Free expansion of an ideal gas and changes in entropy

In summary: I CORRECT THAT THIS IS TRUE ONLY WHEN THE TWO TRANSFORMATIONS ARE REVERSIBLE?Yes, the Clausius theorem only applies when the two transformations are reversible.
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Ahmed1029
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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
 
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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?
 
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Ahmed1029 said:
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
 
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Related to Free expansion of an ideal gas and changes in entropy

What is the free expansion of an ideal gas?

The free expansion of an ideal gas is a process in which a gas expands into a vacuum without any external work being done on it. This means that the gas expands freely and no energy is transferred to or from the surroundings.

What happens to the temperature of an ideal gas during free expansion?

During free expansion, the temperature of an ideal gas remains constant. This is because no energy is transferred to or from the surroundings, so there is no change in the internal energy of the gas.

What is the change in entropy during free expansion?

The change in entropy during free expansion is zero. This is because entropy is a measure of the disorder or randomness of a system, and in free expansion, there is no change in the arrangement of particles, so the entropy remains constant.

How does free expansion affect the volume and pressure of an ideal gas?

During free expansion, the volume of an ideal gas increases while the pressure decreases. This is because the gas is expanding into a larger volume, so the particles have more space to move around, resulting in a decrease in pressure.

What is the relationship between free expansion and the second law of thermodynamics?

The free expansion of an ideal gas is an example of a process that violates the second law of thermodynamics. This is because the second law states that the total entropy of a closed system must increase over time, but in free expansion, the entropy remains constant.

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