Which of the following relationships is correct for an isothermal reversible process involving an ideal gas?

• A. ΔH = ΔU
• B. ΔU = 0
• C. q = 0
• D. w = 0

Answer: (B)

Verified step by step guidance

1. Recall that an isothermal process means the temperature (T) remains constant throughout the process.
2. For an ideal gas, the internal energy (U) depends only on temperature, so if the temperature is constant, the change in internal energy is zero: \(\Delta U = 0\).
3. Enthalpy (H) for an ideal gas also depends only on temperature, so under isothermal conditions, \(\Delta H = 0\), which means \(\Delta H = \Delta U\) is true only if both are zero.
4. Since \(\Delta U = 0\), apply the first law of thermodynamics: \(\Delta U = q + w\), so \$0 = q + w$, meaning heat absorbed by the system equals the work done by the system but with opposite sign.
5. Therefore, neither \(q = 0\) nor \(w = 0\) is generally true for an isothermal reversible process; instead, \(q\) and \(w\) are nonzero but equal in magnitude and opposite in sign.