Question

One mole of an ideal gas expands from 10 dm\(^3\) to 20 dm\(^3\) through an isothermal reversible process. Find \( \Delta U \), \( q \), and \( w \).

• A. \( \Delta U = 0, q = 0, w = 0 \)
• B. \( \Delta U = 0, q \neq 0, w \neq 0 \)
• C. \( \Delta U \neq 0, q = 0, w \neq 0 \)
• D. \( \Delta U \neq 0, q \neq 0, w = 0 \)

Hint:

In an isothermal process, the change in internal energy of an ideal gas is always zero, i.e., \( \Delta U = 0 \). The heat added to the gas is completely used for doing work, so \( q = -w \).

Answer 1

The Correct Option is B

In an isothermal process, the temperature of the gas remains constant. For an ideal gas undergoing an isothermal expansion or compression, the change in internal energy \( \Delta U = 0 \) because internal energy of an ideal gas depends only on temperature, and the temperature does not change. - The first law of thermodynamics is given by: \[ \Delta U = q + w \] Since \( \Delta U = 0 \) for an isothermal process, we have: \[ q = -w \] This means the heat absorbed by the gas \( q \) is equal in magnitude to the work done by the gas \( w \), but with opposite signs. Thus, \( \Delta U = 0 \), \( q \neq 0 \), and \( w \neq 0 \). Therefore, the correct answer is option (2).