An engine made of one mole of an ideal gas in a cylinder with a piston is taken once around the cycle $A\to B\to C\to D\to A$ on a $P$-$V$ diagram, where $AB$ is at constant volume, $BC$ is at constant pressure, $CD$ is adiabatic and $DA$ is at constant pressure. On the diagram $A$ is the lower-left corner and $B$ is directly above it (same small volume, higher pressure); $B\to C$ runs rightward at the high constant pressure to $C$ (larger volume); $C\to D$ is the adiabatic expansion falling to $D$ (still larger volume, lower pressure); and $D\to A$ runs leftward at the low constant pressure back to $A$. Take $C_v=\tfrac{3}{2}R$ (so $C_p=\tfrac{5}{2}R$). Find the heat exchanged by the engine with the surroundings for each section ($AB$, $BC$, $CD$, $DA$) of the cycle, stating whether it is absorbed or released.

> [!TLDR] > Use $Q=nC_v\Delta T$ for the constant-volume leg, $Q=nC_p\Delta T$ for the constant-pressure legs, and $Q=0$ for the adiabatic leg, with $C_v=\tfrac32R$ and $C_p=\tfrac52R$. The result: heat is absorbed in $AB$ and $BC$ (temperature rising), none in $CD$, and released in $DA$ (temperature falling). ### Concept For one mole: - Constant volume: $Q=C_v\,\Delta T$ (no work). - Constant pressure: $Q=C_p\,\Delta T$, with $C_p=C_v+R=\tfrac52R$. - Adiabatic: $Q=0$ by definition. ### Section by section - **$AB$ (constant volume):** pressure rises ($P_B>P_A$) so $T_B>T_A$. $$Q_{AB}=C_v(T_B-T_A)=\tfrac32R(T_B-T_A)>0\ \text{(absorbed).}$$ - **$BC$ (constant pressure):** volume increases at high pressure so $T_C>T_B$. $$Q_{BC}=C_p(T_C-T_B)=\tfrac52R(T_C-T_B)>0\ \text{(absorbed).}$$ - **$CD$ (adiabatic):** $$Q_{CD}=0.$$ …

An engine made of one mole of an ideal gas in a cylinder with a piston is taken once around the cycle A→ B→ C→

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A\to B\to C\to D\to A

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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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Use $Q=nC_v\Delta T$ for the constant-volume leg, $Q=nC_p\Delta T$ for the constant-pressure legs, and $Q=0$ for the adiabatic leg, with $C_v=\tfrac32R$ and $C_p=\tfrac52R$ . The result: heat is absorbed in $AB$ and $BC$ (temperature rising), none in $CD$ , and released in $DA$ (temperature falling).

Concept

For one mole:

Constant volume: $Q=C_v\,\Delta T$ (no work).
Constant pressure: $Q=C_p\,\Delta T$ , with $C_p=C_v+R=\tfrac52R$ .
Adiabatic: $Q=0$ by definition.

Section by section

$AB$ (constant volume): pressure rises ( $P_B>P_A$ ) so $T_B>T_A$ .

Q_{AB}=C_v(T_B-T_A)=\tfrac32R(T_B-T_A)>0\ \text{(absorbed).}

$BC$ (constant pressure): volume increases at high pressure so $T_C>T_B$ .

Q_{BC}=C_p(T_C-T_B)=\tfrac52R(T_C-T_B)>0\ \text{(absorbed).}

$CD$ (adiabatic): $Q_{CD}=0.$ …

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

KEAM 2026Set eng-2026-04194 marksMCQ
$Q. There is no change in internal energy of an ideal gas in an (A) isothermal process (B) adiabatic process (C) isobaric process (D) isochoric process (E) both in adiabatic process and isobaric process$
›Reveal solutionSolution

Ideal-gas internal energy depends only on $T$ ; constant temperature (isothermal) means $\Delta U=0$ .

For an ideal gas, $U$ is a function of temperature alone: $U=nC_VT$ .

Hence $\Delta U=nC_V\Delta T$ . The process in which there is no change in internal energy is the one with $\Delta T=0$ , i.e. the isothermal process. …
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KEAM 2025Set eng-2025-04234 marksMCQ
$Q. Pick out the INCORRECT STATEMENT (A) Internal energy of an ideal gas depends only on its temperature (B) Change in the internal energy in a cyclic process is not zero (C) Change in the internal energy of a gas depends only on its initial and final states (D) Internal energy depends upon state of matter (E) Change in the internal energy in a cyclic process is zero$
›Reveal solutionSolution
The incorrect statement is that the change in internal energy in a cyclic process is not zero.

Concept and Intuition

Internal energy is a state function, so its change over any closed cycle is zero because the system returns to its initial state. The statement claiming otherwise is false.

Step-by-Step Solution

Internal energy $U$ is a state function.

