- +44 141 628 6080
- [email protected]

** Adiabatic Expansion**:

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 = C_{V}dT = nC_{V}‾dT as,

dU = C_{V}‾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 = C_{V}dT = nC_{V}‾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 = C_{V}dT = nC_{V}‾dT, we get for one mole of an ideal gas,

C_{V}‾dT = - pdV

For one mole of the ideal gas,

P = RT/V

By employing this relationship in dU = C_{V}‾dT, we obtain:

C_{V}‾dT = - RTdV/V

On rearranging we obtain,

C_{V}‾ (dT/T) = -R (dV/V)

On Integrating the equation dU = C_{V}‾dT between the temperature limits T_{1} and T_{2} and volume limits V_{1 }and V_{2}, we obtain:

C_{V}‾ _{T1}∫^{T2 }(dT/T) = - R _{V1}∫^{V2} (dT/T)

C_{V} ln (dT/T_{1}) = - R ln (V_{2}/V_{1}) = R ln (V_{1}/V_{2})

= (C_{p}‾ - C_{p}‾) ln (V_{1}/V_{2})

ln (T_{2}/T_{1}) = [(C_{p}‾/C_{p}‾) - 1] ln (V_{1}/V_{2})

= (γ - 1) ln (V_{1}/V_{2})

Here 'γ' is the ratio of the molar heat capacities, C_{p}‾/C_{V}‾

On rearranging the above equation, we obtain,

ln (T_{2}/T_{1}) = ln (V_{1}/V_{2})^{γ-1}

Taking antilogarithms on both sides, we get:

T_{2}/T_{1} = ln (V_{1}/V_{2})^{γ-1}

Or, (T_{2}V_{2})^{γ-1} = (T_{1}V_{1})^{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'

P_{1}V_{1}/T_{1} = P_{2}V_{2}/T_{2}

V_{1}/V_{2} = P_{2}T_{1}/P_{1}T_{2}

Replacing this in the equation T_{2}/T_{1} = ln (V_{1}/V_{2})^{γ-1}, we obtain:

T_{2}/T_{1} = (P_{2}T_{1}/P_{1}T_{2})^{γ-1}

T_{2}(P_{2}T_{2})^{γ-1} = T_{1}(P_{2}T_{1})

That is, T_{2}^{γ }P_{1}^{γ-1 }+ T_{1}^{γ} P_{2}^{γ-1}

Or (T_{2}/T_{1})^{γ} = (P_{2}/P_{1})^{γ-1}

For any reversible adiabatic expansion, T_{2 }can be found out by using the equation T_{2}/T_{1} = ln (V_{1}/V_{2})^{γ-1} or (T_{2}/T_{1})^{γ }= (P_{2}/P_{1})^{γ-1}

As well it is possible to obtain pressure-volume relationship in a reversible adiabatic process by employing the rearranged form of equation V_{1}/V_{2} = P_{2}T_{1}/P_{1}T_{2}

T_{2}/T_{1} = P_{2}V_{2}/P_{1}V_{1}

In equation T_{2}/T_{1} = ln (V_{1}/V_{2})^{γ-1}

On cross multiplying the terms, we get:

P_{2}V_{2}V_{2}^{γ-1 }= P_{1}V_{1}V_{1}

Or P_{1}V_{1}^{γ }= P_{2}V_{2}

Or, PV^{γ }= 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 = C_{V}dT = nC_{V}‾dT get for 1 mol of the ideal gas,

dU = dw = C_{V}‾dT

In case of n mol of the ideal gas,

dU = dw = nC_{V}dT

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 = nC_{V}dT in the temperature limit of T_{1} and T_{2}

ΔU = W = nCv‾ = _{T1}∫^{T2} dT

= nCv‾ (T_{2} - T_{1}) = n C_{V}‾ΔT

Therefore, ΔU and W can be computed if n, C_{V}, T_{1}, and T_{2} are known.

Irreversible Adiabatic Expansion:

If the work is done adiabatically and irreversibly, then the work done on the system is represented by T_{2}/T_{1} = ln (V_{1}/V_{2})^{γ-1}

W = - p_{ext} 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 = - p_{ext }ΔV in equation nC_{V}‾ (T_{2} - T_{1}) = n C_{V}‾ΔT

- p_{ext} ΔV = n C_{V}‾ΔT

Therefore, ΔV = (T_{2} - T_{1}) = - p_{ext} ΔV/n C_{V}‾

The equation above is helpful in computing the final temperature of the ideal gas undergoing adiabatic irreversible expansion as equation T_{2}/T_{1} = ln (V_{1}/V_{2})^{γ-1 }or (T_{2}/T_{1})^{γ} = (P_{2}/P_{1})^{γ-1} is of help in the adiabatic reversible process.

