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

Before you start, keep these anchors in mind:

$Δ U = q + w$ in the chemist convention — anything flowing into the system is positive.
$q$ = heat absorbed by system, $w$ = work done on system, $Δ U$ = change in Internal Energy U.
$Δ U$ is a state function; $q$ and $w$ are path functions.

True or false — justify

TF1. "If a system absorbs heat, its temperature must rise."

False. In isothermal expansion the gas absorbs heat but hands it all straight back out as expansion work, so

Δ U = 0

and

T

stays fixed — heat in does not guarantee warming.

TF2. "

Δ U = q + w

only holds for ideal gases."

False. It is conservation of energy for any system — solids, liquids, real gases, reactions. Only special shortcuts like

U = \frac{3}{2} n R T

assume ideal gas.

TF3. "For a complete cycle,

Δ U = 0

, therefore

q = 0

too."

False.

Δ U_{cycle} = 0

forces

q = - w

, not

q = 0

; a cyclic engine absorbs net heat and does equal net work every loop.

TF4. "Work done by the system is always negative in the chemist convention."

True. Energy leaves the system when it does work on the surroundings, so

w < 0

; "out is minus" is the whole point of this convention.

TF5. "Because

q

and

w

change with the path,

Δ U

must change with the path too."

False. Different paths give different

(q, w)

pairs, but they always sum to the same

Δ U

because

U

depends only on the endpoints.

TF6. "In an adiabatic compression the gas cannot get hotter, since no heat enters."

False.

q = 0

, but

w > 0

(work done on it), so

Δ U = w > 0

— all the compression work becomes internal energy and

T

rises. That is how a diesel engine ignites fuel.

TF7. "A bomb calorimeter measures

Δ H

of combustion."

False. It is rigid, so

Δ V = 0 \Rightarrow w = 0 \Rightarrow Δ U = q

. It measures

Δ U

; you convert to $\Delta H$ afterward using

Δ H = Δ U + Δ (P V)

TF8. "If

q > 0

and

w > 0

, then

Δ U

is definitely positive."

True. Both terms feed energy into the system (heat in and work on it), so their sum — and hence

Δ U

— must be positive.

TF9. "At constant volume, no work of any kind can ever be done on the system."

False. Only PV work vanishes when

Δ V = 0

. A stirrer, electric heater, or electrolysis current still delivers non-PV work, so

Δ U = q + w_{non-PV}

Spot the error

SE1. "Gas expands doing 300 J of work, so

w = + 300

J and

Δ U = q + 300

Error: sign. Expansion is work done by the system, so in the chemist convention

w = - 300

J; the correct line is

Δ U = q - 300

SE2. "Isothermal means

q = 0

because temperature doesn't change."

Error: confusing isothermal with adiabatic. Isothermal fixes

T

(so

Δ U = 0

for an ideal gas) but allows heat flow; it is the adiabatic case that sets

q = 0

SE3. "The reaction released 40 kJ, so

q = + 40

kJ."

Error: sign. Released means the system loses heat, so

q = - 40

kJ. Absorbed is

+

, released is

-

SE4. "A rigid insulated box is heated by an internal electric coil; since

Δ V = 0

w = 0

and since it's insulated

q = 0

, so

Δ U = 0

Error: forgot non-PV work. The coil does electrical work on the system,

w_{elec} > 0

, so

Δ U = w_{elec} > 0

even though PV work and heat are zero.

SE5. "We use

Δ U = q - w

because that's what my physics book says."

Error: mixed conventions. Physics defines

w

as work done by the system (hence the minus). Chemistry defines

w

as work done on it, giving

Δ U = q + w

. Pick one and never mix.

SE6. "Since

Δ U

is a state function,

q

measured in a calorimeter equals

Δ U

regardless of conditions."

Error: hidden assumption.

q = Δ U

only when

w = 0

, i.e. constant volume with no other work. At constant pressure

q_{p} = Δ H \neq = Δ U

in general.

