Real Gas Problems (Advanced)

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Q: 1 2 3
4 5
6

1

Carbon dioxide gas (1.00 mole) at 373 K occupies
536 mL at 50.0 atmosphere pressure. What is the calculated value
of the pressure using
(i) Ideal gas equation
(ii) Van der Waals equation?
[Data  Van der Waals constants for carbon
dioxide:
a = 3.61 L^{2} atm mol^{2}; b = 0.0428 L mol^{1}]
Calculate the % deviation of each value from
that observed.


2 
Predict which of the substances,
NH_{3}, N_{2}, CH_{2}Cl_{2}, Cl_{2}, CCl_{4} has
(i) the smallest van der Waals "a" constant
(ii) the largest "b" constant.


3 
(i) Using Van der Waals equation, calculate
the temperature of 20.0 mole of helium in a 10.0 litre cylinder
at 120 atmosphere pressure.
[Data  Van der Waals constants for helium:
a = 0.0341 L^{2} at mol^{2}; b = 0.0237 L mol^{1}]
(ii) Compare this value with the temperature
calculated from the ideal gas equation.


4 
A particular reaction in the gas phase has an activation energy
of 8.000 kJ mol^{1}.
For 1.00 mole of gas, calculate the number of molecules which exceed
this activation energy at
(a) 300 K
(b) 400 K


5 
At 250 K, the activation energy for a gas phase reaction was determined
to be 6.500 kJ mol^{1}.
What percentage of gaseous molecules would be expected to have less
than this energy at 250 K?


6 
For 1.00 mole of a particular
gas, the average molecular energy is found to be 1.300 kJ mol^{1}
at 298 K. Approximately how many molecules have at least five times
the average molecular kinetic energy.


Real
Gas Problems (Answers) 

Preamble:

Assumptions underlying the concept
of an ideal gas are:
(i) collisions involving
gas phase molecules are perfectly elastic
(ii) the volume of the molecules is insignificant compared with
the volume of their container.
However, for real gases
attractive forces exist between molecules when they collide, giving
rise to "sticky collisions", so that at any instant there
are actually fewer separate particles present than the number of
individual molecules that the ideal gas model assumes. Fewer separate
particles results in less collisions than would otherwise occur
if each molecule acted individually, so therefore lower pressure
is observed than that expected for an ideal gas. Provided that the
temperature is sufficiently high to allow the colliding molecules
to quickly separate again, (i) is reasonably valid for real gases.
However, if the temperature is not sufficiently higher than that
at which the attractive forces allow colliding molecules to stick
together in large enough clumps to coalesce and convert to the liquid
phase, the pressure observed for a real gas will be significantly
less than that expected for an ideal gas.
Provided the pressure of the gas is not high, assumption (ii) is
also reasonably valid but at high pressures, the total actual volume
of the molecules does become significant and the void space not
occupied by the gas is significantly less than the volume, V, of
their container.
The Van der Waals equation
attempts to modify the ideal gas equation to take account of these
two factors as follows:
To increase the observed
pressure P by an amount that compensates for sticky collision, the
term an^{2}/V^{2} is added to P. The factor "a"
is a constant for a given gas and it is a measure of how strong
the attractive forces between the molecules are. Larger "a"
constants reflect stronger attractions between molecules. The n^{2}/V^{2}
component could be regarded as the "density" of collisions
as n/V is the number of moles of gas present per unit volume and
this is squared because each molecule can be both the target and
the missile in each collision.
To reduce the observed volume V to the actual void space in the
container, the term nb is subtracted from V. The constant "b"
is the actual volume of a mole of molecules, larger "b"
values are associated with larger molecules.
These corrections when applied to the ideal gas equation give the
Van der Waals equation for real gas behaviour.
(P + an^{2}/V^{2})(V  nb) = nRT.


