The initial state of a certain gas is (Pᵢ, Vᵢ, Tᵢ). It undergoes expansion till its volume becoms Vf. Consider

NCERT Exemplar · Q22

P_i

(a) the expansion takes place at constant temperature.

(b) the expansion takes place at constant pressure.
Plot the

P

V

diagram for each case. In which of the two cases, is the work done by the gas more?

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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, withdraw money from it, or leave it untouched. The total amount of money in your account changes only when money goes in or comes out. You cannot create money from nothing, and money does not vanish into thin air.

Energy works exactly the same way. In any physical or chemical process, energy is never created and never destroyed. It only moves from one place to another, or changes from one form into another. This is the deepest idea behind the First Law.

Now, in thermodynamics, we focus on a specific "bank account": the internal energy of a system. Internal energy ( $U$ ) is the total energy stored inside a substance — the kinetic energy of its molecules jiggling around, plus the potential energy stored in the bonds between them.

If you want to change how much energy is stored inside a system, you have exactly two ways to do it:

Heat ( $Q$ ) — energy that flows because of a temperature difference. Like putting a cold pan on a hot stove.
Work ( $W$ ) — energy transferred by a force moving something. Like pushing a piston to compress a gas.

That's it. No third option. Every change in internal energy comes from either heat or work.

The Precise Statement

\Delta U = Q - W

Where:

$\Delta U$ = change in internal energy of the system
$Q$ = heat added to the system (positive if heat flows in)
$W$ = work done by the system (positive if the system does work on surroundings)

This sign convention is the standard one used in Indian exams (JEE, NEET, etc.). Heat added to the system is positive. Work done by the system is positive.

Watch out

Some textbooks use $Q = \Delta U + W$ or $\Delta U = Q + W$ with a different sign for work. Always check which convention your exam follows. The one above ( $\Delta U = Q - W$ ) is the most common in Indian syllabi.

What This Equation Really Says

Think of it as a balance sheet:

If you add heat ( $Q > 0$ ), internal energy tends to increase.
If the system does work ( $W > 0$ ), internal energy tends to decrease (because energy leaves the system to do the work).
The net change is simply: what came in minus what went out.

If $\Delta U = 0$ , the system has returned to its original internal energy — but that does not mean nothing happened. Heat could have come in, and exactly the same amount of energy could have left as work. The energy just passed through.

Note

| Process | $Q$ | $W$ | $\Delta U$ |

|---------|-----|-----|------------|

| Gas heated at constant volume | + | 0 | Positive |

| Gas cooled at constant volume | – | 0 | Negative |

A Concrete Example

Take a gas trapped in a cylinder with a movable piston. You place the cylinder on a hot plate.

Heat $Q = +100\ \text{J}$ flows into the gas.
The gas expands, pushing the piston upward, doing work $W = +40\ \text{J}$ on the surroundings.

What happens to the internal energy?

$\Delta U = 100 - 40 = +60\ \text{J}$ …

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Both processes start at $(P_i,V_i)$ and expand to the same final volume $V_f$ . The isobaric path stays at the higher pressure $P_i$ the whole way, while the isothermal path's pressure falls below $P_i$ as volume increases. Since work is the area under the $P$ - $V$ curve, the isobaric (constant-pressure) expansion does more work.

The two paths on a $P$ - $V$ diagram

Case (a) — isothermal ( $T=T_i$ constant). For an ideal gas, $PV=nRT_i=$ constant, so $P=\dfrac{P_iV_i}{V}$ — a curve (a rectangular hyperbola) that starts at $(V_i,P_i)$ and falls smoothly as $V$ increases, reaching $P_f=\dfrac{P_iV_i}{V_f}<P_i$ at $V_f$ .

Case (b) — isobaric ( $P=P_i$ constant). A horizontal line at $P=P_i$ , from $V_i$ straight across to $V_f$ .

Because the isobaric line stays at the full pressure $P_i$ throughout, while the isothermal curve dips below $P_i$ everywhere except right at the start, the isobaric path lies above the isothermal path for the whole expansion (the two only meet at the starting point).

