Question

Which from following is correct expression of first law of thermodynamics for isothermal process?

• A. \(\text{W} = -\text{Q}\)
• B. \(-\Delta \text{U} = -\text{W}\)
• C. \(\Delta \text{U} = \text{Q}_{\text{v}}\)
• D. \(\text{Q}_{\text{p}} = \Delta \text{U}+\text{P}_{\text{ext}}\Delta \text{V}\)

Hint:

In an isothermal ideal gas process the temperature does not change, so the internal energy does not change.

Answer 1

The Correct Option is A

Step 1: Think about what changes:
Temperature is fixed, so the internal energy of an ideal gas is fixed: $\Delta U = 0$.

Step 2: Match each option:
Writing $\Delta U = Q + W$ with $\Delta U = 0$ gives $W = -Q$, exactly option (A).
The other expressions describe other conditions: B has no heat term (adiabatic), C has $W = 0$ (constant volume), D uses constant external pressure.

Final Answer:
The isothermal form of the first law is $W = -Q$, option (A). \[ \boxed{W = -Q} \]