Step 1: Understanding the Question:
The question asks us to identify the single incorrect statement regarding the properties of an isobaric thermodynamic process from the given options.
Step 2: Key Formula or Approach:
An isobaric process is a thermodynamic process in which the pressure remains completely constant ($\Delta P = 0$). We evaluate the remaining variables using the Ideal Gas Law and the First Law of Thermodynamics:
1. Ideal Gas Law: $PV = nRT$
2. First Law of Thermodynamics: $Q = \Delta U + W$
3. Work done at constant pressure: $W = P\Delta V$
Step 3: Detailed Explanation:
Let's analyze each statement systematically:
Option (A): By definition, the prefix "iso-" means equal and "baric" relates to pressure. Thus, pressure remains constant. This statement is correct.
Option (B): The work done is given by $W = P\Delta V$. For mechanical work to be performed ($W \neq 0$), a change in volume ($\Delta V \neq 0$) must occur. This statement is correct.
Option (C): According to Charle's Law for a constant pressure system ($V \propto T$), any modification in volume directly forces a corresponding linear modification in temperature. A process where temperature remains constant ($\Delta T = 0$) is called an isothermal process, not an isobaric one. Thus, this statement is false.
Option (D): The heat energy exchanged ($Q$) splits into two components—performing boundary expansion work ($W$) and altering the internal kinetic energy ($\Delta U$) of the gas molecules. This statement is correct.
Step 4: Final Answer:
The incorrect statement is option (C).