4.7 The Maximum Attainable Work

There is a lesson to be learned here. One can have an expansion only if the internal pressure is greater than the external pressure. And one can have a compression only if the external pressure is greater than the internal pressure. Thus there is a limit on how much work a system can do without an external input of energy.

Let's assume that we have a cylinder containing n moles of a gas at an initial pressure of p₁, temperature T and volume V₁. We'll keep the temperature constant to keep things simple. We are going to let this gas expand to a final volume of V₂. What is the maximum amount of work that we can get from this process?

It is clear that the minimum amount of work we can get is zero. We get that if there is no opposing pressure, i.e. a free expansion.

And it is clear that we can't get an infinite amount of work out of our system. In fact, if the external pressure p_ext is less than the initial internal pressure p₁, no work can be obtained from the system at all since the gas won't expand.

If the external pressure is less than p₁ and greater than zero, we will get work out of the system. So somewhere in between these two pressures there must be a maximum.

The simple answer is that if the gas is going to expand to a final volume of V₂, then the final internal pressure is going to be p₂. So a constant external pressure must be equal to or less than this at all times. If we set the external pressure to p₂, then the work we get from the system is given by:

$$w_0 = - \int_{V_1}^{V_2} p_{ext} dV = - p_{ext} ( V_2 - V_1 )$$ (4.7.18)

If we assume that this process is quasi-static, the situation looks like this:

Figure 4.4: Work done against a constant external pressure

where the curved line is the pressure-volume curve for the gas in the cylinder (assumed ideal) ^4.7 and the marked area at the bottom of the graph represents integral 4.7.1. The work is the negative of the marked area of the graph.

We certainly ought to be able to get more work out of the system than this! Let us represent the external pressure by a pile of infinitesimally small sand grains with no added atmospheric pressure. Now we could pile on some extra sand at the start of the process process and lift it part way to the point where the internal pressure just equals the external pressure exerted by the sand. Now we can slide the extra sand off and then raise the remaining sand the rest of the way.

Figure 4.5: Work done against two piles of sand

The diagram shows this. The intermediate point a is where the extra sand is removed. The areas of the two rectangles under the p-V curve represents the work done.

We could get even more work done by adding yet another pile of sand at the start. We can keep doing this as long as the total pressure exerted by the sand is less than p₁, the initial pressure of the gas in the cylinder.^4.8 Remember the definition of a reversible process? It is one where a very slight change in the external conditions can change the direction of the process. In this case that means a process where the external pressure is always just infinitesimally less than the internal pressure.

In other words, we arrange to start with a sand pile that exerts just infinitesimally less pressure than p₁ and, as the infinitely slow reversible process proceeds, we continually remove one infinitesimal grain of sand at a time. That keeps the external pressure equivalent to the internal pressure at all times.

The work is given by the same integral:

$$w = - \int_{V_1}^{V_2} p_{ext} dV$$ (4.7.19)

but now p_ext isn't constant any more. It is always equal to p, the current internal pressure. In that case we can replace p_ext in the formula above by p to get:

$$w = - \int_{V_1}^{V_2} p dV$$ (4.7.20)

and then, since we've assumed that the gas is ideal, we can replace p by nRT/V. This gives:

$$w = - n R T \int_{V_1}^{V_2} dV / V$$ (4.7.21)

which integrates into:

$$w = - n R T ln \frac{V_2}{V_1}$$ (4.7.22)

The graph below shows this process. Again, the shaded area is the work done.

Figure 4.6: Isothermal reversible work

Notice how equation 4.7.5 takes care of the sign automatically. If the final volume is less than the initial volume (a compression) the fraction is less than one and the logarithm is itself negative. Combined with the minus sign in front of the equation, the work is then positive. For an expansion, the logarithm is greater than one and hence positive, so the work is negative, as it should be.

It should be clear that this is the maximum work we can obtain from the gas in the cylinder. If we were to add even one more grain of sand at any point in the process it would not be lifted. Because the external pressure would then be greater than the internal pressure, the piston would sink rather than rise.

Example: We are not restricted to ideal gases. If, for example, the gas obeys van der Waals equation, the internal pressure is always given by:

$$p = \frac{nRT}{V - nb} - \frac{n^2 a}{V^2}$$

and the integral then becomes:

$$w = - \int_{V_1}^{V_2} \left( \frac{nRT}{V - nb} - \frac{n^2 a}{V^2} \right) dV$$

which integrates to:

$$w = - n R T ln \frac{V_2}{V_1} + n^2a \left( \frac{1}{V_2} - \frac{1}{V_1} \right)$$