Comparing the work done

Work done by the gas is the area under its $P$ - $V$ path:

$W_{isobaric} = P_i(V_f-V_i)$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 39 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. A system releases 10 kJ of heat and performs 15 kJ of work on the surrounding. Hence the change in internal energy is:$
(a) +5 kJ
(b) –5 kJ
(c) +25 kJ
(d) –25 kJ
›Reveal solutionSolution

$\Delta U = q + w = (-10) + (-15) = -25$ kJ.

By the first law of thermodynamics, $\Delta U = q + w$ , where $q$ is heat absorbed by the system and $w$ is work done on the system.

The system releases 10 kJ of heat, so $q = -10$ kJ (heat lost by the system).

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. In an isothermal process, the internal energy of an ideal gas:$
(a) Decreases
(b) Increases
(c) Remains constant
(d) First increases, then decreases
›Reveal solutionSolution

For an ideal gas, $U$ is a function of $T$ alone, so a constant-temperature (isothermal) process leaves $U$ unchanged.

For an ideal gas, the internal energy $U$ arises purely from the kinetic energy of its molecules, which depends only on the absolute temperature $T$ (there are no intermolecular potential energy terms in the ideal-gas model). Formally, $U=nC_vT$ (plus a constant), so $U$ is a function of temperature alone.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Two statements are given below, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes below. Assertion (A): Work and heat are two equivalent form of energy. Reason (R): Work is the transfer of mechanical energy irrespective of temperature difference, whereas heat is the transfer of thermal energy because of temp. difference only.$
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) and (R) both are false.
›Reveal solutionSolution

Both statements are individually correct, but R describes how work and heat differ, not why they're equivalent energy forms — that equivalence instead comes from the first law of thermodynamics.

Assertion (A): Work and heat are both genuinely modes of energy transfer, measured in the same SI unit (joule), and this equivalence (Joule's mechanical-equivalent-of-heat result) underlies the first law of thermodynamics, $\Delta U = Q - W$ . So A is true.

Reason (R): It correctly defines the two: work is the transfer of (mechanical) energy that doesn't require a temperature difference, while heat is energy transferred specifically because of a temperature difference. This is also a true statement.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. First law of Thermodynamics is the law of Conservation of :$
(a) Mass
(b) Energy
(c) Linear Momentum
(d) Angular Momentum.
›Reveal solutionSolution

The First Law of Thermodynamics is the law of conservation of energy applied to heat and work.

The First Law of Thermodynamics is stated as:

ΔQ = ΔU + ΔW

where ΔQ is the heat supplied to the system, ΔU is the change in internal energy, and ΔW is the work done by the system.

…

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CBSE 2026Set sz1 markMCQ
$Q. In a cyclic process:$
(a) the net work done is always positive
(b) the change in internal energy is zero
(c) the net heat transfer is zero
(d) the process is always reversible
›Reveal solutionSolution

In any cyclic process, the internal energy returns to its starting value, so delta U (over the whole cycle) = 0.

Internal energy U is a state function — it depends only on the state (e.g. temperature, for an ideal gas) of the system, not on the path taken to reach that state. In a cyclic process the system is brought back to exactly its initial state after one complete cycle, so U_final = U_initial, giving delta U = 0 for the cycle. This does not mean the net work done or net heat transferred is zero — from the first law, delta Q = delta U …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The first law of thermodynamics is a statement of (A) law of conservation of momentum (B) law of conservation of energy (C) law of conservation of angular momentum (D) None of these$
›Reveal solutionSolution

The first law of thermodynamics is simply the law of conservation of energy applied to thermal systems.

The first law states: $dQ = dU + dW$ , i.e. heat supplied to a system either increases its internal energy or is used to do external work (or both). This is a generalization of the mechanical law of conservation of energy to include heat as a fo …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. A system goes from state i to state f through a process in which internal energy increases by 20 joule. If only the process be replaced by another, the increase in internal energy will be (A) the same (B) less than 20 joule (C) more than 20 joule (D) none of these$
›Reveal solutionSolution

Internal energy change between the same two states is always the same, regardless of the process (path) used.

