Question

We consider a thermodynamic system. If \(\Delta U\) is the increase in internal energy and \(W\) is the work done by the system, which one of the following statement is true?

• A. \(\Delta U=-W\) in an adiabatic process
• B. \(\Delta U=W\) in an isothermal process
• C. \(\Delta U=-W\) in an isothermal process
• D. \(\Delta U=W\) in an adiabatic process

Hint:

Apply the first law, heat equals change in internal energy plus work done by the system, and check what is zero in each process.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The first law states \(Q=\Delta U+W\), where \(W\) is the work done by the system.

Step 2: Adiabatic Process
No heat is exchanged, so \(Q=0\).
Then \(\Delta U=-W\).

Step 3: Isothermal Process
For an ideal gas the temperature is constant, so \(\Delta U=0\). This gives \(Q=W\), not \(\Delta U=\pm W\).

Step 4: Compare the options
Only \(\Delta U=-W\) in an adiabatic process is correct.

Final Answer:
The true statement is that the change in internal energy equals minus the work done in an adiabatic process. \[ \boxed{\text{(A) }\Delta U=-W\ \text{in an adiabatic process}} \]