SE7. "The system did work, so it lost energy, therefore

Δ U

must be negative."

Error: ignored $q$ .

Δ U = q + w

is the net; if enough heat flows in (

q > ∣ w ∣

Δ U

can be positive despite work being done.

Why questions

WQ1. Why do we write

d U

but

δ q

and

δ w

Because

U

is a state function (an exact differential — integrates to

U_{f} - U_{i}

), while

q

and

w

are path-dependent (inexact differentials whose integral depends on the route taken).

WQ2. Why does a rigid container force

w_{P V} = 0

PV work is

w = - P_{ext} Δ V

; a rigid wall keeps

Δ V = 0

, so the product is zero. No volume change means the system pushes nothing outward.

WQ3. Why does an ideal gas have

Δ U = 0

in any isothermal process?

Its internal energy is purely kinetic,

U = \frac{3}{2} n R T

, a function of

T

alone; if

T

is constant,

U

cannot change no matter how

P

V

move.

WQ4. Why do chemists prefer "work done ON is positive" rather than the physicist sign?

Chemists track energy of the system like a bank account: deposits (heat in, work on) are

+

, withdrawals (

-

). It keeps every "into the system" flow positive and consistent.

WQ5. Why does [[Enthalpy|

H = U + P V

]] exist at all?

At constant pressure,

q_{p} = Δ U + P Δ V = Δ (U + P V)

. Defining

H

this way makes the measured heat equal a state-function change,

Δ H = q_{p}

, which only works because

Δ U = q + w

WQ6. Why can Hess's Law add reaction steps algebraically?

Because

Δ U

(and

Δ H

) are state functions, the total change depends only on start and end states — intermediate steps cancel, so enthalpies of steps simply add.

WQ7. Why do heat and work count as the only two ways energy crosses a boundary?

At the molecular level, energy transfer is either disordered (random collisions = heat) or ordered (a coordinated force through a distance = work). There is no third microscopic mode.

Edge cases

EC1. What is

Δ U

for a free expansion of an ideal gas into a vacuum (insulated)?

q = 0

(insulated) and

w = 0

(nothing to push against,

P_{ext} = 0

), so

Δ U = 0

; temperature is unchanged even though the gas expanded.

EC2. A process has

q = + 500

J and

w = - 500

J. What is

Δ U

, and is anything "wrong"?

Δ U = 0

— perfectly normal. All incoming heat leaves as work; the system is unchanged in energy while acting as a pass-through.

EC3. Can

Δ U

be positive while heat is released (

q < 0

Yes — if enough work is done on the system,

w > ∣ q ∣

, e.g. strong compression of an insulated-ish gas that leaks a little heat but still net-gains energy.

EC4. For a truly isolated system (no heat, no work, no mass exchange), what is

Δ U

Δ U = q + w = 0 + 0 = 0

. Internal energy is fixed no matter what internal rearrangements occur.

EC5. Two paths connect the same initial and final states; path A has

q = 200, w = - 50

J and path B has

q = 100, w = 50

J. Consistent?

Yes — both give

Δ U = 150

J. The pair

(q, w)

differs by path, but their sum must match because

Δ U

is a state function.

EC6. Constant-pressure expansion where the gas both absorbs heat and the temperature rises — where does the heat go?

Part raises internal energy (

Δ U > 0

) and part leaves as expansion work (

w < 0

);

q_{p} = Δ U - w = Δ H

splits between warming the gas and pushing back the atmosphere.

EC7. In an adiabatic process, is

Δ U

path-independent even though

q = 0

throughout?

Δ U

is always path-independent (state function); with

q = 0

it happens to equal

w

, and here

w

becomes path-independent for this constraint, but

Δ U

's independence was never in question.

Recall Quick self-test

Cover the answers and re-verdict TF1, SE1, SE4, and EC1 — these four hit the four most common exam traps (heat≠temperature, sign of work, non-PV work, and $P_{ext} = 0$ ).