Initial Differential Form

An atmospheric parcel has (macroscopic) kinetic energy due to its motion and (macroscopic) potential energy due to its elevation above ground level. The parcel also has internal energy due to the translational motion of its molecular constituents (kinetic temperature) and internal motions as a result of the chemical bonds between the atoms of the molecules. These internal motions are of three types: rotation, bond stretching, and bond bending. These internal motions make up what is sometimes called "internal temperature" of the parcel.

The atmospheric parcel is considered to be a "closed system" in that the total amount of matter, in any phase or combination of phases, is constant. For our closed system (the parcel), let $q$ represent the amount of thermal energy (in Joules) it receives from the environment through the energy transfer processes of thermal conduction and/or thermal radiation. Let $w$ represent the work the parcel does on the environment. An example of "work done" on the environment occurs when the parcel expands and "pushes" its imaginary walls - the environment - outward.

The net energy received by the parcel is given by the difference of work in and work out

$\mbox{net energy}$ = $\mbox{work in}$ - $\mbox{work out}$ = $q - w$

If the parcel's kinetic and potential energies do not change and $q - w$ is not zero, then the internal energy of the parcel must change. Let $u_1$ and $u_2$ represent the initial and final values of internal energy, respectively. Then,

$u_2 - u_1 = \Delta u = q - w$

In terms of infinitesimal changes, the last equation may be written

$du = dq - dw$         $(1)$

which is one form (our initial form) of the First Law of Thermodynamics.

Constant pressure - $P$

In the case where pressure $P$ remains constant, the work done by the parcel on the environment may be expressed as

$dw = PdV$

$du = dq - PdV$         $(2a)$

For a one-kilogram mass, the volume $V$ is represented by $\alpha$, the specific volume of the parcel, so that $dV = d\alpha$, and the first law is given as

$du = dq - Pd\alpha$         $(2b)$

In terms of $C_v$ - Specific Heat at constant volume

The specific heat at constant volume of our atmospheric parcel (closed system) is defined as the ratio of energy in and change in temperature

$C_v = \left(\frac{\Delta q}{\Delta T}\right)_{V=constant} = \left(\frac{dq}{dT}\right)_{V=constant}$

From version (2a) of the first law, if $dV = 0$, then $dq = du$, and

$C_v = \left(\frac{du}{dT}\right)_{V=constant}$

For an ideal gas, Joule's law implies that the internal energy $u$ of our parcel depends only on the kinetic temperature and not on the internal temperature, so that

$C_v = \left(\frac{du}{dT}\right)$

regardless of constant volume. This gives the next version of the first law:

$du = dq - Pd\alpha$         $(2b)$
$\Rightarrow dq = du + Pd\alpha$
$\Rightarrow dq = du\cdot\frac{dT}{dT} + Pd\alpha$
$\Rightarrow dq = \left(\frac{du}{dT}\right)dT + Pd\alpha$
$\Rightarrow dq = C_v dT + Pd\alpha$         $(3)$

In terms of $C_p$ - Specific Heat at constant pressure

If energy is added to the parcel and the parcel expands in the constant pressure environment, the temperature rise of the parcel will be less than the case of constant volume because the $P d\alpha$ term in the first law equation (3) is positive. Define the specific heat of constant pressure as

$C_p = \left(\frac{dq}{dT}\right)_{P=constant}$

$d(P\alpha) = \alpha dP + P d\alpha$
$\Rightarrow Pd\alpha = d(P\alpha) - \alpha dP$

so that the first law in equation (3) becomes

$dq = C_vdT + d(P\alpha) - \alpha dP$

From the ideal gas law we have

$P\alpha = RT \Rightarrow d(P\alpha) = d(RT) = RdP     \mbox{(R is a constant)}$

and subbing this result for $d(P\alpha)$ in the previous equation gives

$dq = C_vdT + RdT - \alpha dP$
$\Rightarrow dq = (C_v + R)dT - \alpha dP$         $(4)$

Because $P$ is constant, $dP = 0$ and

$dq = (C_v + R)dT$
$\Rightarrow \frac{dq}{dT} = C_v + R$

But, $\frac{dq}{dT} = C_p$, so we have that $C_p = C_v + R$, so equation (4) becomes

$ dq = C_p dT - \alpha dP$         $(5)$