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

For an isothermal process, the gas's temperature is invariant. During isothermal expansion or compression of an ideal gas, the internal energy change (\( \Delta U \)) is zero. This is because the internal energy of an ideal gas is solely a function of temperature, which remains constant. - The first law of thermodynamics states: \[ \Delta U = q + w \] Given that \( \Delta U = 0 \) for an isothermal process, the equation simplifies to: \[ q = -w \] This implies that the heat exchanged with the gas (\( q \)) has the same magnitude as the work done by the gas (\( w \)), but with opposing signs. Consequently, for an isothermal process, \( \Delta U = 0 \), while \( q eq 0 \) and \( w eq 0 \). Therefore, option (2) is the correct choice.