First law - ΔU = q + w (chemist sign convention): Exercises

Before we start, one picture to fix the sign convention in your eyes — glance back at it whenever a sign feels slippery.

Figure — First law - ΔU = q + w (chemist sign convention)

The system is the box. Two arrows can cross its wall: the heat arrow ( $q$ ) and the work arrow ( $w$ ). If an arrow points inward, that quantity is positive. If it points outward, negative. That is the whole convention.

Level 1 — Recognition

Goal: read a situation and assign the correct sign to $q$ and $w$ before touching arithmetic.

Exercise 1.1

A gas is compressed by a piston; during the push, the gas gives off $250 J$ of heat to the room. State the sign of $q$ and the sign of $w$ (chemist convention). No number for $Δ U$ needed — just the signs and a one-line reason each.

Recall Solution 1.1

Heat: the gas gives off heat → energy leaves the system → $q < 0$ . Specifically $q = - 250 J$ .
Work: the piston compresses the gas → surroundings push in → energy enters → $w > 0$ .

Match this against the arrow picture: the heat arrow points out (negative), the work arrow points in (positive). Signs: $q = -$ , $w = +$ .

Exercise 1.2

For each phrase, give the sign of $w$ : (a) "the gas expands against the atmosphere", (b) "the gas is squeezed to half its volume", (c) "the volume is held fixed while stirring is off".

Recall Solution 1.2

Recall PV work: $w = - P_{ext} Δ V$ .

(a) Expansion: $Δ V > 0$ , so $w = - P_{ext} Δ V < 0$ . Gas does work ON surroundings → energy leaves → $w < 0$ . ✅
(b) Compression: $Δ V < 0$ , so $- P_{ext} Δ V > 0$ → $w > 0$ . ✅
(c) Fixed volume, no other work: $Δ V = 0 \Rightarrow w = 0$ . ✅

Level 2 — Application

Goal: plug correct signs into $Δ U = q + w$ and get a number.

Exercise 2.1

A system absorbs $600 J$ of heat and does $220 J$ of work on its surroundings. Find $Δ U$ .

Recall Solution 2.1

Heat absorbed → $q = + 600 J$ .
Work done BY the system → outward arrow → $w = - 220 J$ . $Δ U = q + w = 600 + (- 220) = + 380 J$ The system kept $380 J$ : it took in $600$ , spent $220$ pushing on the world.

Exercise 2.2

$0.50 mol$ of gas is heated in a rigid flask, absorbing $1.20 kJ$ . Find $Δ U$ and $w$ .

Recall Solution 2.2

Rigid → $Δ V = 0 \Rightarrow w = - P_{ext} Δ V = 0$ .
$q = + 1.20 kJ$ (absorbed). $Δ U = q + w = 1.20 + 0 = + 1.20 kJ$ This is the constant-volume situation: with $w = 0$ , all heat becomes internal energy, so $Δ U = q$ .

Exercise 2.3

A gas releases $450 J$ of heat while being compressed, and $700 J$ of work is done on it. Find $Δ U$ .

Recall Solution 2.3

Heat released → $q = - 450 J$ .
Work done ON system → $w = + 700 J$ . $Δ U = - 450 + 700 = + 250 J$ Even though heat left, the compression added more energy than heat carried away, so $U$ rose.

Level 3 — Analysis

Goal: use physical constraints (ideal-gas facts, process type) to pin down a missing quantity.

Exercise 3.1 — Isothermal ideal gas

$3 mol$ of ideal gas expands isothermally at $310 K$ , doing $1500 J$ of work on the surroundings. Find $Δ U$ and $q$ .

Recall Solution 3.1

Ideal gas internal energy depends only on temperature: $U = U (T)$ . See Isothermal processes. Isothermal → $Δ T = 0 \Rightarrow Δ U = 0$ .
Work done BY system → $w = - 1500 J$ .
First law: $0 = q + (- 1500) \Rightarrow q = + 1500 J$ .

Interpretation: heat flowed in to exactly replace the energy spent as expansion work, keeping $T$ (and $U$ ) fixed.

Exercise 3.2 — Adiabatic compression

A gas is compressed adiabatically; $920 J$ of work is done on it. Find $q$ and $Δ U$ .

Recall Solution 3.2

Adiabatic (Adiabatic processes) → no heat crosses the wall → $q = 0$ .
Work ON system → $w = + 920 J$ . $Δ U = 0 + 920 = + 920 J$ Every joule of work becomes internal energy — the gas heats up (diesel-ignition principle).

Exercise 3.3 — Bomb calorimeter

Combustion of a sample in a rigid bomb releases $22.8 kJ$ to the water bath. Find $Δ U$ of the reaction, and explain why a bomb reports $Δ U$ rather than $Δ H$ .

Recall Solution 3.3

Rigid bomb → $Δ V = 0 \Rightarrow w = 0$ . See Bomb calorimetry.
Exothermic release → $q = - 22.8 kJ$ . $Δ U = q + w = - 22.8 + 0 = - 22.8 kJ$ Because $w = 0$ at constant volume, the measured heat equals $Δ U$ directly. To get $Δ H$ (Enthalpy) you would need the constant-pressure heat, which the bomb does not provide.

