In which of the following scenarios would the change in internal energy (ΔU) of a system be negative according to the First Law of Thermodynamics?

• A. A gas expands against a constant external pressure and does work on the surroundings.
• B. No heat is exchanged, and the system does no work.
• C. A system absorbs heat from the surroundings without doing any work.
• D. A gas is compressed by the surroundings, and heat is absorbed by the system.

Answer: (A)

Verified step by step guidance

1. Recall the First Law of Thermodynamics, which states that the change in internal energy of a system (\( \Delta U \)) is given by the equation: \[ \Delta U = q + w \] where \( q \) is the heat exchanged with the surroundings and \( w \) is the work done on the system.
2. Understand the sign conventions: - Heat absorbed by the system is positive (\( q > 0 \)), heat released is negative (\( q < 0 \)). - Work done on the system is positive (\( w > 0 \)), work done by the system on the surroundings is negative (\( w < 0 \)).
3. Analyze each scenario by identifying the signs of \( q \) and \( w \): - For a gas expanding against constant external pressure and doing work on the surroundings, \( w < 0 \) (work done by system), and if no heat is exchanged, \( q = 0 \). Thus, \( \Delta U = 0 + (negative) = negative \).
4. - For no heat exchange and no work done, both \( q = 0 \) and \( w = 0 \), so \( \Delta U = 0 \).
5. - For heat absorbed without work, \( q > 0 \) and \( w = 0 \), so \( \Delta U > 0 \). - For gas compressed by surroundings (work done on system, \( w > 0 \)) and heat absorbed (\( q > 0 \)), both terms are positive, so \( \Delta U > 0 \).