In an adiabatic process, the heat absorbed is zero, that is, dq = 0
Therefore, from the equation dU = dq + dw, we obtain, dU = 0 + dw = dw
However, for one mole of an ideal gas, dU is represented by dU = CVdT = nCV‾dT as,
dU = CV‾dT
Throughout expansion, dw and therefore dU are negative. That is, as the system does expansion work, its internal energy reduces. This, again, according to equation dU = CVdT = nCV‾dT signifies that dT is negative; that is, temperature reduces. In another words throughout adiabatic expansion, temperature of the system reduces. This principle is employed in Claude's method of liquefaction of gases.
Temperature-Volume Relationship in a Reversible Adiabatic Process:
From the above, we know that dU = dw
Replacing for dw and dU from the equations dw = - pdV and dU = CVdT = nCV‾dT, we get for one mole of an ideal gas,
CV‾dT = - pdV
For one mole of the ideal gas,
P = RT/V
By employing this relationship in dU = CV‾dT, we obtain:
CV‾dT = - RTdV/V
On rearranging we obtain,
CV‾ (dT/T) = -R (dV/V)
On Integrating the equation dU = CV‾dT between the temperature limits T1 and T2 and volume limits V1 and V2, we obtain:
CV‾ T1∫T2 (dT/T) = - R V1∫V2 (dT/T)
CV ln (dT/T1) = - R ln (V2/V1) = R ln (V1/V2)
= (Cp‾ - Cp‾) ln (V1/V2)
ln (T2/T1) = [(Cp‾/Cp‾) - 1] ln (V1/V2)
= (γ - 1) ln (V1/V2)
Here 'γ' is the ratio of the molar heat capacities, Cp‾/CV‾
On rearranging the above equation, we obtain,
ln (T2/T1) = ln (V1/V2)γ-1
Taking antilogarithms on both sides, we get:
T2/T1 = ln (V1/V2)γ-1
Or, (T2V2)γ-1 = (T1V1)γ-1 or (TV)γ-1 = constant
This equation provides the volume-temperature relationship in the reversible adiabatic process.
As well we can get pressure temperature relationship by knowing that, for an ideal 'g'
P1V1/T1 = P2V2/T2
V1/V2 = P2T1/P1T2
Replacing this in the equation T2/T1 = ln (V1/V2)γ-1, we obtain:
T2/T1 = (P2T1/P1T2)γ-1
T2(P2T2)γ-1 = T1(P2T1)
That is, T2γ P1γ-1 + T1γ P2γ-1
Or (T2/T1)γ = (P2/P1)γ-1
For any reversible adiabatic expansion, T2 can be found out by using the equation T2/T1 = ln (V1/V2)γ-1 or (T2/T1)γ = (P2/P1)γ-1
As well it is possible to obtain pressure-volume relationship in a reversible adiabatic process by employing the rearranged form of equation V1/V2 = P2T1/P1T2
T2/T1 = P2V2/P1V1
In equation T2/T1 = ln (V1/V2)γ-1
On cross multiplying the terms, we get:
P2V2V2γ-1 = P1V1V1
Or P1V1γ = P2V2
Or, PVγ = constant
The above equation explains pressure-volume relationship for an ideal gas undergoing the reversible adiabatic expansion or compression.
ΔU and W in a Reversible Adiabatic Process:
The quantities dU and dw for an adiabatic process are associated via dU = dw. By employing this equation and dU = CVdT = nCV‾dT get for 1 mol of the ideal gas,
dU = dw = CV‾dT
In case of n mol of the ideal gas,
dU = dw = nCVdT
The work done on the gas throughout an adiabatic expansion (W) is as well the change in the internal energy (ΔU) can be computed by integrating the equation dU = dw = nCVdT in the temperature limit of T1 and T2
ΔU = W = nCv‾ = T1∫T2 dT
= nCv‾ (T2 - T1) = n CV‾ΔT
Therefore, ΔU and W can be computed if n, CV, T1, and T2 are known.
Irreversible Adiabatic Expansion:
If the work is done adiabatically and irreversibly, then the work done on the system is represented by T2/T1 = ln (V1/V2)γ-1
W = - pext ΔV
As in the case of irreversible isothermal process
We can arrive at the temperature-volume relationship for the adiabatic irreversible process as shown:
By using the equation W = - pext ΔV in equation nCV‾ (T2 - T1) = n CV‾ΔT
- pext ΔV = n CV‾ΔT
Therefore, ΔV = (T2 - T1) = - pext ΔV/n CV‾
The equation above is helpful in computing the final temperature of the ideal gas undergoing adiabatic irreversible expansion as equation T2/T1 = ln (V1/V2)γ-1 or (T2/T1)γ = (P2/P1)γ-1 is of help in the adiabatic reversible process.