1

(Note preamble above.)
At high pressures, it is convenient to quote pressure in atmospheres
rather than kPa. The value of the gas constant R depends on the
units used for P and V.
If P is expressed as atmospheres and V as litres,
R = 0.0821 L atm K^{1} mol^{1}.
(i) Using the Ideal
Gas Equation
V = 0.536 L
n = 1.00 mol
T = 373 K
PV = nRT
P = nRT/V = 1.00 x 0.0821 x 373/0.536
= 57.1 atm
Actual pressure = 50.0
atm
% deviation = (7.1/50) x 100 = 14.2 %
(ii) Using Van der Waals
equation.
(P + an^{2}/V^{2})(V  nb) = nRT.
(P + 3.61 x (1.00/0.536)^{2})(0.536  1.00 x 0.0428) = 1.00
x 0.0821 x 373)
(P + 12.57)(0.493) = 30.62
P + 12.57 = 62.12
P = 49.6 atm
% deviation = (0.4/50) x 100 = 0.8%


2 
(i) N_{2}
The value of the Van der Waals constant, a, of a given gaseous substance
depends on the strength of attractions between its component molecules.
Molecules experiencing the weakest attractive forces will have the
smallest a constant while those with the strongest attractive forces
will have the largest a values.
Of the molecules NH_{3},
N_{2}, CH_{2}Cl_{2}, Cl_{2}, CCl_{4},
the two elements nitrogen and chlorine alone have nonpolar bonds
between their component atoms. This results is weaker attractions
between molecules, so N_{2} and Cl_{2} will have
the smallest a values. The remaining molecules all have polar bonds
between their component atoms and in the case of NH_{3},
hydrogen bonds between molecules. These are factors that lead to
increased a constants. As the N_{2} molecule is smaller
than the Cl_{2} molecule and therefore has less electrons
present, the attractive forces between N_{2} molecules will
be weaker than those between Cl_{2} molecules and N_{2}
will have the smallest a constant.
(ii) CCl_{4}
The value of the Van der Waals b constant is merely the actual volume
of a mole of the molecules and this can be deduced by comparing
the volumes of the molecules in the list. Small molecular volume
results in small b values and a large molecular volume corresponds
to a large b constant. From the compound in the list, CCl_{4} is the
largest and so it will have the greatest b constant.


3 
(i) Using the Van der
Waals equation:
Substituting in
(P + an^{2}/V^{2})(V  nb) = nRT,
P = 120 atm
n = 20.0 mol
V = 10.0 L
(120 + 0.0341 x (20.0/10.0)2)(10.0
 20.0 x 0.0237) = 20.0 x 0.0821 x T
[Note the value of R = 0.0821 because P is in atm and V in L]
(120 + 0.1364)(10.0  0.5) = 1.64 x T
T = 696 K
[Note that the correction to P is not significant as T is well above
the temperature at which helium gas will liquefy while the volume
correction is significant due to the high pressure.]
(ii) Using the Ideal
Gas Equation:
Substituting in
PV = nRT,
120 x 10.0 = 20.0 x 0.0821 x T
T = 1200/1.642
= 731 K


Preamble: 
For any collection of
N molecules, the number N_{E} having an energy E or greater
is given by
the expression
N_{E} = 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, N_{A},
on a molar basis this then becomes
n_{E} = 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,
n_{E} = 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 N_{A}
= 4.05 x 10^{2} x 6.02 x 10^{23}
= 2.4 x 10^{22} 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 N_{A}
= 9.02 x 10^{2} x 6.02 x 10^{23}
= 5.4 x 10^{22} molecules.


5 
The expression in terms of moles for the distribution of molecular
energies,
n_{E} = ne^{E/RT},
can be rewritten to give the fraction of the total moles, (n), that
have energy E or greater, (n_{E}), as
n_{E}/n = e^{E/RT}
At T = 250 K and for E = 6.500 kJ mol^{1} = 6500 J mol^{1},
this becomes
n_{E}/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
n_{E} = 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 n_{E} =
1.00 x e^{6500/(8.314 x 298)}
= 0.0725 mol
Number of molecules with this energy or greater
= 0.0725 x N_{A}
= 0.0725 x 6.022 x 10^{23}
= 4.4 x 10^{22}

