In an isothermal process, the temperature (\( T \)) remains constant: \[ dT = 0 \]
The internal energy (\( U \)) of an ideal gas depends only on temperature. Since \( dT = 0 \), we have: \[ dU = 0 \] Thus, the internal energy remains constant.
From the First Law of Thermodynamics: \[ dQ = dU + dW \] Since \( dU = 0 \), this reduces to: \[ dQ = dW \] Therefore, the heat absorbed (\( dQ \)) is entirely used for work done (\( dW \)).
- If \( dQ > 0 \) (heat is absorbed by the system), then \( dW > 0 \) (work is done by the system).
- This implies both heat transfer and work done are positive in an isothermal process.
The correct answer is (C): (c), (e).
Thermodynamics in physics is a branch that deals with heat, work and temperature, and their relation to energy, radiation and physical properties of matter.
The first law of thermodynamics, also known as the Law of Conservation of Energy, states that energy can neither be created nor destroyed; energy can only be transferred or changed from one form to another.
The second law of thermodynamics says that the entropy of any isolated system always increases. Isolated systems spontaneously evolve towards thermal equilibrium—the state of maximum entropy of the system. More simply put: the entropy of the universe (the ultimate isolated system) only increases and never decreases.
The third law of thermodynamics states that the entropy of a system approaches a constant value as the temperature approaches absolute zero. The entropy of a system at absolute zero is typically zero, and in all cases is determined only by the number of different ground states it has. Specifically, the entropy of a pure crystalline substance (perfect order) at absolute zero temperature is zero