Work done in each process — derivation: Question bank

Prerequisites live in the parent note and in First Law of Thermodynamics, P-V Diagrams, Adiabatic Process and PV^gamma, Isothermal Process and Boyle's Law, Internal Energy and Cv, and Carnot Cycle.

True or false — justify

Isothermal work is zero because the temperature never changes

False.

Δ T = 0

makes

Δ U = 0

, not

W = 0

; the piston still slides out and

W = n R T ln (V_{2} / V_{1}) \neq = 0

. Constant temperature does not mean constant volume.

Work done by a gas depends only on the start and end states

False. Work is path-dependent: two curves joining the same two points on a

P

–

V

diagram enclose different areas, so give different

W

. Only state functions (like

U

) depend on endpoints alone.

In an isochoric process the gas can still do work if pressure changes

False. Work needs a moving piston, i.e.

d V \neq = 0

. If

V

is fixed,

d V = 0

everywhere and

W = \int P d V = 0

no matter how wildly

P

swings.

For the same volume change, isobaric work is always larger than isothermal expansion work

True (for the same start point). In isothermal expansion pressure drops as volume grows, so the area under the isotherm is smaller than the rectangle under the constant (starting) pressure line.

In adiabatic expansion the gas does positive work

True. It expands (

V_{2} > V_{1}

), and with no heat in (

Q = 0

) that work is paid for by internal energy, so the gas cools (

T_{2} < T_{1}

) and

W = \frac{n R ( T _{1} - T _{2} )}{γ - 1} > 0

A gas that is compressed does negative work

True. "Work done by the gas" is

\int P d V

; compression means

V_{2} < V_{1}

, so the integral is negative — the surroundings do work on the gas.

The area enclosed by any cyclic loop equals the net heat absorbed

True (over the full cycle).

Δ U = 0

for a cycle, so the first law gives

Q_{n e t} = W_{n e t} =

enclosed area — but only for the whole loop, not for a single leg.

An anticlockwise cycle produces positive net work

False. Clockwise loops give

W_{n e t} > 0

(engine); anticlockwise loops give

W_{n e t} < 0

(refrigerator/heat pump), work done on the gas.

Spot the error

"Isothermal work is

P Δ V

because that worked for the isobaric case."

The isobaric shortcut only works because

P

is constant and comes out of the integral. In isothermal,

P = n R T / V

varies, so you must integrate

\int n R T d V / V

and get a logarithm, not a product.

"Adiabatic work is

\frac{P _{2} V _{2} - P _{1} V _{1}}{γ - 1}

Sign is flipped. It must be

\frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1}

so that expansion (

P_{1} V_{1} > P_{2} V_{2}

in the cooler final state) gives

W > 0

. The starting-state term comes first.

"For an isobaric process

W = n R Δ T

only if the gas is monatomic."

γ

never enters isobaric work.

W = P Δ V = n R Δ T

follows straight from

P V = n R T

and holds for any ideal gas regardless of atomicity.

"Since

Q = 0

in an adiabatic process, no energy leaves the gas, so its temperature is unchanged."

Energy does leave — as work, not heat. With

Q = 0

the first law gives

Δ U = - W

, so doing positive work drains

U

and the gas cools.

W = \int P d V

needs the gas to be ideal."

The master formula

d W = P d V

comes from force

= P A

and geometry alone — it holds for any substance. Ideality is only used later to write

P

as a specific function of

V

(e.g.

n R T / V

"On a

P

–

V

diagram, work is the area between the curve and the

P

-axis (vertical axis)."

No — work is the area under the curve down to the ==

V

-axis (horizontal)==, because we integrate over

d V

. Area to the

P

-axis would be

\int V d P

, a different quantity.

"In a cyclic process the total work is zero because it returns to the start."

Only

Δ U

and

Δ T

return to zero. Work is a process quantity: the enclosed area is generally nonzero, which is exactly how engines produce net output.

Why questions

Why does isothermal work involve a logarithm while isobaric does not

Because in isothermal

P = n R T / V

depends on

V

, and

\int d V / V = ln V

; in isobaric

P

is constant and factors out, leaving a plain

\int d V = Δ V

Why can we pull

P

out of the integral for an isobaric process but not an isothermal one

A constant may leave the integral; a varying quantity may not. Isobaric

P

is fixed by definition, but isothermal

P

falls as the gas spreads, so it must stay inside.

Why does an adiabatically expanding gas cool while an isothermally expanding gas does not

The isothermal gas absorbs heat from a reservoir to replace the energy it spends as work, keeping

T

fixed. The adiabatic gas is sealed off (

Q = 0

), so it spends its own internal energy and

T

drops.

Why is work called path-dependent when energy conservation still holds

Conservation applies to

U

(a state function):

Δ U = Q - W

. But

Q

and

W

individually depend on the route taken; only their combination

Q - W

is fixed by the endpoints.

Why does the steeper adiabatic curve lie below the isotherm through the same point during expansion

Because

P V^{γ} =

const falls faster (

γ > 1

) than

P V =

const. So for the same expansion the adiabat drops lower, enclosing less area and giving less work than the isotherm.

Why does an isochoric process send all its heat into internal energy

With

d V = 0

W = 0

, so the first law

Δ U = Q - W

becomes

Δ U = Q

; every joule of heat raises the internal energy (and temperature) directly.

Why does the direction of traversal (clockwise vs anticlockwise) decide the sign of cyclic work

The enclosed area is swept out during expansion (bottom of loop, moving right,

+ W

) versus compression (top, moving left,

- W

). Clockwise makes the expansion leg dominate, so net work is positive.

Edge cases

What is the work if the initial and final volumes are equal but the path was not isochoric (a loop that returns to the same

V

)

Not necessarily zero.

\int P d V

over a curve that leaves and returns to the same

V

can enclose area; only a truly fixed-

V

(vertical-line) path gives

W = 0

What happens to isothermal work as

V_{2} \to V_{1}

(vanishing expansion)

ln (V_{2} / V_{1}) \to ln 1 = 0

, so

W \to 0

smoothly — no volume change, no work, consistent with the master formula.

What is adiabatic work in the limit

T_{2} \to T_{1}

(negligible cooling)

W = \frac{n R ( T _{1} - T _{2} )}{γ - 1} \to 0

. If temperature barely changes, almost no internal energy is spent, so almost no work is done.

Can work be negative even during an expansion

For a single gas pushing out, no —

d V > 0

and

P > 0

force

\int P d V > 0

. Negative work requires compression (

d V < 0

) somewhere along the path.

What is the work along a vertical line on a

P

–

V

diagram (pressure changes at fixed volume)

Zero. A vertical segment has

d V = 0

throughout, so

\int P d V = 0

— this is exactly an isochoric leg, e.g. a side of many cycles.

Is the free expansion of a gas into vacuum an example of

W = n R T ln (V_{2} / V_{1})

No. In free expansion there is no opposing pressure (

P_{e x t} = 0

), so no force acts through the moving boundary and

W = 0

, even though the gas volume increases.

What is

W

for a horizontal line on a

P

–

V

diagram

A horizontal segment is constant pressure, so

W = P (V_{2} - V_{1})

— the plain isobaric rectangle area under that line.

Recall Quick self-test

Path-dependent means work depends on... ::: the curve taken, not just the endpoints. $Δ U = 0$ implies... ::: $Q = W$ , not $W = 0$ . Free expansion into vacuum does work equal to... ::: zero (no opposing pressure). Anticlockwise cyclic loop gives net work that is... ::: negative (done on the gas).