Here, we want to expand these concepts to a thermodynamic system and its environment.

Specifically, we elaborated on the concepts of heat and heat transfer in the previous two chapters. Here, we want to understand how work is done by or to a thermodynamic system; how heat is transferred between a system and its environment; and how the total energy of the system changes under the influence of the work done and heat transfer.

## Work Done by a System

A force created from any source can do work by moving an object through a displacement. Then how does a thermodynamic system do work? A gas confined to a cylinder that has a movable piston at one end. If the gas expands against the piston, it exerts a force through a distance and does work on the piston. If the piston compresses the gas as it is moved inward, work is also done—in this case, on the gas. The work associated with such volume changes can be determined as follows: Let the gas pressure on the piston face be p. Then the force on the piston due to the gas is pA, where A is the area of the face. When the piston is pushed outward an infinitesimal distance dx, the magnitude of the work done by the gas is

*dW=Fdx=pAdx.*

Since the change in volume of the gas is * dV=Adx*, this becomes

**dW=pdV.**

For a finite change in volume from * V_{1}* to

*, we can integrate this equation from*

**V**_{2}*to*

**V**_{1}*to find the net work:*

**V**_{2}\begin{equation}\ref{eq1}W=\int_{V1}^{V2}pdV\end{equation}

The work done by a confined gas in moving a piston a distance *dx* is given by * dW=Fdx=pdV*.

This integral is only meaningful for a quasi-static process, which means a process that takes place in infinitesimally small steps, keeping the system at thermal equilibrium. (We examine this idea in more detail later in this chapter.) Only then does a well-defined mathematical relationship (the equation of state) exist between the pressure and volume. This relationship can be plotted on a pV diagram of pressure versus volume, where the curve is the change of state. We can approximate such a process as one that occurs slowly, through a series of equilibrium states. The integral is interpreted graphically as the area under the pV curve (the shaded area). Work done by the gas is positive for expansion and negative for compression.

When a gas expands slowly from V1toV2, the work done by the system is represented by the shaded area under the *pV* curve.

The paths ABC, AC, and ADC represent three different quasi-static transitions between the equilibrium states A and C.

The expansion is isothermal, so T remains constant over the entire process. Since n and R are also constant, the only variable in the integrand is V, so the work done by an ideal gas in an isothermal process is

W=nRT\(\int_{V1}^{V2}\)\(\frac{dV}{V}\)=nRTln\(\frac{V2}{V1}\)

Notice that if * V_{2}>V_{1}* (expansion), W is positive, as expected.

The straight lines from A to B and then from B to C represent a different process. Here, a gas at a pressure p1 first expands isobarically (constant pressure) and quasi-statically from

*to*

**V**_{1}*, after which it cools quasi-statically at the constant volume V2 until its pressure drops to p2. From A to B, the pressure is constant at p, so the work over this part of the path is*

**V**_{2}W=\(\int_{V1}^{V2}\)pdV=p1\(\int_{V1}^{V2}\)dV=p_{1}(V_{2}-V_{1})

From B to C, there is no change in volume and therefore no work is done. The net work over the path ABC is then**W=p _{1}(V_{2}−V_{1})+0=p_{1}(V_{2}−V_{1}).**

A comparison of the expressions for the work done by the gas in the two processes of the Figure shows that they are quite different. This illustrates a very important property of thermodynamic work: It is path dependent. We cannot determine the work done by a system as it goes from one equilibrium state to another unless we know its thermodynamic path. Different values of the work are associated with different paths.

Isothermal Expansion of a van der Waals Gas

Studies of a van der Waals gas require an adjustment to the ideal gas law that takes into consideration that gas molecules have a definite volume (see The Kinetic Theory of Gases). One mole of a van der Waals gas has an equation of state

\begin{equation}(p+\frac{a}{V^2})(V−b)=RT\end{equation}

To evaluate this integral, we must express p as a function of V. From the given equation of state, the gas pressure is

\begin{equation}W=\int_{V1}^{V2}(\frac{RT}{V−b}−\frac{a}{V^2})dV=∣RTln(V−b)+\frac{a}{V}∣\stackrel{V2}{V1}=RTln\frac{(V2−b)}{(V1−b)}+a(\frac{1}{V2}−\frac{1}{V1})\end{equation}

### Significance

The internal energy * E_{int}* of a thermodynamic system is, by definition, the sum of the mechanical energies of all the molecules or entities in the system. If the kinetic and potential energies of molecule i are

*and*

**K**_{i}*, respectively, then the internal energy of the system is the average of the total mechanical energy of all the entities:*

**U**_{i}

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