Dealing with infinitesimals

Introduction: the pV work example

In thermodynamics we often come up with formulas like this one for the work of changing the system volume by dV

$\begin{displaymath}dw ~=~ -p_{ex} dV ~~~~~~(pV~work:~general~formula, ~differential~form)\end{displaymath}$

where $p_{ex}$ is the external pressure. We have to be able to understand and manipulate such equations to understand the material and solve problems. See the Work section for details on pV work.

The meaning of the formula.
Since something is changing (e.g. volume by dV) we are dealing with a process. This process involves a tiny, minuscule or infinitesimally small change of state from

$\begin{displaymath}initial~state:~ T, ~V, ~p \end{displaymath}$

$\begin{displaymath}final~state:~~~ T+dT, ~V+dV, ~p+dp \end{displaymath}$

(If this is hard for you to imagine, think of a measurably large change of state to $T+\Delta T, ~V+\Delta V, ~p+\Delta p$ and then break it up into many smaller steps - we are looking at one of those tiny little steps.)

In our case of pV work, only dV enters the formula, so we usually do not think of dT and dp. They may be important for other properties - e.g. dT for calculating heat transfer.

The integral form.
To apply our differential formula in practice, we have to consider a finite change of state

$\begin{displaymath}initial~state:~ T_i, ~V_i, ~p_i \end{displaymath}$

$\begin{displaymath}final~state:~~~ T_f=T_i+\Delta T, ~V_f=V_i+\Delta V, ~p_f=p_i+\Delta p \end{displaymath}$

Let us break up this large, finite change of state into a series of steps. We assume that within each step the external pressure is approximately constant, equal to $p_{ex,k}$ and the volume change is $\Delta V_k$ . The work done may be expressed as

$\begin{displaymath}w ~\approx~ -\sum_k p_{ex,k} \Delta V_k \end{displaymath}$

where the approximation comes in because the pressure may actually vary within some of the steps. If we make the number of steps infinitely large, we end up with an exact formula containing an integral

$\begin{displaymath}w ~=~ \int_i^f ~dw ~=~ -\int_{V_i}^{V_f} ~p_{ex} dV ~~(pV~work:~general~formula, ~integral~form)\end{displaymath}$

This is the formula we will be using to calculate e.g. the isothermal reversible expansion of an ideal gas. The integral formula is complicated, the differential formula simple. They mean the same thing.

Heat capacities

The same ideas hold for this formula for heating/cooling of a body at constant composition and constant pressure

$\begin{displaymath}dq ~=~ C_P dT ~~~~(heating~at~c.c.~and~p=const) \end{displaymath}$

with being the constant-pressure heat capacity. Again we have to think of this formula as describing the effects of a tiny change of state

$\begin{displaymath}initial~state:~ T, ~V, ~p \end{displaymath}$

$\begin{displaymath}final~state:~~~ T+dT, ~V+dV, ~p+dp \end{displaymath}$

and remember that for practical applications we have to use the integral form

$\begin{displaymath}q ~=~ \int_{T_i}^{T_f} C_P dT ~~~~(heating~at~c.c.~and~p=const) \end{displaymath}$

For example, if the heat capacity does not vary with temperature we can easily calculate this integral

$\begin{displaymath}q ~=~ \int_{T_i}^{T_f} C_P dT ~=~ C_P\int_{T_i}^{T_f}dT ~=~ C_P(T_f ~-~ T_i) ~=~ C_P\Delta T \end{displaymath}$

The differential formula is simple, and generally applicable - although not practical. The integral formula is unwieldy but practically useful. The differential and integral formulae are completely equivalent. See the Heat section for details.