Question

Heat is given to an ideal gas in an isothermal process. Then

• A. internal energy of the gas will decrease.
• B. internal energy of the gas will increase.
• C. internal energy of the gas will not change.
• D. the gas will do negative work.

Hint:

In an isothermal process, the internal energy of an ideal gas does not change because temperature is constant, even though heat is added or work is done by the gas.

Answer 1

The Correct Option is C

Step 1: Definition of isothermal process.
An isothermal process is a thermodynamic process in which the temperature of the system remains constant. In this case, heat is added to the gas, but the temperature does not change. According to 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.

Step 2: Internal energy of an ideal gas.
For an ideal gas, the internal energy depends only on the temperature. Since the temperature remains constant in an isothermal process, the internal energy of the gas does not change:
\[ \Delta U = 0. \]

Step 3: Work done by the gas.
In an isothermal process, the gas may do work on its surroundings as it expands or contracts. The work done by an ideal gas in an isothermal expansion is given by:
\[ W = nRT \ln \left( \frac{V_f}{V_i} \right), \]
where:
- \( n \) is the number of moles of gas,
- \( R \) is the gas constant,
- \( T \) is the temperature,
- \( V_f \) and \( V_i \) are the final and initial volumes of the gas.

Step 4: Conclusion.
Since the internal energy does not change in an isothermal process (as temperature is constant), the correct answer is:
\[ \boxed{\text{(C) internal energy of the gas will not change.}}. \]