Answer to Question #93501 in Molecular Physics | Thermodynamics for Amit Rajaram Shinde

Question

In a isothermal process temperature remains constant. According to first law of thermodynamics Q=W+∆U. here ∆U is internal energy. For ideal gas ∆U is function of temperature only hence ∆U=0 for isothermal process. Hence we get Q=W for isothermal process. But according to kelvin-planck statement of second law of thermodynamics heat is not completely convertible into work. Then how this Q=W is possible for isothermal process.

Answer 1

An ideal gas by definition has no interactions between particles, no intermolecular forces, so pressure change at constant temperature does not change internal energy.

Real gases have intermolecular interactions, attractions between molecules at low pressure and repulsion at high pressure. Their internal energy changes with change in pressure, even if temperature is constant.

For an ideal gas, in an isothermal process, Q=W.