1. The problem statement, all variables and given/known data In a reversible isothermal expansion of an ideal gas, as the gas expands, heat is supplied to it, so that the temperature remains constant. Thus, temperature and hence kinetic energy of the molecules does not change but the 'disorder' of the gas increases as it occupies a greater volume. But then how can any isothermal process be reversible (since del(S) in reversible process=0)? Such isothermal process may be quasi-static, but a quasi-static process need not be reversible. Moreover, the above process is not spontaneous, since it says a reversible isothermal process is not spontaneous (though has delS>0), so increase in entropy does not necessarily imply spontaneity? 2. Relevant equations delS = delS(transfer) + delS(generated) 3. The attempt at a solution I asked a professor and he said that I was "mixing up the calculation of the entropy change with the experimental measurement. The change in entropy in expansion of ideal gas from V to 2 V is R ln(2). This is calculated from the expression of the entropy of ideal gas. It is the same if the process is done isothermally or in isolated system . All that matters is the initial and the final states. And you do not have to tell how you reached from the initial to the final state. The story is different when you talk about measuring the change in entropy, then it depends on how you do the process, and whether work is done or not. It also matters if you do it quasi-statically or not." CAN ANYONE EXPLAIN THIS THING THAT HE SAID??