Level 4 — Synthesis

Goal: combine PV-work calculation, unit care, and the first law in one problem.

Exercise 4.1 — Compute the PV work yourself

A gas expands against a constant external pressure of $1.00 atm$ , from $2.0 L$ to $6.0 L$ , while absorbing $600 J$ of heat. Find $w$ (in joules) and $Δ U$ . (Use $1 L\cdotatm = 101.325 J$ .)

Recall Solution 4.1

$Δ V = 6.0 - 2.0 = + 4.0 L$ .
PV work (see PV work): $w = - P_{ext} Δ V = - (1.00 atm) (4.0 L) = - 4.0 L\cdotatm$ .
Convert: $w = - 4.0 \times 101.325 = - 405.3 J$ (work done BY the gas, so negative — good). $Δ U = q + w = 600 + (- 405.3) = + 194.7 J$ The system netted about $+ 195 J$ : absorbed $600$ , spent $405$ pushing the atmosphere back.

Exercise 4.2 — Non-PV work at constant volume

In a sealed rigid vessel a stirrer does $150 J$ of shaft work on the fluid, and the fluid loses $40 J$ of heat to the walls. Find $Δ U$ . Is $w = 0$ here? Explain.

Recall Solution 4.2

Volume is fixed, so PV work is zero — but the stirrer supplies non-PV (shaft) work, which is NOT zero: $w_{shaft} = + 150 J$ (energy pushed into the fluid).
Heat lost → $q = - 40 J$ . $Δ U = q + w = - 40 + 150 = + 110 J$ So $w = 0$ is a claim about PV work only. Whenever a stirrer, current, or other mechanism acts, you must include that work term.

Level 5 — Mastery

Goal: chain steps, exploit the state-function property, and close a cycle.

Exercise 5.1 — Path independence of $Δ U$

A system goes from state A to state B by two different routes:

Route 1: absorbs $800 J$ of heat, does $300 J$ of work on surroundings.
Route 2: absorbs $q_{2}$ of heat, has $150 J$ of work done ON it.

Find $Δ U$ from Route 1, then find $q_{2}$ that Route 2 must have.

Recall Solution 5.1

Route 1: $q = + 800$ , $w = - 300$ (work done BY system). $Δ U = 800 - 300 = + 500 J$ Because $U$ is a state function, $Δ U$ between the same A and B is $+ 500 J$ on any route.

Route 2: $w_{2} = + 150$ (work done ON system). Solve $Δ U = q_{2} + w_{2}$ : $500 = q_{2} + 150 \Rightarrow q_{2} = + 350 J$ Different $(q, w)$ pairs — $(800, - 300)$ vs $(350, + 150)$ — but they sum to the same $Δ U$ . That is the whole point of a state function.

Exercise 5.2 — Closing a cycle

A gas runs a cyclic process A→B→C→A. The step energies are:

A→B: $Δ U_{1} = + 250 J$
B→C: $Δ U_{2} = - 400 J$

Find $Δ U_{3}$ for C→A, and the net work over the cycle if the total heat absorbed over the whole cycle is $q_{cycle} = + 90 J$ .

Recall Solution 5.2

Step 1 — internal energy closes. Over any complete cycle the system returns to its start, so $Δ U_{cycle} = 0$ (see Carnot cycle for the same idea in an engine). $Δ U_{1} + Δ U_{2} + Δ U_{3} = 0$ $250 + (- 400) + Δ U_{3} = 0 \Rightarrow Δ U_{3} = + 150 J$ Step 2 — first law over the cycle. With $Δ U_{cycle} = 0$ : $0 = q_{cycle} + w_{cycle} \Rightarrow w_{cycle} = - q_{cycle} = - 90 J$ Net work is $- 90 J$ : the system did $90 J$ of net work on the surroundings (outward arrow), fuelled exactly by the $90 J$ of net heat it absorbed. Energy books balance around the loop.

Exercise 5.3 — Full accounting

$1 mol$ of ideal gas is taken through: (i) isothermal expansion at $400 K$ absorbing $2000 J$ , then (ii) adiabatic compression with $1300 J$ of work done on it. Find $Δ U$ for step (i), $Δ U$ for step (ii), and total $Δ U$ .

Recall Solution 5.3

Step (i), isothermal: $Δ T = 0 \Rightarrow Δ U_{i} = 0$ . (Heat and work cancel: $q = + 2000$ , so $w = - 2000 J$ .) Step (ii), adiabatic: $q = 0$ , $w = + 1300 J$ : $Δ U_{ii} = 0 + 1300 = + 1300 J$ Total: $Δ U_{total} = Δ U_{i} + Δ U_{ii} = 0 + 1300 = + 1300 J$ . The isothermal leg changes $U$ by nothing; only the adiabatic compression raises internal energy.

Recall Self-test cloze

In the chemist convention, work done ON the system is positive. For an isothermal ideal-gas process, $Δ U =$ ::: $0$ For an adiabatic process, $q =$ ::: $0$ Around any complete cycle, $Δ U_{cycle} =$ ::: $0$ At constant volume the PV work is zero, but non-PV (shaft/electrical) work may not be.