A gas enclosed in a cylinder expands irreversibly against a constant external pressure of $2\ ext{atm}$, its volume increasing from $5\ ext{L}$ to $10\ ext{L}$. Calculate the work done on the gas.

Question

A gas enclosed in a cylinder expands irreversibly against a constant external pressure of $2\ 	ext{atm}$, its volume increasing from $5\ 	ext{L}$ to $10\ 	ext{L}$. Calculate the work done on the gas.

Accepted Answer

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}

The gas's internal energy increases by 60 J. The remaining 40 J of the heat you supplied left as work. Energy is conserved — it just got split between storage (higher temperature) and mechanical output (lifting the piston).

Why This Matters for Exams

The First Law is not just a formula to plug numbers into. It is a logical constraint. Every thermodynamic process — isothermal, adiabatic, isobaric, isochoric, cyclic — is just a special case of this one equation with one of the terms set to zero or held constant.

Isothermal process (constant temperature): $\Delta U = 0$ , so $Q = W$ .
Adiabatic process (no heat exchange): $Q = 0$ , so $\Delta U = -W$ .
Isochoric process (constant volume): $W = 0$ , so $\Delta U = Q$ .
Cyclic process (returns to initial state): $\Delta U = 0$ , so $Q = W$ .

Important

The First Law is a statement of energy conservation. It tells you that energy cannot be created or destroyed — only transferred as heat or work. This is the foundation upon which all of thermodynamics is built.

If you've searched "First Law of Thermodynamics class 11 physics notes" or "First Law of Thermodynamics NCERT solutions", this page covers exactly that ground — the concept is a standard part of the Class 11 Physics NCERT/CBSE syllabus. It also carries real weight in JEE Main, NEET and state engineering/medical entrance exams, where questions on first law of thermodynamics test both conceptual understanding and calculation speed.

For expansion against a constant external pressure, the work done on the system (IUPAC convention) is $w = -P_{ext}\Delta V$ . Here $P_{ext} = 2\ \text{atm}$ and $\Delta V = V_2 - V_1 = 10 - 5 = 5\ \text{L}$ . So $w = -(2\ \text{atm})(5\ \text{L}) = -10\ \text{L atm}$ . Converting to joules using $1\ \text{L atm} = 101.325\ \text{J}$ : $w = -10 \times 101.325 = -1013.25\ \text{J}$ . The negative sign indicates that the gas, in expanding, does work on the surroundings, so energy leaves the system as work — consistent with the IUPAC convention where $w$ is work done on the system. [!ANSWER] $w = -1013\ \text{J}$ (i.e. $-1.013\ \text{kJ}$ ); the negative sign shows the gas does work on the surroundings.

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).

The system performs 15 kJ of work ON the surroundings, so work done ON the system is $w = -15$ kJ.

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

The internal energy of the system decreases by 25 kJ.

✓Final answer
$(d) -25 kJ.$
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.

An isothermal process is, by definition, one in which the temperature $T$ does not change ( $dT=0$ ). Since $U$ depends only on $T$ , it follows that $dU=0$ throughout an isothermal process — the internal energy remains constant, even though heat may flow in/out and work may be done, because that heat and work exactly balance ( $dQ=dW$ , from the first law with $dU=0$ ).

✓Final answer
$(c) Remains constant.$
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.

However, R is a definition that distinguishes work from heat (how the energy gets transferred), not an explanation of why they are nonetheless equivalent, interchangeable forms of energy. The equivalence itself follows from experiments (Joule's paddle-wheel experiment) and is embodied in the first law of thermodynamics — a separate idea from R's distinction. So R, though true, is not the correct explanation of A.

✓Final answer
$(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).$
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.

This says that energy supplied as heat can only go into increasing the internal energy of the system and/or doing work on the surroundings — no energy is created or destroyed, it is only converted between forms (heat, internal energy, mechanical work). This is exactly the principle of conservation of energy.

✓Final answer
$The correct option is (b) Energy.$
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 + delta W = delta W, so the net heat absorbed over a cycle equals the net work done, and neither needs to be zero (nor is the process necessarily reversible).

✓Final answer
$The correct option is (b) the change in internal energy is zero.$
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 form of energy — no energy is created or destroyed, only converted between heat, internal energy, and work.

✓Final answer
$(B) law of conservation of energy.$
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.

Therefore, if the system goes from state $i$ to state $f$ , the change in internal energy $\Delta U = U_f - U_i$ is fixed by the initial and final states alone. Even if a completely different process (path) is used to go from $i$ to $f$ — involving different amounts of heat exchanged and work done along the way — $\Delta U$ must still come out to the same value, 20 J.

(Note: heat $Q$ and work $W$ individually do depend on the path — only their combination $\Delta U = Q - W$ is path-independent.)

✓Final answer
$(A) the same.$
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)

This is nothing but the law of conservation of energy extended to include heat as a form of energy -- energy is neither created nor destroyed, only converted between heat, internal energy, and mechanical work.

✓Final answer
$(c) Conservation of energy.$
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.

It depends on ln(V2/V1), the expansion ratio -- a larger ratio of final to initial volume means more work is done, even at the same temperature.

So neither 'temperature only' nor 'expansion ratio only' alone is correct; the work done depends on both factors together.

✓Final answer
$The correct option is (c) Both of them.$
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 $T$ , and hence $U$ , does change; in isochoric/isobaric processes both $T$ and $U$ generally change.)

✓Final answer
$(b) isothermal$
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.

Substituting into the first law: 0 = ΔU + ΔW, i.e., ΔU + ΔW = 0, or ΔU = -ΔW. This means any work done by the gas comes entirely at the expense of its internal energy (and vice versa).

✓Final answer
$The correct option is (d) ΔU + ΔW = 0.$
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:

The Zeroth law, which deals with thermal equilibrium and defines temperature.

The Second law, which is about the direction of heat flow / entropy and the impossibility of a 100% efficient heat engine.

Newton's third law, which is about action-reaction force pairs in mechanics, unrelated to heat.

✓Final answer
$This is the First Law of Thermodynamics — option (b).$

Q.A gas enclosed in a cylinder expands irreversibly against a constant external pressure of 2 atm2\ \text{atm}2 atm, its volume increasing from 5 L5\ \text{L}5 L to 10 L10\ \text{L}10 L. Calculate the work done on the gas.

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Q.A gas enclosed in a cylinder expands irreversibly against a constant external pressure of $2\ \text{atm}$ , its volume increasing from $5\ \text{L}$ to $10\ \text{L}$ . Calculate the work done on the gas.

Unlock everything free for 14 days