FL: Reversible and Irreversible Processes

Reversible and Irreversible Processes


A reversible process is one which can be "undone" without a affecting the surroundings. Alternatively, it is a process that can reverse direction at any point due to an infinitesimal change in external conditions (driving force).


A reversible process is an ideal one that yields the maximum possible work attainable.


Consider the following isothermal expansions/compressions of an ideal gas. Calculate the work for each step when the total mass of the blocks is 1020kg, the area of the piston surface is 0.1 $m^2$, the initial height of the piston is 0.4$m$ and the amount of gas present is 1 mol.

1 step
$\Delta U + \Delta E_k + \Delta E_p = Q + W$
$W = -Q$
$W = - Q = -\int P_E dV = -P_E \Delta V$

$P_E$ is larger for the compression than for the expansion! The work obtained from the expansion is less than that required for the compression.

1 step
$W = - Q = -\int P_E dV = -(P_{E}\Delta V)_{1 \to i} - (P_{E}\Delta V)_{i \to 2}$

The difference between $(P_{E}\Delta V)_{i \to 1(2)}$ and $(P_{E}\Delta V)_{1(2) \to i}$ (the forward vs the reverse) is smaller than for the 1 step process. The difference in the work obtained/required is also smaller.

1 step
$W = - Q = -\int P_E dV = -(P_{E}\Delta V)_{1 \to i_1} -(P_{E}\Delta V)_{i_1 \to i_2} - (P_{E}\Delta V)_{i_2 \to i_3} - (P_{E}\Delta V)_{i_3 \to 2}$

Again, the difference decreases further for the multi-step process. If the step gets even smaller, the external pressure at each step eventually becomes essentially equal to the internal pressure at each step ($P_E \to P$).

$W = - Q = -\int P_E dV = -\int P dV = -\int nRT \frac{dV}{V} = -nRT\ln\left (\frac{V_2}{V_1}\right )$

Explain the difference between a reversible and irreversible process.


Because a reversible process is the ideal/limiting case, we will use it as the reference point for efficiency calculations.