Reversible Processes and Work

Reversible Process 

We are interested in processes that take us from in equilibrium state to another. I  general this will be done irreversibly. But the wonder of state functions is that we can describe or imagine an ideal process that does the job. A simple definition of reversible is that it should be able to return to its original state but this is not all. This process must leave its surroundings unchanged.
Consider the following system: A pendulum and gravitational field. We can apply a force to swing it so we are the environment or the surrounding acting on the system. Imagine pushing the pendulum with an infinitesimal amount of force to move it from one angle to another. We do this to ensure that the pendulum is stays in a state of equilibrium along its path. Ensuring this is what is called a quasi-static process. Now imagine reducing this force so that the pendulum swings back and does work on us and return to its original state. The amount of work it does on us(environment) will be equal to the amount we applied as long as there are no dissipative forces otherwise some work will have to be done to fight against the dissipative forces. Therefore the pendulum has left its surrounding unchanged.
Therefore moving through equilibrium state is a quasi-static process and therefore a reversible process is a quasi-static process where no dissipative forces are present.

Work 

Let us consider, the all to common system involving a gas in cylinder covered by a piston one side. The gas is kept in by a balancing force from the frictionless piston. If the piston is let go slowly enough so that the expansion is quasi-static, the gas will start to expand. All the work done by the gas goes to the environment since there are no dissipative forces. In this simplified and ideal case we can easily calculate the work done by the gas.
\begin{eqnarray}
  dW &=& F dx \\
        &=& P A dx\\
        &=& P dV  \hspace{10mm} \text{reversible processes}
\end{eqnarray}

Therefore we have that:
\begin{eqnarray}
 W = \int_{v_1}^{v_2} P\, dV
\end{eqnarray}

This equation has been derived for reversible process but there are special irreversible processes where it can still be applied because there are no finite pressure drops. So, imagine a cylinder and piston system but this time put small reacting solids that slowly produce gas. This gas does work on the piston which is frictionless. Pressure in cylinder is only infinitesimally smaller than the surrounding and piston is always in mechanical equilibrium and surrounding has pressure  \( P_o\)  so

\begin{equation}
W = \int_{v_1}^{v_2} P\, dV = \int_{v_1}^{v_2} P_o \,dV
\end{equation}

Notice the process is not reversible since the chemical reaction that produces gas is not reversible.
Looking at a P-V diagram we clearly see that the total work done is path dependent as it is the integral under the curve. Therefore dW is and in-exact differential .
Note: The sign 

Example Calculations

1. Suppose we have an ideal gas at T= 300K and isothermally compressed from 1 atm to 10 atm. What is the work done on the gas.

We have that \( \int_{v_1}^{v_2} P \,dv \) but we are not told the volumes but instead told the pressure so obviously we need to change our integration variable. We the use the ideal gas law (PV = nRT) to get the relation between dV and dP . So we have \( dP = - \frac{nRT}{V^2} dV\). So we have that  \(  \int _{P_1}^{P_2} \frac{PV^2}{nRT} \,dP\) we can replace V using the ideal gas law to finally have the integral \(  \int_{P_1}^{P_2} \frac{nRT}{P} dP \) and therefore the formula for work done is \( nRT \ln(\frac{P_2}{P_1}) \). This gives the answer of 57431 J.

2. During and adiabatic irreversible process \( PV^ {\gamma} =c \) all through out the process (not just in this question but in general). Calculate the work done by the gas. \( \gamma \) and c are constants.

The process is reversible so we can use \( \int_{v_1}^{v_2} P \,dV \). We can plug in what P is in terms of and do the integral to get c\( (V_2^{-\gamma+1}-V_1^{-\gamma +1}) \frac{1}{-\gamma +1} \). We have that  \( P_1 V_1 ^{\gamma}=c\) and \( P_2 V_2 ^{\gamma}=c \). We pull back c into the two terms to get  \( W= \frac{P_1V_1 - P_2V_2}{\gamma-1} \)

3. Imagine a gas of n moles at high pressure in a container with volume V and with diathermal walls. The gas is let to leak out slowly through a valve to the surrounding at atmospheric pressure \( P_o\).

This is an irreversible process but it is one of those irreversible processes that is special. The point is that the gas is leaking out very slowly so that there is no finite pressure drop. So we can still use our beloved formula to get \( W= P_o(nv_o -V) \). Here \( v_o \) is the molar volume of the gas outside the container.