** Joule-Thomson Effect**:

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 p_{1} and p_{2 }in the two compartments. It will be noted that p_{1} is greater than p_{2}. The left hand side piston is then slowly pushed inwards in such a way that, without changing the value of p_{1}, volume V_{1} 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 V_{2}. Accurate temperature measurements are made in both the compartments.

Fig: Joule-Thomson Experiment

The total work done on the system is represented by:

W = - (p_{2}V_{2} - p_{1}V_{1}) = p_{1}V_{1} - p_{2}V_{2 }

It must be remembered that p_{2}V_{2} is the work done by the system and p_{1}V_{1} the work on it. The conditions are adiabatic and therefore q = 0. For finite process, the equation dw = - p dV and W = - p_{1}V_{1} - p_{2}V_{2} can be combined and represented as:

ΔU = W = (p_{1}V_{1 }- p_{2}V_{2})

Or ΔU + (p_{2}V_{2} - p_{1}V_{1}) = 0

By using the equation ΔH = ΔU + (p_{2}V_{2} - p_{l}V_{l})

From the equation ΔH = ΔU + (p_{2}V_{2} - p_{l}V_{l}) and ΔU + (p_{2}V_{2} - p_{1}V_{1}) = 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.

** Kirchhoff's equation**:

For a particular reaction, Δ_{r}H and Δ_{r}U 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 Δ_{r}H and Δ_{r}U at any other temperature. The variation of Δ_{r}H and Δ_{r}U by temperature is illustrated by Kirchhoff's equation. Let us now derive this equation.

When C_{p} is the heat capacity of a substance, then for a temperature increase dT, the increase in enthalpy is represented by equation dH = C_{p}dT

In case of enthalpy of a reaction (Δ_{r}H), we can rewrite:

d(Δ_{r}H) = ΔC_{p}dT

Here, ΔC_{p }= (sum of C_{p }values of products) - (sum of C_{p} values of reactants)

As well, d(Δ_{r}H) is the change in enthalpy of reaction due to the reason of change in temperature, dT.

On integration d(Δ_{r}H) = ΔC_{p}dT gives,

Δ_{r}H_{2} - Δ_{r}H_{1} =_{ T1}∫^{T2 }ΔC_{p}dT

Here, Δ_{r}H_{2} and Δ_{r}H_{1} are the enthalpies of reaction at temperatures T_{2} and T_{1}, correspondingly. The equation above is known as the Kirchhoff's equation. Likewise we can as well get the expression,

Δ_{r}U_{2} - Δ_{r}U_{1} = _{T1}∫^{T2 }ΔC_{V}dT

Here, Δ_{r}U_{1} and Δ_{r}U_{2 }are the changes in internal energy of the reaction at temperatures T_{1} and T_{2}, and ΔC_{V} 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 Δ_{r}H_{2} - Δ_{r}H_{1} =_{ T1}∫^{T2 }ΔC_{p}dT

1) If ΔC_{P} = 0, then Δ_{r}H_{2 }- Δ_{r}H_{1} implying thus that the enthalpy of reaction doesn't change.

2) If ΔC_{P} is constant that is, it doesn't differ with temperature, then

Δ_{r}H_{2} = Δ_{r}H_{1 }+ ΔC_{p} (T_{2} - T_{1})

That is, Δ_{r}H either deceases or increases regularly by temperature. For most of the reactions, above equation is valid for small range of temperatures.

3) If ΔC_{p} changes by temperature, then equation Δ_{r}H_{2} - Δ_{r}H_{1} =_{ T1}∫^{T2 }ΔC_{p}dT has to be integrated via expressing C_{p} as a function of temperature. The variation in C_{p} is generally expressed in the following manner:

C_{p}‾ = a + bT + cT^{2} +.... 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, ΔH_{atom}, and is for all time positive. For illustration: the enthalpy of the given reaction is the enthalpy of atomization of the ethane gas:

C_{2}H_{6 }(g) → 2C (g) + 6H (g) ΔH_{atom} = - 2828.38 KJ mol^{-1}

On the analysis of ΔH_{atom} 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)} ΔH_{atom} = 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 ΔH^{o}atom/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:

H_{2}O (g) → OH (g) + H (g) D_{1} = 501.9 kJ mol^{-1}

OH (g) → O (g) + H (g) D2 = 423.4 kJ mol^{-1}

The quantities D_{1 }and D_{2} 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 protected]

1940177

Questions

Asked

3689

Tutors

1483370

Questions

Answered

Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!

Submit AssignmentÂ©TutorsGlobe All rights reserved 2022-2023.

## Mechanism of photosynthesis

tutorsglobe.com mechanism of photosynthesis assignment help-homework help by online photosynthesis tutors