We are already familiar with the van der Waals expression for a single mole of gas. Now, we will need to use the partial derivatives section of Appendix C in order to generate the corresponding relationship.
Starting with the van der Waals Equation for 1 mole of gas:
( )
The total differential can be written from Appendix C as,
m
Since we have already found the total differential for the pressure, we can determine the values of these two partial derivatives.
( )
( )
2 3Now we can substitute these values into the total derivative to reproduce the derived equation given above.
( ) ( )
( ) ( )
2 3
2 3
2 and
Substitution of these into
gives,
For the first two segments of the expression above, the substitution will be as follows;
( )
The last segment is slightly tricky because we need to separate the equation again and this will create four segments rather than the original
Now everything can be put together to get:
2
2.66. If a substance is burned at constant volume with no heat loss, so that the heat evolved is all used to heat the product gases, the temperature attained is known as the adiabatic flame temperature. Calculate this quantity for methane burned at 25 °C in the amount of oxygen required to give complete combustion to CO2 and H2O. Use the data in Appendix D and the following approximate expressions for the heat capacities:
CP,m (CO2)/J K–1 mol–1 = 44.22 + 8.79 × 10–3 T/K CP,m (H2O)/J K–1 mol–1 = 30.54 + 1.03 × 10–2 T/K
Solution:
Given: Appendix D, CP,m (CO2)/J K–1 mol–1 = 44.22 + 8.79 × 10–3 T/K CP,m (H2O)/J K–1 mol–1 = 30.54 + 1.03 × 10–2 T/K Required: Tadiabatic flame temperature
With any problem of this type, it is always important to begin by writing down all of the reactions that will be useful. The balanced reaction for the complete combustion of methane gas is as follows:
4 2 2 2
CH (g)+2O (g)→CO (g)+2H O(g)
Remember from previous problems that we can determine the standard enthalpy change by using the enthalpies of formation for all species involved.
O
2 2 4
(products) (reactants)
(CO , g) 2 (H O, g) (CH , g)
f f
f f f
H H H
H H H H
∆ ° = ∆ ° − ∆ °
∆ ° = ∆ + ∆ ° − ∆ °
∑ ∑
By using the enthalpies of formation found in Appendix D, we can calculate this value.
(
1) (
1) (
1)
1
393.51 kJ mol 2 241.826 kJ mol 74.6 kJ mol 802.562 kJ mol
H H
− − −
−
∆ ° = − + × − − −
∆ ° = −
We have been given the expressions for the heat capacities for both carbon dioxide and gaseous water so we can combine them in order to get the total heat capacity for the products.
( ) ( )
Since we are working under constant volume, we will need to determine the appropriate values for the heat capacity. Recall that:
, ,
We can use the expression for the heat absorbed in order to determine the final temperature. Remember that under these conditions, the heat absorbed by the gas will be equal to the standard internal energy change for the reaction. Since we are using the heat capacity given in terms of constant pressure conditions, we will be making the appropriate arrangements to involve the value we solved for initially.
( )
802 562 96.9855 298.15 1.4695 10 88 893.422 5
K K
802 562 96.9855 28 916.226 83 1.4695 10 1306.288 844
1.4695 10 96.9855 832 784.
T
This forms a quadratic equation which can then be solved to determine the adiabatic flame temperature for methane. Let each term
96.9855 96.9855 4 1.4695 10 832 784.515 7
K 2 1.4695 10
96.9855 58 357.261 04
K 2.939 10
Using the positive value for the square root we obtain,
2
2
2
96.9855 58 357.261 04 2.939 10
The value as calculated may vary by about 20 K lower depending upon how many significant figures we used in the calculation. This value may be reduced by about 170 K under constant pressure condition.
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2.67. Two moles of a gas are compressed isothermally and reversibly, at 300 K, from an initial volume of 10 dm3 to a final volume of 1 dm3. If the equation of state of the gas is P(Vm – b) = RT, with b = 0.04 dm3 mol–1, calculate the work done on the system, ∆U, and ∆H.
Solution:
Given: n=2 mol, T =300 K, Vi =10 dm , 3 Vf =1 dm3
(
m–)
, with 0.04 dm mol3 –1 P V b =RT b=Required: w, , ∆U ∆H
We have seen many problems like this one already. Remember that under isothermal conditions, there is no change in temperature. It is also important to note that the process is reversible. Lastly, we are working with a Real Gas which means that we cannot use the Ideal Gas Law.