Our main concentration so far is centered on ideal gases. It was illustrated earlier that the internal energy of an ideal gas is independent of volume or pressure. This, though, is not true for real gases as intermolecular forces exist among their molecules. Therefore whenever a real gas is expanded, work has to be done in overcoming such forces. If no energy is supplied from the external source, then the internal energy of the gas is employed up in doing this work. This yields in a fall in the temperature of the gas. Though, some gases exhibit rise in temperature. This phenomenon of change in temperature whenever a gas is made to expand adiabatically from a high pressure area to a low pressure area is termed as the Joule-Thomson effect. The phenomenon can be comprehended if we consider the apparatus illustrated in the figure shown below. It comprises of an insulated tube fitted by a porous plug and two airtight pistons one on either side of the plug. The gas is kept under pressure p1 and p2 in the two compartments. It will be noted that p1 is greater than p2. The left hand side piston is then slowly pushed inwards in such a way that, without changing the value of p1, volume V1 of gas is introduced via the plug into the other compartment. This outcome in the outward movement of the other piston and as well in the volume increase. Assume that the final volume be V2. Accurate temperature measurements are made in both the compartments.
Fig: Joule-Thomson Experiment
The total work done on the system is represented by:
W = - (p2V2 - p1V1) = p1V1 - p2V2
It must be remembered that p2V2 is the work done by the system and p1V1 the work on it. The conditions are adiabatic and therefore q = 0. For finite process, the equation dw = - p dV and W = - p1V1 - p2V2 can be combined and represented as:
ΔU = W = (p1V1 - p2V2)
Or ΔU + (p2V2 - p1V1) = 0
By using the equation ΔH = ΔU + (p2V2 - plVl)
From the equation ΔH = ΔU + (p2V2 - plVl) and ΔU + (p2V2 - p1V1) = 0 we notice that,
ΔH = 0
Therefore in the Joule-Thompson experiment, ΔH = 0 or enthalpy is constant.
As, in the Joule-Thomson experiment, we evaluate the temperature change with change in pressure at constant enthalpy, we state Joule-Thomson coefficient, μJT, as
μJT = (∂T/∂P)
Whenever μJT is positive, the expansion causes cooling and if μJT is negative, the expansion causes heating. However, if μJT is equivalent to zero, there is neither cooling nor heating because of Joule-Thomson expansion. The temperature at which μJT = 0 is termed as the inversion temperature (Ti) of the gas. If the gas is expanded above its inversion temperature, it is heated; if it is expanded beneath its inversion temperature, it is cooled. In order to reduce the temperature of a gas and then to liquefy by Joule-Thomson process, it is necessary to bring its temperature beneath its inversion temperature.
Inversion temperature of the hydrogen gas is much below room temperature.
Thus, it is hazardous to open a compressed hydrogen gas cylinder under atmospheric conditions. As hydrogen gas is discharged from the cylinder, it expands, gets headed and as well combines by oxygen present in the air; the latter reaction causes the explosion.
For a particular reaction, ΔrH and ΔrU usually differ with temperature. It is of great significance to study such variations quantitatively in such a way that these might be computed for any temperature from the known values of ΔrH and ΔrU at any other temperature. The variation of ΔrH and ΔrU by temperature is illustrated by Kirchhoff's equation. Let us now derive this equation.
When Cp is the heat capacity of a substance, then for a temperature increase dT, the increase in enthalpy is represented by equation dH = CpdT
In case of enthalpy of a reaction (ΔrH), we can rewrite:
d(ΔrH) = ΔCpdT
Here, ΔCp = (sum of Cp values of products) - (sum of Cp values of reactants)
As well, d(ΔrH) is the change in enthalpy of reaction due to the reason of change in temperature, dT.
On integration d(ΔrH) = ΔCpdT gives,
ΔrH2 - ΔrH1 = T1∫T2 ΔCpdT
Here, ΔrH2 and ΔrH1 are the enthalpies of reaction at temperatures T2 and T1, correspondingly. The equation above is known as the Kirchhoff's equation. Likewise we can as well get the expression,
ΔrU2 - ΔrU1 = T1∫T2 ΔCVdT
Here, ΔrU1 and ΔrU2 are the changes in internal energy of the reaction at temperatures T1 and T2, and ΔCV is the difference in heat capacities between the products and reactants at constant volume. Let us now take three of the special cases of the equation ΔrH2 - ΔrH1 = T1∫T2 ΔCpdT
1) If ΔCP = 0, then ΔrH2 - ΔrH1 implying thus that the enthalpy of reaction doesn't change.
