Problem 161
Question
An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If \(\mathrm{T}_{\mathrm{i}}\) is the initial temperature and \(\mathrm{T}_{\mathrm{f}}\) is the final temperature, which of the following statements is correct? [2006] (a) \(\left(\mathrm{T}_{\mathrm{f}}\right)_{\text {irrev }}>\left(\mathrm{T}_{\mathrm{i}}\right)_{\mathrm{rev}}\) (b) \(\mathrm{T}_{\mathrm{f}}>\mathrm{T}_{\mathrm{i}}\) for reversible process but \(\mathrm{T}_{\mathrm{f}}=\mathrm{T}_{\mathrm{i}}\) for irreversible process (c) \(\left(\mathrm{T}_{\mathrm{f}}\right)_{\text {irrev }}=\left(\mathrm{T}_{\mathrm{i}}\right)_{\mathrm{rev}}\) (d) \(\mathrm{T}_{\mathrm{f}}=\mathrm{T}_{\mathrm{i}}\) for both reversible and irreversible processes
Step-by-Step Solution
VerifiedKey Concepts
Isolated System
- There is no heat (Q = 0) transfer, meaning no energy can flow in or out.
- No work (W = 0) is done by or on the system, which means it cannot change its volume by interacting with the surroundings.
- Internal energy (E = 0) remains constant since it is not influenced by outside factors.
First Law of Thermodynamics
- When heat is added (Q > 0), it increases the system's internal energy unless it's entirely used to do work.
- When the system does work (W > 0), it loses energy as some of it is used to perform work on its surroundings.
Reversible vs Irreversible Processes
- Reversible Process: This is an idealized process that happens so slowly that the system is always in equilibrium. In practice, it means every state of the system can be reversed without any net change in the system and surroundings.
- Irreversible Process: This occurs when the process moves the system out of equilibrium and cannot be undone by reversing the steps. Most real-world processes are irreversible due to factors like friction or rapid changes.