Q.State first law of thermodynamics.
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — First Law of Thermodynamics
First Law of Thermodynamics
The Intuition: Energy is a Bank Account
Imagine you have a bank account. You can deposit money into it, or withdraw money from it. The total amount of money in your account changes only by the net of what goes in and what comes out. You cannot create money from nothing, nor can you destroy it — it just moves.
Energy works the same way. In any physical or chemical process, energy is never created or destroyed. It is only transferred from one place to another, or converted from one form to another. This is the First Law of Thermodynamics — the law of conservation of energy, applied to systems where heat and work are the currencies.
When you heat a gas in a piston, the gas expands and pushes the piston up. The energy you put in as heat doesn't vanish — part of it stays inside the gas (raising its temperature), and part of it leaves as work done on the piston. The total energy of the universe remains constant.
The Precise Statement
ΔU=Q−W
Where:
- ΔU = change in the internal energy of the system (the energy stored inside — kinetic energy of molecules, potential energy in bonds, etc.)
- Q = heat added to the system (positive if heat flows into the system)
- W = work done by the system on the surroundings (positive if the system expands and pushes against something)
The sign convention is crucial. Many textbooks use Q+W with work done on the system. The version above (Q−W) is the most common in Indian exam syllabi (CBSE, JEE, NEET). Stick to one convention and be consistent.
Common Mistake
Students often forget the sign of work. If a gas expands, it does positive work on the surroundings — so W is positive, and ΔU=Q−W becomes smaller. If a gas is compressed, work is done on the gas — so W is negative, and ΔU=Q−(−∣W∣)=Q+∣W∣, which increases internal energy.
What Each Term Means Physically
Internal energy (U) is the total microscopic energy of the system. For an ideal gas, it depends only on temperature — higher temperature means higher U. For real substances, it also depends on volume and phase.
Heat (Q) is energy transferred due to a temperature difference. If you put a hot pan on a cold stove, heat flows from pan to stove. In thermodynamics, we always ask: who is the system? If the system is the gas, then Q is positive when heat flows into the gas.
Work (W) in thermodynamics is usually pressure-volume work: W=∫PdV. When a gas expands against a piston, it does work on the piston. When you compress a gas, you do work on it.
A Simple Example
Take a cylinder with a movable piston, containing 1 mole of an ideal gas. You supply 500 J of heat to the gas. The gas expands and does 200 J of work on the piston.
- Q=+500 J (heat enters the system)
- W=+200 J (work done by the system)
ΔU=500−200=300 J …
The first law of thermodynamics is essentially the law of conservation of energy applied to a thermodynamic system, relating the heat absorbed and the work done to the resulting change in internal energy. …
The First Law of Thermodynamics is the law of conservation of energy applied to a thermodynamic system.
Statement: Energy can neither be created nor destroyed; it can only be transformed from one form to another. The total energy of an isolated system (system + surroundings) remains constant.
Mathematical form: For a closed system,
ΔU=q+w
where:
- ΔU = change in internal energy of the system
- q = heat absorbed by the system from the surroundings
- w = work done on the system
(Sign convention: heat absorbed by the system and work done on the system are taken as positive.)
…
- CBSE 2026Set ANNUAL1 markQ.Write the relation between Cp and Cv for an ideal gas.
›Reveal solutionSolution
For an ideal gas, Cp - Cv = R (the universal gas constant), per mole.
Cv is the heat required to raise the temperature of one mole of gas by 1 K at constant volume (all the heat goes into raising internal energy, since no work is done: qv = ΔU = CvΔT). Cp is the heat required to raise the temperature of one mole of gas by 1 K at constant pressure; at constant pressure some of the heat also does work (PΔV) as the gas expands, so more heat is needed for the same temperature rise: …
- CBSE 2026Set ANNUAL1 markMCQQ.The change in internal energy (ΔU) of an ideal gas in a cyclic process is(a) equal to the heat absorbed(b) equal to the work done(c) (3/2)kB T(d) zero
›Reveal solutionSolution
Internal energy is a state function, so dU = 0 in a cyclic process. Answer (D).
Internal energy U depends only on the state (temperature) of the gas, not on the path. In a cyclic process the system returns to its initial state, so the initial and final internal energies are equal:
…
- CBSE 2026Set ANNUAL1 markMCQQ.Which of the following is an extensive variable of a thermodynamic system ?(a) Pressure(b) Temperature(c) Density(d) Volume
›Reveal solutionSolution
Volume is extensive; pressure, temperature and density are intensive. Answer (D).
An extensive property depends on the amount of matter (it doubles if the system is doubled); an intensive property does not.
…
- CBSE 2026Set ANNUAL1 markMCQQ.The work done by a gas in an isochoric process is(a) equal to the product of pressure and volume(b) zero(c) both (A) and (B)(d) none of these
›Reveal solutionSolution
Isochoric process: dV = 0, so W = 0. Answer (B).
Work done by a gas is W = integral of P dV. In an isochoric process the volume is held constant, so dV = 0 throughout. Therefore the work done is zero, and by the …
- CBSE 2026Set ANNUAL1 markMCQQ.The relation between enthalpy and internal energy is(a) ΔE = ΔH + PΔV(b) ΔH = ΔE + PΔV(c) ΔH = − ΔE − PΔV(d) ΔH = ΔE + VΔP
›Reveal solutionSolution
ΔH = ΔE + PΔV.
Enthalpy is defined as H = E + PV (E = internal energy). At constant pressure the change is ΔH = ΔE + PΔV. For reactions involving ga …
- CBSE 2026Set ANNUAL1 markMCQQ.For the reaction, C(s) + O2(g) → CO2(g), which of the following is correct ?(a) ΔH = ΔE(b) ΔH < ΔE(c) ΔH > ΔE(d) None of these
›Reveal solutionSolution
Δn(gas) = 0, so ΔH = ΔE.
ΔH = ΔE + Δn(g)RT. Count gaseous moles: reactants have 1 mol gas (O2); products have 1 mol gas (CO2). Solid carbon is not counted. So Δn(g) = 1 − 1 = …
- CBSE 2019Set ANN1 markQ.According to the first law of thermodynamics, for an isolated system, Δu = .........
›Reveal solutionSolution
For an isolated system, no matter or energy can cross its boundary, so by the first law, the internal energy change is exactly zero.
The first law of thermodynamics states: ΔU = q + w, where q is heat exchanged with the surroundings and w is work done on/by the system.
An isolated system is, by definition, one that can exchange neither matter nor energy with its surroundings. This means:
- No heat can flow in or out: q = 0
- No work can be done on or by the system: w = 0
Substituting into the first law:
ΔU = q + w = 0 + 0 = 0
…
- CBSE 2019Set ANNUAL1 markMCQQ.At constant pressure:(a) ΔH = qV(b) ΔH = 0(c) ΔH = ΔU − PΔV(d) ΔH = qp
›Reveal solutionSolution
Enthalpy is defined precisely so that, at constant pressure, the heat absorbed or released by a process equals ΔH: ΔH=qp.
From the first law of thermodynamics, ΔU=q+w, where w=−PΔV for expansion work at constant external pressure. So ΔU=qp−PΔV, i.e. qp=ΔU+PΔV. Enthalpy is defined as H=U+PV, so at constant pressure ΔH=ΔU+PΔV. Comparing the two expressions shows qp=ΔH — heat excha …
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.