Questions: Thermodynamic Processes and the PV Diagram

Question 1

An ideal gas undergoes isothermal expansion. Which relationship correctly describes the heat Q, work W, and internal energy change ΔU for this process?

• A. Q = 0 and W = 0, so ΔU = 0
• B. ΔU = 0 and Q = W, so heat absorbed equals work done by the gas
• C. Q = 0 and W = −ΔU, so work comes entirely from internal energy
• D. ΔU > 0 and Q = 0, so all work input goes into internal energy

Answer: (B)

For an ideal gas, internal energy depends only on temperature. If temperature is constant (isothermal), ΔU = 0. The first law then gives 0 = Q − W, so Q = W. The gas does absorb heat — that heat is entirely converted into work done on the surroundings. This is why isothermal ≠ Q = 0.

Question 2

An adiabatic expansion of an ideal gas necessarily occurs at constant temperature.

• T. True
• F. False

Answer: False

Adiabatic means Q = 0, not T = constant. With no heat input, the first law gives ΔU = −W. When the gas expands and does positive work on its surroundings, its internal energy decreases — and for an ideal gas, that means temperature drops. Adiabatic and isothermal are distinct processes; conflating them is a common error.

Question 3

On a PV diagram, why is an adiabatic curve steeper than an isothermal curve passing through the same point?

Model answer: For an isothermal process, PV = constant, giving a slope of −P/V. For an adiabatic process, PV^γ = constant (where γ = Cp/Cv > 1), giving a slope of −γP/V. Since γ > 1, the adiabatic curve falls more steeply. Physically, in an adiabatic expansion the temperature drops (no heat input), so pressure falls faster than it would at constant temperature.

The steepness difference is a direct consequence of the adiabatic temperature drop. On an isothermal curve, T stays fixed so the ideal gas law keeps pressure proportional to 1/V. On the adiabatic, the gas cools as it expands, compounding the pressure drop — hence the steeper descent.