Question

In a process, \(\Delta U=0\) and \(q\neq 0\). Then the process must involve

• A. No work is done during the process.
• B. Heat is transferred, but no mechanical work occurs.
• C. Work done by or on a system exactly equal in magnitude and opposite in sign to a heat transfer.
• D. There is a change in internal energy of the system throughout the process.

Hint:

Apply the first law, delta U = q + w, with delta U equal to zero.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The first law of thermodynamics states \(\Delta U=q+w\).

Step 2: Apply the condition
With \(\Delta U=0\), we get \(q+w=0\), so \(w=-q\).
Since \(q\neq0\), the work is also nonzero.

Step 3: Compare the options
(A) and (B) say no work is done, which contradicts \(w=-q\).
(D) says internal energy changes, which contradicts \(\Delta U=0\).
(C) says work is equal and opposite to the heat transfer, which is correct.

Final Answer:
Work done equals the heat transferred in magnitude with opposite sign. \[ \boxed{\text{(C) }\text{Work done equals heat transfer in magnitude, with opposite sign}} \]