[thompsjc]

Entropy

This regards entropy change for an irreversible isothermal expansion of an ideal gas.  I want to understand why entropy change is positive and cannot understand the textbook explanations.  It is clear to me that, for this process, the entropy change for the system is nRln(V2/V1), because entropy change is calculated for the corresponding reversible process, which entails work, yet no internal energy change, so dq = dw.  It also seems clear that the change for the surroundings in the REVERSIBLE case is –nRln(V2/V1) because of heat flowing out of them.  For the REVERSIBLE process, it is clear to me that delta S total is zero.  However, in the textbook explanations for the IRREVSIBLE process, it is claimed that there is no work done, nor heat exchanged.  I understand no work is done, but here is my confusion:

1. How can there be no heat exchange if the process is isothermal within the system (not the surroundings)?  When a gas expands out of a compressed state it cools.  If it does not cool, it seems heat must have been provided to it from the surroundings!

2. Why, for the system, is the stipulation made that delta S is calculated from the corresponding reversible process, when for the surroundings this requirement seems to be abandoned?

I always thought I understood thermodynamics, but a re-read has got me wondering.

John Thompson

[Kemi]

1. Ideal-gas molecules do not interact with each other (except during collisions), so a change in volume at constant temperature does not change internal energy. (On the other hand, a real gas does not always cool during expansion.) If ΔU = 0 and w = 0, we get q = 0. You cannot use this q for calculating the entropy of the system because, by definition, you need to use q_rev.

2. The surroundings are so enormous that their pressure does not change during any process. At constant pressure q = ΔH. Because enthalpy is a state function, it does not depend on how the change is brought about, so we can write q_sur,rev = ΔH_sur = q_sur.

You may like a quote attributed to Arnold Sommerfeld: http://en.wikiquote.org/wiki/Arnold_Sommerfeld.

[juanrga]

Entropy

This regards entropy change for an irreversible isothermal expansion of an ideal gas.  I want to understand why entropy change is positive and cannot understand the textbook explanations.  It is clear to me that, for this process, the entropy change for the system is nRln(V2/V1), because entropy change is calculated for the corresponding reversible process, which entails work, yet no internal energy change, so dq = dw.  It also seems clear that the change for the surroundings in the REVERSIBLE case is –nRln(V2/V1) because of heat flowing out of them.  For the REVERSIBLE process, it is clear to me that delta S total is zero.  However, in the textbook explanations for the IRREVSIBLE process, it is claimed that there is no work done, nor heat exchanged.  I understand no work is done, but here is my confusion:

1. How can there be no heat exchange if the process is isothermal within the system (not the surroundings)?  When a gas expands out of a compressed state it cools.  If it does not cool, it seems heat must have been provided to it from the surroundings!

2. Why, for the system, is the stipulation made that delta S is calculated from the corresponding reversible process, when for the surroundings this requirement seems to be abandoned?

I always thought I understood thermodynamics, but a re-read has got me wondering.

John Thompson

The isothermal change of entropy is given by the production of entropy ##d_iS## plus the flow of entropy ##d_eS## due to heat flow with surrounds
$$dS = d_iS + d_eS = \left(\frac{p - p_{ext}}{T}\right)dV +  \frac{dQ}{T}$$
For ideal gas ##U=U(T)##; therefore, for an isothermal process ##\Delta U = 0## and the first law implies ##dQ = -dW##.

If the expansion is free ##dW = 0## and the above expression reduces to
$$dS =  \frac{p}{T}dV$$
which is positive (by the second law).