Energy added as heat Q minus work done W equals increase in internal energy ΔU. Sign conventions and examples.

What is the First Law of Thermodynamics? — ΔU = Q - W Explained

Question

State the First Law of Thermodynamics and explain the sign convention for heat (Q) and work (W). If 500 J of heat is supplied to a gas and the gas does 200 J of work on its surroundings, what is the change in internal energy?

Solution — Step by Step

The First Law of Thermodynamics states:

\Delta U = Q - W

Here, $\Delta U$ is the change in internal energy, $Q$ is the heat added to the system, and $W$ is the work done by the system. This is just the law of conservation of energy — energy that enters as heat either increases internal energy or leaves as work.

The NCERT convention (which boards follow) is:

$Q > 0$ when heat is absorbed by the system
$Q < 0$ when heat is released by the system
$W > 0$ when work is done by the system (gas expands)
$W < 0$ when work is done on the system (gas compressed)

This is the chemistry/NCERT sign convention. JEE occasionally uses $\Delta U = Q + W$ where $W$ is work done on the system — watch which convention the question uses.

Given: $Q = +500\ \text{J}$ (heat supplied to gas), $W = +200\ \text{J}$ (gas does work on surroundings).

\Delta U = Q - W = 500 - 200 = 300\ \text{J}

$\Delta U = +300\ \text{J}$ means the internal energy of the gas increased by 300 J. The gas absorbed 500 J but “spent” 200 J doing expansion work — the remaining 300 J went into raising the kinetic energy of molecules (temperature increased).

Final Answer: $\Delta U = +300\ \text{J}$

Why This Works

The First Law is simply energy accounting. Think of internal energy like a bank balance — heat is money deposited, and work is money withdrawn. Whatever is left stays in the account.

Internal energy $U$ depends only on the state of the gas (temperature, for an ideal gas). It doesn’t matter whether we reach a final temperature by supplying heat, doing compression work, or some combination — the change $\Delta U$ depends only on the initial and final states. This makes it a state function, unlike $Q$ and $W$ which are path-dependent.

For an ideal gas, $\Delta U = nC_v\Delta T$ — so if you know $\Delta U$ , you immediately know the temperature change. This connection appears repeatedly in JEE problems linking the First Law to specific heat ratios.

Alternative Method — Using the Work-Done-On Convention

Some JEE problems write the law as $\Delta U = Q + W'$ where $W'$ is work done on the system. This is the same physics, just a sign flip.

If the gas does 200 J on surroundings, then work done on the gas is $W' = -200\ \text{J}$ .

\Delta U = Q + W' = 500 + (-200) = 300\ \text{J}

Same answer. When you see $\Delta U = Q + W$ in a JEE solution, it’s using this convention — don’t mix the two in the same problem.

In NCERT and board exams, always use $\Delta U = Q - W$ where $W$ = work done by the system. In JEE, read the question statement — it usually specifies. If not, default to the NCERT convention.

Common Mistake

Flipping the sign of W. Students see “gas does 200 J of work” and write $\Delta U = 500 + 200 = 700\ \text{J}$ , adding instead of subtracting. The logic to remember: when the gas expands and does work, it uses up some of the heat it absorbed — so $W$ is subtracted. Energy going out as work means less energy stays inside.

A related mistake: confusing work done by the gas with work done on the gas when reading problem statements. “Work done against external pressure” = work done by the gas = positive $W$ in the NCERT formula. “Work done on the gas” (compression) = negative $W$ in the NCERT formula.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the Work-Done-On Convention

Common Mistake

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