Thermodynamics: Conceptual Doubts Cleared (6)

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Question

A student claims: “In an adiabatic process, Q=0Q = 0, so the internal energy of the gas does not change.” Another student says: ”ΔU\Delta U depends on temperature change, and adiabatic processes do change temperature, so ΔU0\Delta U \neq 0.” Who is right? Resolve the confusion using the first law.

Solution — Step by Step

The first law of thermodynamics says:

ΔU=QW\Delta U = Q - W

where QQ is heat added to the system and WW is work done by the system. Some textbooks write ΔU=Q+W\Delta U = Q + W where WW is work done on the system — same physics, different sign convention. We’ll use the JEE/NCERT convention: ΔU=QW\Delta U = Q - W.

In an adiabatic process, no heat is exchanged with the surroundings: Q=0Q = 0. So the first law becomes:

ΔU=W\Delta U = -W

The change in internal energy equals the negative of the work done by the gas. If the gas expands (W>0W > 0), ΔU<0\Delta U < 0 — internal energy drops, temperature falls. If the gas is compressed (W<0W < 0), ΔU>0\Delta U > 0 — temperature rises.

Student 1 is wrong. The condition Q=0Q = 0 does not imply ΔU=0\Delta U = 0. It only means heat doesn’t flow in or out. The internal energy still changes because the gas does work (or has work done on it).

Student 2 is right. For an ideal gas, ΔU=nCVΔT\Delta U = n C_V \Delta T. Adiabatic compression heats the gas, adiabatic expansion cools it — both change ΔU\Delta U.

Why This Works

The first law has two ways to change internal energy: heat transfer (QQ) and work (WW). Setting one to zero (Q=0Q = 0 for adiabatic) doesn’t kill the other. This is the conceptual trap NEET examiners exploit constantly.

The sister case is isothermal: ΔT=0\Delta T = 0, so ΔU=nCVΔT=0\Delta U = n C_V \Delta T = 0, which means Q=WQ = W. Heat flows in but immediately becomes work, leaving internal energy unchanged.

Memorize this 2x2 matrix:

ProcessQQWWΔU\Delta U
Isothermal=W= W0\neq 000
Adiabatic000\neq 0W-W
Isobaric0\neq 0PΔVP\Delta VnCVΔTnC_V\Delta T
Isochoric0\neq 000=Q= Q

Every NEET thermodynamics MCQ tests one cell of this table.

Alternative Method

Reason from the molecular picture. In adiabatic expansion, gas molecules push against the piston and lose kinetic energy (no heat is replenished). Lower KE means lower temperature, which means lower internal energy.

The biggest student error: assuming Q=0Q = 0 means ΔU=0\Delta U = 0. Always write the first law explicitly before concluding.

Final answer: Student 2 is right. ΔU0\Delta U \neq 0 in adiabatic processes — work changes internal energy even when heat doesn’t.

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