First law of thermodynamics — dU = dQ − dW, sign conventions: Exercises

For the parent theory see the main note and its neighbours below.

Level 1 — Recognition

L1·1 — Name the sign

A gas absorbs $250 J$ of heat. State $Q$ with its sign, and say in words which channel this energy entered by.

Recall Solution

WHAT: "Absorbs" = heat flows into the gas, so $Q = + 250 J$ . WHY: In physics convention $Q > 0$ means heat added TO the gas. PICTURE: heat lining up outside and stepping in — the queue in the "QUWe by" mnemonic. $Q = + 250 J$

L1·2 — Which term is zero?

A gas is heated inside a rigid sealed box (its volume cannot change). Which of $Q$ , $W$ , $Δ U$ is forced to be zero, and why?

Recall Solution

WHAT: $W = 0$ . WHY: $W = \int P d V$ . Rigid box $\Rightarrow d V = 0 \Rightarrow W = 0$ . This is an isochoric (constant-volume) process — see Isothermal, adiabatic, isobaric, isochoric processes. PICTURE: the piston is bolted in place; the gas cannot push anything, so it does no work. Then $Δ U = Q$ : all heat becomes internal energy.

Level 2 — Application

L2·1 — Heat in, gas expands

A gas absorbs $600 J$ of heat and expands, doing $220 J$ of work on its piston. Find $Δ U$ .

Recall Solution

Signs: $Q = + 600$ (added), $W = + 220$ (gas expands → work by gas). Apply: $Δ U = Q - W = 600 - 220 = + 380 J$ . PICTURE: of $600 J$ that walked in, $220 J$ walked back out through the piston; $380 J$ stayed → hotter gas.

L2·2 — Compression with heat loss

A gas is compressed; the surroundings do $180 J$ of work on it, and it releases $70 J$ of heat. Find $Δ U$ .

Recall Solution

Signs: work on gas $= 180 J \Rightarrow W_{by} = - 180 J$ . Heat released $\Rightarrow Q = - 70 J$ . Apply: $Δ U = Q - W = (- 70) - (- 180) = + 110 J$ . PICTURE: $180 J$ pumped in by compression, $70 J$ leaked out as heat → net $+ 110 J$ stored, gas warms.

L2·3 — Isobaric work

A gas expands at constant pressure $P = 2.0 \times 1 0^{5} Pa$ from $V_{1} = 1.0 \times 1 0^{- 3} m^{3}$ to $V_{2} = 3.0 \times 1 0^{- 3} m^{3}$ . Find the work $W$ done by the gas.

Recall Solution

WHY constant $P$ is easy: $W = \int_{V_{1}}^{V_{2}} P d V$ ; with $P$ constant it slides outside the integral, leaving $W = P (V_{2} - V_{1}) = P Δ V$ . Apply: $W = (2.0 \times 1 0^{5}) (3.0 \times 1 0^{- 3} - 1.0 \times 1 0^{- 3}) = (2.0 \times 1 0^{5}) (2.0 \times 1 0^{- 3}) = 400 J$ . PICTURE (below): on a $P$ – $V$ diagram the isobar is a flat horizontal line; work is the rectangle under it.

Figure — First law of thermodynamics — dU = dQ − dW, sign conventions

Level 3 — Analysis

L3·1 — Isothermal expansion, find $Q$

An ideal gas expands isothermally ( $T$ constant) and does $520 J$ of work. Find $Q$ and $Δ U$ .

Recall Solution

Key fact: For an ideal gas $U = U (T)$ only (see Internal energy and degrees of freedom). Constant $T \Rightarrow Δ U = 0$ . Apply: $0 = Q - W \Rightarrow Q = W = 520 J$ . PICTURE: every joule of heat that enters walks straight out as work; the internal "warmth" level never moves.

L3·2 — Adiabatic compression, find the trend

A gas is compressed adiabatically; $310 J$ of work is done on it. Find $Δ U$ and state whether it heats or cools.

Recall Solution

Key facts: Adiabatic $\Rightarrow Q = 0$ . Work on gas $\Rightarrow W_{by} = - 310 J$ . Apply: $Δ U = Q - W = 0 - (- 310) = + 310 J > 0 \Rightarrow$ gas heats up. WHY: With no heat channel, the only place the compression energy can go is internal energy. This is why a bicycle pump gets warm.

L3·3 — Cyclic process, net heat

A gas is taken around a closed cycle and returns to its exact starting state. Over the cycle the total work done by the gas is $+ 150 J$ . Find $Δ U_{cycle}$ and the net heat $Q_{cycle}$ .

Recall Solution

Key fact: $U$ is a state function. Start state = end state $\Rightarrow Δ U_{cycle} = 0$ , no matter the path. Apply: $0 = Q_{cycle} - W_{cycle} \Rightarrow Q_{cycle} = W_{cycle} = + 150 J$ . PICTURE (below): on a $P$ – $V$ diagram a cycle is a closed loop; the enclosed area is the net work, and here that area supplies the net heat absorbed.

