# Free expansion of an ideal gas and changes in entropy

• A
Ahmed1029
For a freely expanding ideal gas(irreversible transformation), the change in entropy is the same as in a reversible transformation with the same initial and final states. I don't quite understand why this is true, since Clausius' theorm only has this corrolary when the two transformations are reversible. Anyone can help?
Edit : the transformations are also isothermal, no change in temperature during any of the transformations

Last edited:

Mentor
Do you think that the change in entropy for a gas experiencing an irreversible transformation should be Q/T? What is your understanding of how to determine the change of entropy for a gas that has experienced an irreversible transformation? Is entropy a property of the gas or a feature of the applied process?

• Ahmed1029 and Lord Jestocost
Homework Helper
For a freely expanding ideal gas(irreversible transformation), the change in entropy is the same as in a reversible transformation with the same initial and final states. I don't quite understand why this is true, since Clausius' theorem only has this corollary when the two transformations are reversible. Anyone can help?
The reason it is true is because that is how entropy is defined. Change in entropy is the integral of dQ/T over a reversible process from the initial to final states:

##\Delta S = \int_i^f\frac{dQ_{rev}}{T}##

In a free expansion of an ideal gas, there is no work done and no heat flow in getting from the initial to final states. So internal energy (T) is unchanged (first law). The reversible path between the initial and final states consists of:
1. a quasi-static expansion doing work against an external pressure. The external pressure is kept an arbitrarily small amount less than the internal gas pressure
2. positive heat flow into the gas during expansion to maintain constant temperature.

Since the process does not change internal energy, Q-W=0 where Q is the heat flow into the gas and W is the work done BY the gas: ie. Q = W

Therefore: ##\Delta S = \frac{Q_{rev}}{T}=\frac{W_{rev}}{T}>0##

AM

Last edited:
• Ahmed1029