An ideal gas undergoes isothermal expansion, absorbing 500 J of heat from a reservoir. How much work does the gas do on its surroundings?
ALess than 500 J — some heat goes into increasing internal energy
BExactly 500 J — all absorbed heat is converted to work
CMore than 500 J — the gas uses stored internal energy plus the absorbed heat
DZero — isothermal processes do no work because temperature doesn't change
For an ideal gas, internal energy depends only on temperature: U = nC_vT. If T is constant (isothermal), then ΔU = 0. By the first law: ΔU = Q − W, so 0 = Q − W, meaning W = Q = 500 J. All absorbed heat is converted to mechanical work. This is not a violation of thermodynamics — the temperature stays constant only because the gas continuously draws heat from the reservoir. The common mistake is thinking heat must 'go somewhere' besides work; for an ideal gas at constant T, the only destination is work.
Question 2 Multiple Choice
A gas expands isothermally from V_i to V_f. The same expansion is performed two ways: reversibly (infinitely slowly) and irreversibly (rapidly against lower external pressure). Which produces more work?
ABoth produce the same work — the initial and final states are identical
BThe irreversible process — faster expansion generates more kinetic energy
CThe reversible process — it maintains the maximum possible opposing force at every step
DNeither — isothermal processes always produce W = nRT ln(V_f/V_i) regardless of path
The reversible isothermal expansion maximizes work because it keeps the external pressure just infinitesimally below the gas pressure at every moment, extracting the maximum possible work against the largest possible opposing force throughout. An irreversible expansion uses a lower external pressure (or free expansion into vacuum with zero opposing force), so less work is extracted. The work done depends on the path, not just the endpoints. Option A and D are wrong for this reason — W = nRT ln(V_f/V_i) is the formula for the reversible path specifically.
Question 3 True / False
For an ideal gas undergoing reversible isothermal expansion, the internal energy increases because heat is flowing into the gas.
TTrue
FFalse
Answer: False
For an ideal gas, internal energy depends only on temperature. If the process is isothermal (constant T), then ΔU = 0 — internal energy does not change, regardless of how much heat flows in. The absorbed heat Q is entirely converted to work W (by Q = W from the first law with ΔU = 0). This surprises students who expect that adding heat must raise energy, but that is only true if temperature rises. Here the gas absorbs heat and immediately does an equal amount of work, keeping T (and U) constant.
Question 4 True / False
A reversible isothermal expansion produces the maximum possible work for any expansion between the same two equilibrium states.
TTrue
FFalse
Answer: True
The reversible path is the theoretical maximum work output for an expansion. Any irreversibility (finite pressure differences, friction, unresisted expansion) reduces the work extracted below nRT ln(V_f/V_i). In the extreme of free expansion into vacuum, W = 0 for the same change in volume. The reversible path achieves the maximum by maintaining the system at every instant in equilibrium with the largest possible opposing force, extracting work at every step rather than letting any pressure differential go to waste as thermal dissipation.
Question 5 Short Answer
What makes an isothermal expansion 'reversible,' and why does this condition result in maximum work output?
Think about your answer, then reveal below.
Model answer: A reversible expansion is performed infinitely slowly, with the external pressure kept just infinitesimally below the gas pressure at every moment. This keeps the system in a continuous series of equilibrium states. Because the external pressure is always nearly equal to the gas pressure, the gas does work against the maximum possible opposing force at every step, extracting the most work possible. Any faster expansion sets the external pressure lower than the gas pressure, wasting the pressure difference as unrecovered energy rather than useful work.
The concept of reversibility is key to thermodynamic efficiency. A reversible process is one that can be run in reverse along the same path — infinitesimal perturbations can reverse direction. This requires quasi-static conditions (infinitely slow). The Carnot cycle's efficiency derivation depends on both isothermal steps being reversible; if they were irreversible, the engine could not achieve the theoretical maximum efficiency η = 1 − T_C/T_H.