Question:medium

In an isothermal expansion of an ideal gas, the heat supplied to the gas is:

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For an ideal gas: \[ \Delta U = 0 \] whenever temperature remains constant. Hence, in isothermal expansion, all supplied heat is used to perform work.
Updated On: May 29, 2026
  • entirely used to increase internal energy
  • entirely used to do work
  • partly used to do work and partly to increase internal energy
  • none of the above
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
An isothermal process is a thermodynamic process in which the temperature of the system remains constant throughout (\( \Delta T = 0 \)).
For an ideal gas, the internal energy (\( U \)) is a function of temperature only. This is because ideal gas molecules have no intermolecular forces, so their internal energy consists only of kinetic energy, which depends on temperature.
Therefore, if temperature does not change, internal energy does not change.
Step 2: Key Formula or Approach:
We use the First Law of Thermodynamics:
\[ \Delta U = q - w \]
Where:
- \( \Delta U \) is the change in internal energy.
- \( q \) is the heat added to the system.
- \( w \) is the work done by the system on the surroundings.
Step 3: Detailed Explanation:
1. Since the process is isothermal (\( T = \text{constant} \)), then \( \Delta T = 0 \).
2. For an ideal gas, \( \Delta U = n C_v \Delta T \). Since \( \Delta T = 0 \), then \( \Delta U = 0 \).
3. Substituting \( \Delta U = 0 \) into the First Law:
\[ 0 = q - w \implies q = w \]
This mathematical equality means that every bit of thermal energy (heat) supplied to the gas from the surroundings is converted into mechanical energy (work) performed by the gas as it expands.
The gas cannot use this heat to increase its internal energy because doing so would raise its temperature, violating the isothermal condition.
Step 4: Final Answer:
In an isothermal expansion, the heat supplied is equivalent to the work done.
The correct answer is option (B).
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