Level 4 — Synthesis

L4·1 — Two-step path, same endpoints

A gas goes from state A to state B by Path 1: it absorbs $900 J$ of heat and does $350 J$ of work. By Path 2 (different route, same A→B), the gas does only $120 J$ of work. How much heat $Q_{2}$ does Path 2 involve?

Recall Solution

Step 1 — get $Δ U$ from Path 1. $Δ U = Q_{1} - W_{1} = 900 - 350 = + 550 J$ . Step 2 — reuse it. $Δ U$ depends only on endpoints A and B, so Path 2 has the same $Δ U = + 550 J$ . Step 3 — solve for $Q_{2}$ . $550 = Q_{2} - 120 \Rightarrow Q_{2} = 670 J$ . WHY this works: $U$ is a state function — this is the whole point of the lake analogy. Change the path, and $Q$ and $W$ change together, but their difference is pinned.

L4·2 — Free expansion into vacuum

An ideal gas in a chamber is suddenly connected to an evacuated chamber and rushes in (free expansion). The whole system is insulated. Find $W$ , $Q$ , $Δ U$ , and $Δ T$ .

Recall Solution

$W = 0$ : the gas expands against vacuum — there is nothing pushing back, so $\int P_{ext} d V = 0$ . No piston is moved, no surroundings are lifted. $Q = 0$ : insulated walls, so no heat crosses. $Δ U = Q - W = 0 - 0 = 0$ . $Δ T = 0$ : ideal gas $U = U (T)$ ; $Δ U = 0 \Rightarrow T$ unchanged. WHY it surprises people: the gas expands yet does no work and doesn't cool — because "expanding" only costs energy when you push against something.

Level 5 — Mastery

L5·1 — Cycle leg-by-leg energy audit

An ideal gas runs a 3-step cycle A→B→C→A:

A→B (isochoric heating): absorbs $Q_{A B} = 600 J$ .
B→C (isothermal expansion): absorbs $Q_{B C} = 450 J$ .
C→A (isobaric compression): releases $Q_{C A} = - 720 J$ , and $250 J$ of work is done on the gas.

Find (a) $W$ for each leg, (b) total work by the gas over the cycle, (c) verify $Δ U_{cycle} = 0$ .

Recall Solution

Leg A→B (isochoric): $d V = 0 \Rightarrow W_{A B} = 0$ . Then $Δ U_{A B} = Q_{A B} - W_{A B} = 600 J$ .

Leg B→C (isothermal, ideal gas): $Δ U_{B C} = 0$ (constant $T$ ). So $W_{B C} = Q_{B C} - Δ U_{B C} = 450 - 0 = 450 J$ .

Leg C→A (isobaric compression): work $250 J$ done on gas $\Rightarrow W_{C A} = - 250 J$ . Then $Δ U_{C A} = Q_{C A} - W_{C A} = (- 720) - (- 250) = - 470 J$ .

(b) Net work by gas: $W_{cycle} = W_{A B} + W_{B C} + W_{C A} = 0 + 450 - 250 = + 200 J$ .

(c) Check $Δ U_{cycle}$ : $Δ U_{A B} + Δ U_{B C} + Δ U_{C A} = 600 + 0 - 470 = + 130 J$ .

This is NOT zero — the numbers are inconsistent! A true cycle must have $Δ U_{cycle} = 0$ . Cross-checking with heat: $Q_{cycle} = 600 + 450 - 720 = + 330 J$ , but $W_{cycle} = 200 J$ , and a cycle requires $Q_{cycle} = W_{cycle}$ . The mismatch ( $330 \neq = 200$ ) exposes an over-specified, unphysical data set.

The mastery lesson: the First Law is a consistency check. On a genuine cycle you may specify heats and works freely on individual legs, but they must satisfy $\sum Δ U = 0$ and $Q_{cycle} = W_{cycle}$ . Always audit the loop before trusting the numbers.

L5·2 — Repair the cycle

Keep everything in L5·1 except the heat released on C→A. Find the value of $Q_{C A}$ that makes the cycle physically consistent, then state the cycle's net heat and net work.

Recall Solution

Requirement: $Δ U_{A B} + Δ U_{B C} + Δ U_{C A} = 0$ . We keep $W_{C A} = - 250 J$ , so $Δ U_{C A} = Q_{C A} - (- 250) = Q_{C A} + 250$ . $600 + 0 + (Q_{C A} + 250) = 0 \Rightarrow Q_{C A} = - 850 J .$ Net heat: $Q_{cycle} = 600 + 450 - 850 = + 200 J$ . Net work: unchanged, $W_{cycle} = 0 + 450 - 250 = + 200 J$ . Check: $Q_{cycle} = W_{cycle} = 200 J$ ✓ and $Δ U_{cycle} = 0$ ✓. Now it's a legal engine cycle putting out $200 J$ of work per loop.

Connections

Work done in thermodynamic processes (PV diagrams)
Isothermal, adiabatic, isobaric, isochoric processes
Internal energy and degrees of freedom
Heat capacities Cp and Cv
Conservation of energy (mechanics)
Second law of thermodynamics