JEE/NEET Physics · Thermodynamics series · Part 1 of 6 · All parts →
- Zeroth law: bodies in contact equalize temperature — that’s what thermometers use
- Internal energy U = the total molecular jiggle + interaction energy of the gas
- First law: ΔQ = ΔU + W — heat in = stored energy + work out
- Sign convention: Q positive entering, W positive done BY the gas (physics style)
- The first law is energy conservation wearing thermal clothes
Put ₹100 of heat into a gas: some is stored as molecular jiggle, some is spent pushing a piston. The first law is simply the receipt — heat in = energy stored + work done. Nothing more, and nothing less. Part 1 of the Thermodynamics series — the law that runs every engine.
- The zeroth law: thermometers’ licence
- Internal energy: the gas’s bank balance
- The first law receipt
- Sign conventions that save problems
- Solved examples
- Common mistakes
- This physics in your daily life
- Practice set
- Recap
The Zeroth Law: Thermometers’ Licence
Two bodies in contact equalize temperature (zeroth law) — and if A matches a thermometer and B matches the same thermometer, A and B match each other. This unglamorous law is what makes temperature MEASURABLE at all.
Internal Energy: The Gas’s Bank Balance
U is the sum of all molecular kinetic energy (the jiggling) plus interaction energy. For an ideal gas, U depends only on temperature: heat it, U rises; cool it, U falls — pressure and volume don’t get a vote. ΔU = nC_vΔT always.
The First Law Receipt
| Letter | What it means (plain words) | Value / unit |
|---|---|---|
| Q | heat supplied TO the gas | J; positive in, negative out |
| U | internal energy — the gas’s jiggle account | J; ideal gas: U(T) only |
| W | work done BY the gas | J = area under P-V curve; positive on expansion |
Sign Conventions That Save Problems
| Event | Q | W | ΔU |
|---|---|---|---|
| Gas heated, expands | + | + | >/≠0 |
| Gas compressed, cooled | − | − | usually − |
| Gas expands, doing work, no heat (adiabatic) | 0 | + | − (pays from savings) |
| Heat added at fixed volume | + | 0 | + |
Solved Examples
ΔU = 200 − 50 = +150 J.
✔
Answer: +150 J
Work done BY gas = −300; ΔU = Q − W = (−100) − (−300) = +200 J.
The gas heated up — pumping a bicycle pump does exactly this. ✔
Answer: +200 J
Nothing to push against: W = 0. Insulated: Q = 0.
ΔU = 0 → temperature unchanged (ideal gas).
Expansion without cooling — the strange free lunch. ✔
Answer: ΔU = 0, T constant
- Sign of W. Physics convention: W = work done BY the gas (positive in expansion). Chemistry uses the opposite — pick one, state it, stay consistent.
- Assuming expansion always cools. Only adiabatic expansion cools; isothermal expansion keeps T (heat flows in to pay the work bill).
- U depending on pressure/volume. For ideal gases, U is a temperature-only account — a deep and examinable fact.
- Confusing heat Q with temperature. Q is energy in transit; T is the state of the gas. Ice melting takes big Q at constant T.
This Physics in Your Daily Life
- Bicycle pump warms up — your muscles’ work becomes the gas’s internal energy: first law felt in your palm.
- Rice cooker with lid rattling — steam does work lifting the lid: heat → internal energy + work, the full receipt in your kitchen.
- Your body runs on the first law — food energy = stored + work done + heat out: dieting is literally balancing this equation.
- Pressure cookers and steam engines — controlled heat-in, work-out conversions: civilization ran on this one-line receipt for 200 years.
- A hot drink cooling in your hands — Q leaves the drink, enters your palms: the ledger balancing itself in both directions at once.
A shopkeeper’s cashbox: money comes in (heat), some stays in the box (internal energy), some pays out for services (work). The first law is the daily audit: cash in must equal cash stored plus cash spent. Energy shops never cheat — a perfect accountant wrote the universe’s books.
Add 500 J to a gas that does 200 J of piston work: box gains 300 J — temperature rises by ΔU/nC_v. Do the same with no piston: all 500 J stays, gas heats more. Same deposit, different split: the piston is the ‘expense’ the energy can choose to have or not.
Draw the gas as a box with three arrows: Q in from below (a flame), U inside (a battery icon), W out to the right (a piston). The first law is the diagram’s arithmetic: inflow = storage + outflow. Every thermodynamic process is this same picture with different arrow sizes.
Practice set (answers hidden — try first)
(NEET-level) Q = 150 J in, W = 40 J by gas: ΔU =
(JEE Main-level) 500 J work done ON gas, no heat exchange: ΔU =
(NEET-level) Isothermal expansion of ideal gas: ΔU =
(Concept) The zeroth law justifies:
(JEE Main-level) Free expansion of ideal gas: temperature
- zeroth law: contact equalizes temperature
- U of ideal gas depends on T only
- ΔQ = ΔU + W (physics signs: W by gas +)
- W = area under P-V curve
- first law = energy conservation + heat
- 🔁 zeroth law role
- 🔁 U(T) for ideal gases
- 🔁 first law with signs
- 🧠 Chant: ‘in = store + spend’.
- 🧠 Sign anchor: ‘expansion = gas pays out (W+)’.
- 🏠 Daily: bicycle pump warmth = work → U.
- 🏠 Daily: your body’s calorie equation is the first law.
Quick revision
- Zeroth law: bodies in contact equalize temperature — that’s what thermometers use
- Internal energy U = the total molecular jiggle + interaction energy of the gas
- First law: ΔQ = ΔU + W — heat in = stored energy + work out
- Sign convention: Q positive entering, W positive done BY the gas (physics style)
- The first law is energy conservation wearing thermal clothes
- The zeroth law: thermometers’ licence
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