Question

Find the Entropy change in an isothermal expansion of one mole of an ideal gas from volume of ${V_1}$ to ${V_2}$ .

Answer 1

In an isothermal process, temperature remains constant such that $\Delta T = 0$ . According to Joule’s law, if q = 0 i.e. if heat supplied is equal to zero then there is no change in the internal energy, and thus, change in entropy directly depends on change in volume of the gas.

Complete step-by-step answer:

We have studied this in thermodynamics unit that the change in entropy in an isothermal irreversible process is given by
$q = - w = nRT\ln \dfrac{{{V_f}}}{{{V_i}}}$
We can derive this expression using other terms of thermodynamics.
It is known that the change in internal energy of a system can be written as:
$\Delta U = q + w$
Where, $\Delta U$ is the change in internal energy, q is the heat given by the system and w can be the work done on the system. As determined by the Joule’s experiment, if q = 0, then $\Delta T = 0$ . For isothermal reversible and irreversible changes, we can write this expression as:
Isothermal reversible change: $q = - w = {p_{ex}}({V_f} - {V_i})$
Isothermal irreversible change: $q = - w = nRT\ln \dfrac{{{V_f}}}{{{V_i}}}$
Now, from the second law of thermodynamics, we know that during entropy change, the amount of heat released or absorbed isothermally and reversibly is divided by the absolute temperature. Entropy formula can be written as
$\Delta S = \dfrac{{{q_{rev,iso}}}}{T}$
Using the first law of thermodynamics which we discussed above, we know that $\Delta U = q + w$ and for the isothermal expansion of an ideal gas, $\Delta T = 0$ , we got $q = - w = nRT\ln \dfrac{{{V_f}}}{{{V_i}}}$ .
Therefore, substituting the value of q in entropy equation, we get
$q = - w = nRT\ln \dfrac{{{V_f}}}{{{V_i}}}$
We are given the question that n is equal to 1 mole of a gas and the initial volume is ${V_1}$ and the final volume is ${V_2}$ .
Substituting this value of n and these terms of volume in the change in entropy formula, we get
$\Delta S = 2.303\,\,R\,\,\log \dfrac{{{V_2}}}{{{V_1}}}$
Where R is the gas constant. So, this is the Entropy change in an isothermal expansion of one mole of an ideal gas from volume of ${V_1}$ to ${V_2}$ .

Note: The process after which the system and the surroundings return back to their original states, is called a reversible process. Whereas the process in which permanent changes occur in the system and surroundings both on completion is called an irreversible process. Whether the process is reversible or irreversible, the amount of work done is always zero in free expansion of gases.