2) If ΔCP is constant that is, it doesn't differ with temperature, then
ΔrH2 = ΔrH1 + ΔCp (T2 - T1)
That is, ΔrH either deceases or increases regularly by temperature. For most of the reactions, above equation is valid for small range of temperatures.
3) If ΔCp changes by temperature, then equation ΔrH2 - ΔrH1 = T1∫T2 ΔCpdT has to be integrated via expressing Cp as a function of temperature. The variation in Cp is generally expressed in the following manner:
Cp‾ = a + bT + cT2 +.... the coefficients a, b, c ... and so on, are the characteristic of a particular substance.
Bond Enthalpies and Estimation of Enthalpies of formation:
Bond enthalpy is a helpful theory in Thermochemistry. It finds out application in the computation of standard enthalpy of formation and standard enthalpy of the reaction of most of the compounds.
In a molecule, atoms are linked via chemical bonds. Whenever a molecule decomposes into atoms, the bonds are broken and the enthalpy rises. This is as well stated as the enthalpy of atomization, ΔHatom, and is for all time positive. For illustration: the enthalpy of the given reaction is the enthalpy of atomization of the ethane gas:
C2H6 (g) → 2C (g) + 6H (g) ΔHatom = - 2828.38 KJ mol-1
On the analysis of ΔHatom for a large number of such reactions, it has been found out that specific values of bond enthalpies might be assigned to various kinds of bonds (Table shown below). Such bond enthalpies correspond to the decomposition of the molecule in the gaseous state to atoms in the gaseous state. Some substances in the solid state whenever sublimed are converted into the gaseous atoms. Therefore, graphite whenever heated is converted into gaseous atoms, and the heat needed for one mole can be known as the molar enthalpy of atomization of graphite which is equivalent to. 717 kJ mol-1. If graphite is taken as the reference state for carbon, then the atomization can be represented as follows:
C(graphite) → C(g) ΔHatom = 717 KJ mol-1
Table: Bond Enthalpies
a - in alkenes
b - in aromatic compounds
Enthalpies of the atomization of, some more elements that become atomized on sublimation are given:
Substance ΔHoatom/KJ mol-1
C (graphite) 717
Na (s) 108
K (s) 90
Cu (s) 339
It must be made clear that the bond enthalpy is not bond dissociation energy. This could be comprehended if we consider bond dissociation energy of water:
H2O (g) → OH (g) + H (g) D1 = 501.9 kJ mol-1
OH (g) → O (g) + H (g) D2 = 423.4 kJ mol-1
The quantities D1 and D2 are the first and second bond dissociation energies and are different from the bond enthalpy given for O-H in the table of Bond Enthalpies. Again, bond enthalpy is some type of average of a large amount of experimental data. These are of immense value in predicting the standard enthalpy of formation of a large number of Compounds being synthesized and as well for estimating the standard enthalpy of reactions comprising these latest molecules.
The given steps will assist you in the computation of standard enthalpy of formation from the bond enthalpies and enthalpies of atomization of elements:
1) Primarily, write the stoichiometric equation; then write (the most acceptable) Lewis structure of each of the reactants and product.
2) Make use of bond enthalpies from the table of Bond Enthalpies and enthalpies of atomization from the second table to compute the heat needed to break all the bonds in the reactants and the heat released whenever the atoms form the product. The bond enthalpy of X-X bond can be represented as B(X-X) in arithmetic expressions.
3) The standard enthalpy of formation
= (Heat needed to break all the bonds in reactants) - (heat discharged if the atoms forth the product)
4) Bond enthalpy value can be applied to compounds only if these are in the gaseous state; if the compounds are in solid or liquid state, molar enthalpies of sublimation or vaporization as well should be considered.
Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)
Expand your confidence, grow study skills and improve your grades.
Since 2009, Tutorsglobe has proactively helped millions of students to get better grades in school, college or university and score well in competitive tests with live, one-on-one online tutoring.
Using an advanced developed tutoring system providing little or no wait time, the students are connected on-demand with a tutor at www.tutorsglobe.com. Students work one-on-one, in real-time with a tutor, communicating and studying using a virtual whiteboard technology. Scientific and mathematical notation, symbols, geometric figures, graphing and freehand drawing can be rendered quickly and easily in the advanced whiteboard.
Free to know our price and packages for online Chemistry tutoring. Chat with us or submit request at email@example.com
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!