In a cyclic process initial and final states coincide, so $\Delta U = 0$ .

Statement (B) says $\Delta U \neq 0$ in a cyclic process — this is incorrect.

Common Mistakes …
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KEAM 2025Set eng-2025-04264 marksMCQ
$Q. Which one is INCORRECT statement? (A) In an isochoric process, volume remains constant (B) In an adiabatic process, there is a heat exchange with the surrounding (C) In an isobaric process, pressure remains constant (D) In an isothermal process, temperature remains constant (E) In a cyclic process, the change in internal energy is zero$
›Reveal solutionSolution
By definition an adiabatic process is one in which no heat flows in or out ( $Q=0$ ). Statement (B) claims there is heat exchange, so it is the incorrect statement asked for.

Checking each statement:

(A) Isochoric $\to$ constant volume. ✓ Correct.

(B) Adiabatic $\to$ "heat exchange with the surrounding." ✗ Wrong — an adiabatic process has $Q=0$ ; work is done at the expense of internal energy with no heat transfer.

(C) Isobaric $\to$ constant pressure. ✓ Correct. …
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KEAM 2025Set pha-2025-0429F4 marksMCQ
$Q_1 - Q_2$
›Reveal solutionSolution

At constant temperature the internal energy is unchanged, so the net work equals the net heat: $W = Q_1 - Q_2$ .

By the first law of thermodynamics, $\Delta U = Q_{net} - W$ . For a cyclic/isothermal process at constant temperature, the change in internal energy is zero, so the work done by the system …
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KEAM 2024Set eng-2024-06064 marksMCQ
$Q. When heat is supplied to the gas in an isochoric process, the supplied heat changes its (A) volume only (B) internal energy and volume (C) internal energy only (D) internal energy and temperature (E) temperature only$
›Reveal solutionSolution

In an isochoric process $W=0$ , so all heat raises the internal energy, which for a gas means its temperature also increases.

In an isochoric (constant-volume) process the gas does no work, $W=P\,\Delta V=0$ . By the first law of thermodynamics $Q=\Delta U+W=\Delta U$ , so all the supplied heat increases the internal energy. Because the internal energy of a gas is a function of temperat …
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KEAM 2024Set eng-2024-06074 marksMCQ
$Q. The work done by a gas on the system is zero in (A) adiabatic process (B) isothermal compression (C) isochoric process (D) isobaric process (E) isothermal expansion$
›Reveal solutionSolution

Work $=\int P\,dV$ ; with no volume change the work is zero.

An isochoric (constant-volume) process has $dV=0$ , so …
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KEAM 2023Set eng-2023-P1-A14 marksMCQ
$\Delta Q$
›Reveal solutionSolution
Since the gas returns to the same $P$ and $V$ , $\Delta U=0$ and therefore $\Delta Q=\Delta W$ .

Concept and Intuition

Internal energy is a state function: equal initial and final $(P,V)$ means equal temperature, so $\Delta U=0$ . The first law $\Delta Q=\Delta U+\Delta W$ then gives $\Delta Q=\Delta W$ .

Step-by-Step Solution

Same $P$ and $V$ initial/final $\Rightarrow$ same state $\Rightarrow \Delta U=0$ .

First law: $\Delta Q=\Delta U+\Delta W=0+\Delta W$ . …
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KEAM 2022Set eng-2022-P1-A14 marksMCQ
$Q. A process in which the amount of heat supplied to the system goes fully to change its internal energy and temperature is (A) adiabatic process (B) cyclic process (C) isobaric process (D) isothermal process (E) isochoric process$
›Reveal solutionSolution
Heat going entirely to internal energy (no work) is a constant-volume, i.e. isochoric, process.

Concept and Intuition

First law: $\Delta Q = \Delta U + W$ . If all the heat raises internal energy and temperature with no work done, then $W = 0$ . Work is zero when volume is constant, which defines an isochoric process.

Step-by-Step Solution

$W = \int P\,dV = 0$ requires $dV = 0$ (constant volume).

Then $\Delta Q = \Delta U$ , all heat becomes internal energy. …
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KEAM 2021Set eng-2021-P1-A14 marksMCQ
$Q. A process in which there is no flow of heat between the system and surroundings is a/an (A) adiabatic process (B) cyclic process (C) isobaric process (D) isochoric process (E) isothermal process$
›Reveal solutionSolution
No heat flow means the process is adiabatic.

Concept and Intuition

The defining feature of an adiabatic process is $Q=0$ : the system is thermally isolated, so any change in internal energy comes purely from work.

Step-by-Step Solution

Given: no heat exchange, $Q=0$ .

Isothermal keeps $T$ constant (heat can flow), isobaric keeps $P$ constant, isochoric keeps $V$ constant, cyclic returns to the initial state — none require $Q=0$ . …
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