Internal energy $U$ is a state function — it depends only on the state (pressure, volume, temperature) of the system, not on the history or path by which that state was reached. This is a direct consequence of the first law of thermodynamics.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. First law of thermodynamics leads to$
(a) Conservation of mass
(b) Conservation of linear momentum
(c) Conservation of energy
(d) Conservation of angular momentum
›Reveal solutionSolution

The first law of thermodynamics states that heat supplied to a system equals the increase in internal energy plus the work done by the system, which is precisely a statement of energy conservation for thermal processes.

The first law is written as:

dQ = dU + dW

(heat supplied = change in internal energy + work done by the system)

…

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CBSE 2025Set hz1 markMCQ
$Q. The work done in an isothermal expansion of a gas depends upon:$
(a) Temperature only
(b) Expansion ratio only
(c) Both of them
(d) None of them
›Reveal solutionSolution
For an ideal gas undergoing isothermal expansion, W = nRT ln(V2/V1) -- this formula explicitly contains both the temperature T and the expansion ratio V2/V1, so the work done depends on both.

For an ideal gas at constant temperature T expanding reversibly from volume V1 to V2, using PV = nRT (so P = nRT/V):

W = integral from V1 to V2 of P dV = integral from V1 to V2 of (nRT/V) dV = nRT ln(V2/V1)

This single expression shows two dependencies at once:

It is directly proportional to T -- a higher (constant) temperature means more work is done for the same expansion ratio. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. The thermodynamic process in which internal energy remains constant is$
(a) adiabatic
(b) isothermal
(c) isochoric
(d) isobaric
›Reveal solutionSolution

For an ideal gas the internal energy $U$ is a function of temperature alone ( $U=nC_vT$ ).

In an isothermal process the temperature $T$ is held constant throughout, so $\Delta U=0$ , i.e. internal energy remains constant. (Contrast: in an adiabatic process $Q=0$ but $T$ , an …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. For an adiabatic process, the true statement is -$
(a) ΔQ = ΔU + ΔW
(b) ΔQ = ΔW
(c) ΔQ = ΔU
(d) ΔU + ΔW = 0
›Reveal solutionSolution

For an adiabatic process, ΔQ = 0, so the first law of thermodynamics gives ΔU + ΔW = 0.

The first law of thermodynamics states: ΔQ = ΔU + ΔW, where ΔQ is heat supplied to the system, ΔU is the change in internal energy, and ΔW is the work done by the system.

An adiabatic process is, by definition, one in which no heat enters or leaves the system, i.e., ΔQ = 0. …

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CBSE 2024Set ANN1 mark
$Q. "The heat given to a system is used to increase the internal energy and doing external work". This is$
(a) Zeroth law in thermodynamics
(b) First law in thermodynamics
(c) Second law in thermodynamics
(d) Newton's third law in motion
›Reveal solutionSolution
The statement describes heat splitting into a rise in internal energy plus external work done — this is precisely the First Law of Thermodynamics, a statement of energy conservation.

The First Law of Thermodynamics says that when an amount of heat ΔQ is given to a system, it can go into two things:

Increasing the internal energy of the system, ΔU

Being used by the system to do external work, ΔW (e.g. a gas expanding and pushing back the surroundings)

Mathematically:

ΔQ = ΔU + ΔW

This is essentially a statement of the law of conservation of energy applied to thermal processes. It is distinct from: …
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Q.The initial state of a certain gas is (PiP_iPi​, ViV_iVi​, TiT_iTi​). It undergoes expansion till its volume becoms VfV_fVf​. Consider the following two cases:

The two paths on a PPP-VVV diagram

Comparing the work done

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Q.The initial state of a certain gas is ( $P_i$ , $V_i$ , $T_i$ ). It undergoes expansion till its volume becoms $V_f$ . Consider the following two cases:

The two paths on a $P$ - $V$ diagram

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