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: Understanding the Concept:
The first law is \(\Delta U = Q + W\), where \(W\) is the work done on the system. In an isothermal process \(T\) is constant, so for an ideal gas \(\Delta U = 0\).

Step 2: Key Formula or Approach:
Substitute \(\Delta U = 0\) into \(\Delta U = Q + W\).

Step 3: Detailed Explanation:
\(0 = Q + W\), so \(W = -Q\).
This is option (A): the work done on the system equals the heat given out, or the heat absorbed is converted fully into work by the system.
Option (B), \(-\Delta U = -W\), means \(\Delta U = W\) with no heat exchange, which is the adiabatic case.
Option (C), \(\Delta U = Q_v\), holds for a constant volume process.
Option (D), \(Q_p = \Delta U + P_{ext}\Delta V\), is the form for a constant pressure process.

Final Answer:
For an isothermal process \(W = -Q\), option (A). \[ \boxed{W = -Q} \]