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  Heat Change, Chemistry tutorial

Heat change under constant volume:

The reactions are carried out under constant volume or under constant pressure conditions. Let us now arrive at an expression helpful in computing the heat change in a system under the constant volume conditions.

Let us observe the equation dU = dq + dw. Let us suppose that the work done on the system is only pressure-volume work, while electrical, magnetic or other kinds of work are not involved. Then from the equation F = PextA and dw = - pdV

dU = dq - pdV

or dq = dU + pdV

If the procedure is carried out at constant volume, then

dV - pdV = 0

Therefore, dqv - dU

For the finite changes in internal energy the equation dqv - dU becomes,

qv = ΔU

That is, heat absorbed via a system at constant volume is precisely equivalent to its internal energy change.

Now try to correlate the internal energy change by the heat capacity at constant volume supposing that there is no phase change or chemical reaction. From the equation dqv - dU and qv = ΔU

dU = CVdT = nCV‾dT

This holds good for n mol of the ideal gas.

This can be represented as,

CV = (∂U/∂T)

That is, heat capacity at constant volume is equivalent to change in internal energy per 1 K increase in temperature at constant volume.

In order to get ΔU whenever an ideal gas is heated from temperature T1 to T2 at constant volume, the integrated form of dU = CVdT = nCV‾dT is to be employed.

That is, ΔU = T1∫T2 CVdT = T1∫T2 CV‾dT

Therefore, by knowing Cv‾ over the temperatures T1 to T2 it is possible to get the value of ΔU.

Enthalpy and Enthalpy changes:

The enthalpy of a system is stated by the relation,

H = U + pV

Here, U, p and V are the internal energy, pressure and volume of the system. As U, p and V are state variables, H as well is a state function. That is, the enthalpy of a system in a particular state is fully independent of the manner in which that state has been accomplished, If H1 and H2 are the enthalpies of the initial and final states of a system, then the enthalpy change accompanying the procedure is represented by,

ΔH = H2 - H1

      = (U2 + p2V2) - (U1 + p1V1)

      = ΔU + (p2V2 - p1V1)

In case of a constant pressure process (pl = p2 = p), the above equation can be represented as,

ΔH = ΔU + p (V2 - V1)

ΔH = ΔU + pΔV

Rewriting the equation above dq = dU + pdV for a finite change, we obtain

qp = ΔU + pΔV

Using the above equation in ΔH = ΔU + pΔV, qp = ΔH

The subscript p in qp represents the constant pressure condition.

In another words, the enthalpy change is equivalent to the heat absorbed by the system at constant pressure.

For a small change in enthalpy, we can write:

dqp = dH

Supposing that there is no phase change or chemical reaction we encompass:

dH = CpdT = nCp‾dT

In order to get ΔH value if an ideal gas is hated from temperature T1 to T2, at constant pressure, the integrated form of dH = CpdT = nCp‾dT is to be employed.

ΔH = T1∫T2 CVdT = T1∫T2 nCp‾dT

As most of the laboratory processes are taken out at constant pressure (that is, atmospheric pressure), the enthalpy change of a system is of great importance. It might be observed that as the absolute value of the internal energy of a system is not acknowledged, it is as well not possible to know the absolute enthalpy of the system. Luckily, for most of the processes we are just concerned by the changes in enthalpy which might be measured by taking any appropriate reference states of elements. 

Such process in which the heat is supplied to the system is termed as endothermic and, ΔH is given a positive sign; and exothermic processes (in which heat is evolved), ΔH is negative.

Enthalpy changes connected by certain typical methods are given special names. For illustration, enthalpy of vaporization or evaporation is the enthalpy change accompanying the conversion of one mole of a liquid to its vapor. Likewise, enthalpy of fusion and sublimation are the enthalpy changes accompanying fusion or sublimation of one mole of a substance. For a chemical reaction, the enthalpy of reaction is the difference in the enthalpies of the products and the reactants as per Stoichiometry given in the chemical equation.

Relation between Cp‾ and Cv‾ of an Ideal gas:

The internal energy of an ideal gas based only on its temperature and is independent of pressure or volume. This is quite understandable as in an ideal gas, there are no intermolecular interactions; no repulsive or attractive forces have to be overcome throughout expansion. Though, the enthalpy of the gas changes considerably whenever a gas contracts or expands.

For one mole of an ideal gas,

H = U + pV = U + RT

On differentiating we obtain,

dH = dU + RdT [As 'R' is a constant]

By using the equation dU = CVdT = nCV‾dT and dH = CpdT = nCP‾dT one mole of an ideal gas

Cp‾dT = Cv‾dT + RdT

Cp‾ = Cv‾ + R

And therefore, Cp‾ - Cv‾ = R

As well, for n mol, Cp‾ - Cv‾ = nR

This signifies that Cp‾ is for all time greater than Cv‾ for an ideal gas. This is due to the reason if the temperature of a gas is increased at a constant pressure, then there will be expansion of the gas. This will need some additional amount of heat (as compared to heating an ideal gas under constant volume conditions). Therefore, more heat will be needed in increasing the temperature- of the gas via 1 K under constant pressure conditions than under constant volume.

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