Processes at constant volume (isochoric) and at constant pressure (isobaric) are of interest because the first one gives a way of understanding how the change of internal energy $\Delta U$ can solely be related to the thermal energy or heat $Q$.
On the other hand, isobaric processes are of interest because these lead us to another important state function: the enthalpy $H$. This is key for thermochemical calculations and several operations of chemical engineering.
Another form of the first law of thermodynamics
From the first law of thermodynamics,
$\Delta U = Q + w$ Eq. (01)
it would mean very little because $V$, $P$ and $T$ are not explicitly shown. If we take the work $w$ in Eq. (01) as,
$w=-PV$ Eq. (02)
which would intrinsically mean that the work is done by the system on the surroundings or that an expansion of the system would only be considered. This last assumption makes sense since many chemical reactions tend to release a gas and pressurise the container of the reactants.
Thus, after substitution of $w$ given in Eq. (02) into Eq. (01) we arrive to,
$\Delta U = Q - PV$ Eq. (03)
Equation (03) is another form of the first la¿w of thermodynamics.
Isochoric processes (or at constant volume)
Since, for a given thermodynamical process, the volume $V$ of the system remains constant, its pressure $P$ and temperature $T$ are allowed to change. Now if the system were to be studied for very small changes (infinitesimal changes in fact), Eq. (03) could be expressed, using derivatives, as follows
$dU = DQ - P dV$ Eq. (04)
where $d$ is used to indicate an exact differential operator and $D$ to indicate an inexact differential operator.
You may notice that a term $V dP$ would be missing in Eq. (04) but it is because the work, as exposed here, is represented by changes in volume.
Now, for ischoric processes Eq. (04) reduces to,
$dU = \left( DQ \right)_V$ Eq. (05)
because $V$ is constant. The subscript $V$ as been introduced outside the parenthesis to indicate that $V$ is constant. Also, from the mathematical point of view Eq. (05) would be inconsistent because this would state that an exact differential equals an inexact differential. Then, it would be better rewrite Eq. (05) as,
$dU = \left( dQ \right)_V$ Eq. (06)
Now, integration of Eq. (06) lead us to,
$\Delta U = U_2 - U_1 = Q_V$ Eq. (06)
Equation (06) can be interpreted as follows. For a process, such as a chemical reaction occurring inside a reactor, at constant volume the change of internal energy $\Delta U$ is equal to the thermal energy supplied, as heating through a coil for example.
Isobaric process (or at constant pressure)
In this case, pressure $P$ is kept constant while volume $V$ and temperature $T$ are allowed to change. Again, a perspective from the first law of thermodynamics is required. We start with Eq. (04) and reorder it as follows,
$DQ = dU + P dV$ Eq. (04a)
and since $P$ is constant $DQ$ can be rewritten as an exact differential. Equation (4a) then becomes,
$\left( dQ \right)_P = dU + P dV$ Eq. (05)
Further integration of Eq. (05) from a state-1 to a state-2 would give,
$Q_P = \left( U_2 - U_1 \right) + P \left( V_2 - V_1\right)$ Eq. (06)
Next, Eq. (06) can be reordered so that state-1 and state-2 can be more easily identified. Thus,
$Q_P = \left( U_2 + P V_2 \right) - \left( U_1 + P V_1\right)$ Eq. (07)
where the terms $U+PV$ are known as the enthalpy $H$. Then, Eq. (07) is rewritten as,
$Q_P = H_2 - H_1$ Eq. (08)
The enthalpy can then be understood from the point of view supplied to the system due to a work at constant pressure. In other words, the enthalpy would be related to the energy required for a chemical rection to occur, for example. Also, $H$ is a state function and as a consequence $Q_P$ becomes a state function too.
You should notice that the relationship between $Q$ and $H$ given in Eq. (08) is only valid for $PV$ work and the ssumptions made here.
Any question? Write in the comments and I shall try to help.
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