Since the gas is undergoing a compression, the reversible work done on the system is given by:
( )
The change in internal energy for a Real Gas is given by Eq. 2.124 and Eq. 2.125:
2
Since the constant a is not involved in the equation of state for this unknown real gas, there will be no change in the internal energy throughout the reaction.
0
∆ =U
Eq. 2.127 will then enable us to determine the change in enthalpy.
( )
H U PV
∆ = ∆ + ∆
We have not been given the pressures corresponding to the system so they must be calculated using the equation of state.
( )
5.029 233 871 bar 2.0 mol
54.228 260 87 bar Pf =
We can now find the value for the P-V work done and subsequently the change in enthalpy for the reaction.
(
5.029 233 871 bar 10.0 dm) (
3)
PVi i =
5 5 3 54.228 260 87 bar dm 54.228 260 87 bar
( ) 5422.826 087 5029.233 871 J ( ) 393.592 216 J
The application of the correct number of significant figures gives, 4 10 J2
∆ = ×H
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2.68. Three moles of a gas are compressed isothermally and reversibly, at 300 K, from an initial volume of 20 dm3 to a final volume of 1 dm3. If the equation of state of the gas is
2
2 m
m
P n a V nRT V
+ =
with a = 0.55 Pa m6 mol–1, calculate the work done, ∆U, and ∆H.
Solution:
Given:
2
3 3
3 mol, 300 K, i 20 dm , f 1 dm , 2 m m
n T V V P n a V nRT
V
= = = = + =
Required: w, , ∆U ∆H
We will solve this problem in the exact same way as problem 2.67 was done. Since the two problems are extremely similar, no additional explanation will be provided.
rev
22 417.21439 J 4702.5 J 17 714.714 39 J
2
3.0 mol
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2.69. One mole of a van der Waals gas at 300 K is compressed isothermally and reversibly from 60 dm3 to 20 dm3. If the constants in the van der Waals equation are
a = 0.556 Pa m6 mol–1 and b = 0.064 dm3 mol–1 calculate wrev, ∆U, and ∆H.
Solution:
Given: van der Waals gas: n=1 mol, T =300 K, Vi =60 dm , 3 Vf =20 dm3 a=0.556 Pa m mol , 6 –1 b = 0.064 dm mol3 –1 Required: wrev, , ∆U ∆ H
The reversible work for a van der Waals gas is given by Eq. 2.122 and Eq. 2.123:
( )
using Van der Waals Eq:
ln 1
Recall that from Eq. 2.124 and Eq. 2.125 we can solve for the change in internal energy:
2
2 1 1
41 459.669 13 Pa 1.0 mol
123 727.877 2 Pa Pf =
( ) ( )
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2.70. Show that the Joule-Thomson coefficient μ can be written as:
calculate ∆H for the isothermal compression of 1.00 mol of the gas at 300 K from 1 bar to 100 bar.
Solution:
Given: n=1.00 mol, T =300 K, Pi =1 bar, Pf =100 bar Required: µ, H∆
From Eqs. 2.108 and 2.110 we can see that;
H
T T
P P
µ =∂∂ ≈∆∆
This value is equal to zero for an Ideal Gas but it may be either positive or negative for a Real Gas. When there is expansion taking place, the change in pressure will be negative (cooling expansion: here the change in temperature is also negative allowing the Joule-Thomson coefficient to be positive).
Conversely, a negative µ corresponds to a rise in temperature upon expansion. This is interesting to note because most gases under regular temperatures will cool when they are able to expand. Since the Joule-Thomson expansion occurs at constant enthalpy, the total differential will be:
It follows that:
,
We have seen previously that for an ideal gas, the Joule-Thomson coefficient is equal to zero. This also indicates that the enthalpy is independent of pressure. For real gases, we will see that the enthalpy shows some variation with pressure.
Since by using the expression for the total differential. Take note that in Chapter 1 there is a table (Table 1.5) which provides the Van der Waals constants for many gases. For simplicity, we will assume that we are working with hydrogen gas. (This is a good choice since μ for H2 is positive.)
6 2
Assuming to be independent of temperature, 2