How do you calculate the work done when an object is moved against a force?
What work is done by a gas at pressure, P, when it changes its volume (reversibly)?
Why is only possible to use this expression if the change in volume is done reversibly?
Why can the pressure be calculated given the values of the volume and the temperature?
Work in Newtonian physics
To lift a heavy object you have to expend some effort. By contrast, when you drop it you expend zero effort - it will accelerate towards the earth because it is acted upon by the force of gravity. Early on in your scientific education, you will have learnt that when lifting an object you have to do work on it against the force of gravity or that you have to give the box more potential energy. If you think about the meaning of words in this statement what you are saying it is saying that you are giving the box the potential to accelerate down to the earth at some point in the future i.e. when it is released. When you were told this you should have been told that the work done in moving an object upwards is given by the following expression:
$\Delta w = \int_{x}^{x+\Delta x} \mathbf{F}(x')\textrm{d}x'$
In this expression, the integrand is a vector quantity called the force, which is a function of the position of the box. If the force is constant the expression above reduces to the familiar:
$\Delta w = mg\Delta x$
Expansion of gasses
Suppose that I have a piston with a heavy head that contains some gas as shown in the figure. For the gas to expand it must do some work in order to lift up the piston head. This work comes about because the gas exerts a force on the piston head. Furthermore, we know that the pressure, P, is defined as the force F over the area of the piston head, A - i.e. P=F / A. We can thus write that the work done when the piston head moves is given by:
$\Delta w = -\int_{x}^{x+\Delta x} \mathbf{F}(x) \textrm{d}x' = -\int_{x}^{x+\Delta x} \frac{\mathbf{F}(x)}{A} A \textrm{d}x' =-\int_{v}^{v+\delta v} P(V,T) \textrm{d}V$
The third equality follows because the cross sectional area multiplied by the change in the position of the piston head - is equal to the change in the volume of the gas. Notice furthermore that this equation only holds when the the change in gas's volume takes place reversibly. This means that in our model we are assuming that the intensive quantity - the pressure - is distributed uniformly throughout the system. If this were not the case the change would take place irreversibly as processes would be occurring for which we have no mathematical description in the model. A key point here is that, because the transition is reversible, the state of the system can be characterised using a small set of thermodynamic variables at all points during the transition. If the transition were taking place irreversibly the pressure would be poorly defined. We thus would be unable to calculate its value using the function of state, P(V,T) as we have in the above integral.
<aside> 📌 SUMMARY: Work done can be calculated by integrating the force along the path that the system changed along.
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