Thermodynamics · 22.1
The First Law & Gas Processes
Squeeze a gas, heat a gas, let a gas push a piston — every change obeys one budget line: ΔU = Q − W. The p–V diagram turns that budget into a picture, and the area under the curve is the work.
Build the model
Connect the measurement to the mechanism.
A gas holds internal energy U. Heat flowing in raises it; work done by the gas spends it: ΔU = Q − W.
On a p–V diagram the work done by the gas is the area under the path, so different routes between the same two states cost different amounts of heat. Four special processes — constant pressure, constant volume, constant temperature, and no heat flow — cover almost every problem.
- Simple definition
- The first law of thermodynamics says the change in a gas's internal energy equals the heat put in minus the work the gas does: ΔU = Q − W.
- Example
- A gas absorbs 500 J of heat and pushes a piston with 200 J of work — its internal energy rises by exactly 300 J, never more, never less.
Internal energy is a bank balance: heat in is a deposit, work done by the gas is a withdrawal. The signs matter — work done on the gas counts as a deposit.
Q = heat into the gas; W = work done by the gas
A piston of area A pushed a distance d does work pA·d = pΔV. Add up the strips under any curve and the area is the work — different paths, different work.
Constant pressure: W = pΔV
Each process pins one quantity: isobaric holds pressure, isochoric holds volume so no work is done, isothermal holds temperature so internal energy is unchanged, adiabatic lets no heat through.
p fixed · V fixed (W = 0) · T fixed (ΔU = 0) · Q = 0
U is a state variable — Q and W are not
The internal energy of an ideal gas depends only on where you are on the p–V diagram (in fact only on T). Heat and work depend on the route taken between two states, which is why 'the heat in the gas' is a meaningless phrase.
Reading the four paths
On the p–V diagram an isobaric process is a horizontal line, an isochoric one is vertical, an isotherm is the curve pV = constant, and an adiabat is a steeper curve pV^γ = constant. Steeper because compression work heats the gas instead of leaking away.
Adiabatic means insulated — or fast
Adiabatic is Greek for 'nothing gets through': no heat enters or leaves, so ΔU = −W. Compressing air in a bicycle pump warms it; air rising and expanding in the atmosphere cools. Rapid processes are nearly adiabatic because heat has no time to flow.
Change one variable at a time
Make the relationship visible.
Choose a process
The shaded area under the path is the work done by the gas. Watch how the same volume change costs different heat on different paths — and how the adiabat drops more steeply than the isotherm.
Held fixedpressure
Work by gas W300 J
ΔU (monatomic)450 J
Heat in Q = ΔU + W750 J
T₂/T₁ = p₂V₂/p₁V₁2
Catch the common trap
Explain before calculating.
An ideal gas expands isothermally, doing 400 J of work on the piston. How much heat flows into the gas?
Choose an answer to test the model.
Worked examples
State the rule, substitute, then check units.
EasyA gas absorbs 500 J of heat and does 200 J of work on a piston. Find the change in its internal energy.
- First law: ΔU = Q − W.
- ΔU = 500 − 200 = 300 J.
AnswerΔU = +300 J
MediumA gas at a constant pressure of 1.0 × 10⁵ Pa expands from 2.0 × 10⁻³ m³ to 5.0 × 10⁻³ m³ while absorbing 750 J of heat. Find the work done by the gas and the change in internal energy.
- Isobaric work: W = pΔV = 1.0 × 10⁵ × 3.0 × 10⁻³ = 300 J.
- First law: ΔU = Q − W = 750 − 300.
- ΔU = 450 J — the rest of the heat became piston work.
AnswerW = 300 J; ΔU = +450 J
HardA gas is first heated at constant volume with 400 J of heat, then expands at a constant 2.0 × 10⁵ Pa from 1.0 × 10⁻³ m³ to 2.5 × 10⁻³ m³ while absorbing another 750 J. Find the total work done and the total change in internal energy.
- Isochoric leg: V fixed, so W₁ = 0 and ΔU₁ = Q₁ = 400 J.
- Isobaric leg: W₂ = pΔV = 2.0 × 10⁵ × 1.5 × 10⁻³ = 300 J; ΔU₂ = 750 − 300 = 450 J.
- Totals: W = 0 + 300 = 300 J; ΔU = 400 + 450 = 850 J.
AnswerW = 300 J; ΔU = +850 J
ChallengingAn insulated cylinder holds 0.080 mol of monatomic ideal gas (U = 3/2 nRT). A piston compresses it, doing 300 J of work on the gas. Find ΔU and the temperature rise.
- Adiabatic: Q = 0. Work done BY the gas is −300 J, so ΔU = Q − W = 0 − (−300) = +300 J.
- For a monatomic gas ΔU = 3/2 nRΔT, so ΔT = 300 ÷ (1.5 × 0.080 × 8.31).
- ΔT = 300 ÷ 0.997 ≈ 300 K — squeezing an insulated gas heats it dramatically.
AnswerΔU = +300 J; ΔT ≈ 300 K