Question

Explain and discuss another experiment than (Verification of the Ideal Gas Equation using an isothermal expansion process Experiment) that can be conducted to verify the idea gas equation.

Answer 1

This is a famous example, which illustrates the calculation of entropy change and the difference between reversible and irreversible processes.

A sample of 1 mol of ideal gas expands isothermally, doubling its volume. Calculate the entropy change in the system and surroundings when the expansion is done a) reversibly and b) irreversibly, into vacuum.

a) Reversible expansion of IG, T=const

For ideal gas at T = const

Delta U = q + w = 0      thus      q = -w

Process is reversible so we can use the actual heat transfer to calculate entropy change in system

[\begin{displaymath}q = q_{rev} = -w = nRT \ln \frac{V_{f}}{V_{i}} = nRT \ln 2 \end{d
Apply formula for entropy change at T=const
[\begin{displaymath}\Delta S ~=~ \frac{q_{rev}}{T} ~=~ nR \ln~\frac{V_{f}}{V_{i}} ~=~ nR \ln 2 \end{displaymath}]
(Same result as from statistical treatment.)

Calculation of the entropy change in the surroundings is usually easy. The heat transfer in the surroundings is

[\begin{displaymath}q' = -q \end{displaymath}]
i.e. heat entering system leaves surroundings, and vice-versa. For a system at T=const the relation between system temperature T and surroundings temperature T' is
[\begin{displaymath}T = T' = const \end{displaymath}]
Thus the surroundings are also at constant temperature and we can use the actual heat transfer to calculate the entropy change in the surroundings
[\begin{displaymath}\Delta S' = \frac{q'_{rev}}{T'} = - \frac{q}{T} \end{displaymath}]
Thus
[\begin{displaymath}\Delta S' = -\Delta S = -nR\ln 2 \end{displaymath}]

The total entropy change for the reversible process

[\begin{displaymath}\Delta S_{tot} = \Delta S + \Delta S' = 0   (reversible)\end{displaymath}]

b) Isothermal expansion into vacuum

This is a spontaneous, irreversible process.

For expansion into vacuum [$p_{ex} = 0$] , thus

[\begin{displaymath}w = 0 \end{displaymath}]
For an isothermal process in an ideal gas
[\begin{displaymath}\Delta U = q + w = 0      thus      q = -w = 0 \end{displaymath}]

There are two arguments for why the entropy change of the surroundings must be zero in this case. First argument. Since w = q = 0, there is no interaction between system and surroundings, the state of the surroundings does not change, and all thermodynamic functions of the surroundings remain constant, including entropy S'. Second argument. To maintain the condition of constant temperature in the system, we have

[\begin{displaymath}T = T' = const \end{displaymath}]
and additionally the surroundings need to have special properties which allow us to consider that heat transfer q' is independent of path. Then we can use the actual value q' for this process to calculate
[\begin{displaymath}\Delta S' ~=~ \frac{q'_{rev}}{T'} ~=~ \frac{q'}{T'} ~=~ -\frac{q}{T} ~=~ 0 \end{displaymath}]
where, as always, heat leaving system enters surroundings: q' = -q.

The final state of the system

[\begin{displaymath}V_f = 2V_i      T_f = T_i \end{displaymath}]
is the same as in the reversible case!

Since the change of state of the system is the same, the entropy change in the system must be the same, as S is a state function

[\begin{displaymath}\Delta S = \frac{q_{rev}}{T} = nR \ln 2 \end{displaymath}]

Thus, for the irreversible process

[\begin{displaymath}\Delta S_{tot} = \Delta S + \Delta S' = nR\ln 2 > 0   (irreversible)\end{displaymath}]