Question

For an ideal gas undergoing an isothermal change, there is:

• A. a decrease in Internal energy of the system and heat released by the system is equal to the work done by the system
• B. an increase in Internal energy of the system and heat absorbed by the system is greater than the work done on the system
• C. no change in Internal energy of the system and heat released by the system is equal to the work done by the system
• D. no change in Internal energy of the system and heat absorbed by the system is equal to the work done by the system

Hint:

For an ideal gas: Isothermal = Constant Temperature = Zero Internal Energy Change ($\Delta U = 0$). According to the first law, this simplifies the energy balance to $\Delta Q = \Delta W$, meaning any heat absorbed must be completely converted into work performed by the system.

Answer 1

The Correct Option is D

Understanding the Concept: The internal energy ($U$) of an ideal gas depends solely on its absolute temperature ($U \propto T$). According to the First Law of Thermodynamics, the total energy balance in a closed system must satisfy the equation: \[ \Delta Q = \Delta U + \Delta W \] where $\Delta Q$ is heat absorbed, $\Delta U$ is internal energy change, and $\Delta W$ is work done by the system.
Step 1: Analyze the internal energy change ($\Delta U$).
An isothermal process occurs at a constant temperature ($\Delta T = 0$). Because the temperature of the ideal gas remains fixed, its internal energy cannot change: \[ \Delta U = 0 \]
Step 2: Apply the first law energy balance to determine work and heat tracking.
Substituting $\Delta U = 0$ into the thermodynamic balance equation gives: \[ \Delta Q = 0 + \Delta W \implies \Delta Q = \Delta W \] This shows that any heat energy absorbed by the gas system ($\Delta Q>0$) is completely transformed into mechanical work performed by the gas on its surroundings ($\Delta W>0$). This matches statement (D) perfectly.