzmann constant = 1.381 x 10-23 J K-1
and T is the temperature in Kelvin.
Alternatively, by multiplying both sides of the expression by the Avogadro number, NA, on a molar basis this then becomes
nE = ne-E/RT
where R = 8.314 J K-1 mol-1
Note that in order for the units of E to be consistent with those of R (J K-1 mol-1), E must be expressed in J rather than kJ.
4
From the preamble (above)S,
nE = ne-E/RT
(a) At 300 K:
n = 1.00 mol
E = 8.000 kJ mol-1 = 8000 J mol-1
Therefore n = 1.00 x e-8000/ (8.314 x 300)
= 4.05 x 10-2 mol
The number of molecules, N = 4.05 x 10-2 x NA
= 4.05 x 10-2 x 6.02 x 1023
= 2.4 x 1022 molecules.
(b) At 400 K:
n = 1.00 mol
E = 8.000 kJ mol-1 = 8000 J mol-1
Therefore n = 1.00 x e-8000/ (8.314 x 400)
= 9.02 x 10-2 mol
The number of molecules, N = 9.02 x 10-2 x NA
= 9.02 x 10-2 x 6.02 x 1023
= 5.4 x 1022 molecules.
5
The expression in terms of moles for the distribution of molecular energies,
nE = ne-E/RT,
can be rewritten to give the fraction of the total moles, (n), that have energy E or greater, (nE), as
nE/n = e-E/RT
At T = 250 K and for E = 6.500 kJ mol-1 = 6500 J mol-1,
this becomes
nE/n = e-6500/(8.314 x 250)
[Note that E must be expressed as J to match the units of R]
= 0.044 or 4.4%.
Therefore the % of molecules that have less than 6.500 kJ mol-1 energy =
100.0 - 4.4 = 95.6%.
6
The expression in terms of moles for the distribution of molecular energies is
nE = ne-E/RT
The average value of energy = 1.300 kJ mol-1
Therefore five times this energy = 5 x 1.300 = 6.500 kJ mol-1
E = 6.500 kJ mol-1 = 6500 J mol-1
T = 298 K
Then nE = 1.00 x e-6500/(8.314 x 298)
= 0.0725 mol
Number of molecules with this energy or greater
= 0.0725 x NA
= 0.0725 x 6.022 x 1023
= 4.4 x 1022
Answer:
8000 J mol-1
Explanation:
Given: Take nitrogen to be a van der Waals gas with a=1.352 dm6atm mol−2andb=0.0387 dm3mol−1, and calculate ∆Hm when the pressure on the gas is decreased from 500 atm to 1.00 atm at 300 K. For a van der Waals gas, μ={(2a/RT)−b}/Cp,m. Assume Cp,m=7–2R
At 300 K:
n = 1.00 mol
E = 8.000 kJ mol-1 = 8000 J mol-1
Therefore n = 1.00 x e-8000/ (8.314 x 300)
= 4.05 x 10-2 mol
The number of molecules, N = 4.05 x 10-2 x NA
= 4.05 x 10-2 x 6.02 x 1023
= 2.4 x 1022 molecules.
Now, from the equation nE = ne-E/RT
At 300 K:
n = 1.00 mol
E = 8.000 kJ mol-1 = 8000 J mol-1
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