For the process to occur under adiabatic conditions, the correct condition is: (i) Δ T = 0 (ii) Δ p = 0 (iii)

Exercises · 5.2

Q. For the process to occur under adiabatic conditions, the correct condition is:

(i)

\Delta T = 0

(ii)

\Delta p = 0

(iii)

q = 0

(iv)

w = 0

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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, 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

\Delta U = Q - W

Where:

$\Delta U$ = 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.

Watch out

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 $\Delta U = Q - W$ becomes smaller. If a gas is compressed, work is done on the gas — so $W$ is negative, and $\Delta U = Q - (-|W|) = Q + |W|$ , 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: $W = \int P \, dV$ . 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)

\Delta U = 500 - 200 = 300 \text{ J}

The internal energy of the gas increases by 300 J. The remaining 200 J left the system as work. Energy is conserved — the 500 J you put in went partly to raising the gas's temperature (300 J) and partly to moving the piston (200 J).

What the First Law Does NOT Tell You

The First Law only tells you about the quantity of energy. It does not tell you whether a process is possible or in which direction it will occur. That is the domain of the Second Law. For example, the First Law allows heat to flow from a cold body to a hot body — as long as the total energy is conserved. But we know that never happens spontaneously. That's where the Second Law steps in.

Important

Key Takeaway

The First Law is the energy balance sheet. For any thermodynamic process: Energy added as heat minus energy lost as work equals the change in the system's internal energy. Energy is never lost — it only changes form or moves elsewhere.

A quick search for "First Law of Thermodynamics class 11 chemistry" or "NCERT chemistry syllabus first law of thermodynamics" will confirm what's true here: this concept is a standard, curriculum-aligned part of Class 11 Chemistry. Given how often it's tested in JEE Main, NEET and state CET Chemistry papers, it's worth revisiting this explanation until the reasoning feels automatic, not just the final formula.

Adiabatic means "no heat exchange with surroundings," which translates directly to $q = 0$ . The correct condition is (iii).

The word "adiabatic" comes from the Greek adiabatos, meaning "impassable." In thermodynamics, an adiabatic process is one where the system's boundary is thermally insulated—heat cannot pass in or out. This is a statement about energy transfer, not about the state variables themselves.

Let's examine each option through the lens of what adiabatic actually means:

Option (i): $\Delta T = 0$

This describes an isothermal process, where temperature remains constant. During an adiabatic process, temperature typically does change. For instance, when you compress a gas adiabatically (like in a bicycle pump), it heats up; when it expands adiabatically, it cools down. The First Law tells us that $\Delta U = q - w$ , and for an adiabatic process $q = 0$ , so $\Delta U = -w$ . For an ideal gas, $\Delta U = nC_V\Delta T$ , which means $\Delta T \neq 0$ whenever work is done. This is incorrect.
Option (ii): $\Delta p = 0$

This describes an isobaric process, where pressure stays constant. Adiabatic processes—especially for gases—almost always involve pressure changes. Think of the rapid compression in a diesel engine: the gas is compressed so fast that heat doesn't escape (adiabatic), but the pressure skyrockets. This is incorrect.
Option (iii): $q = 0$

This is the defining condition. An adiabatic process is one in which no heat is exchanged between the system and its surroundings. Mathematically, $q = 0$ . The First Law then simplifies to $\Delta U = -w$ : any change in internal energy comes solely from work done on or by the system. This is correct.
Option (iv): $w = 0$

This would mean no work is done, which describes a process at constant volume (isochoric) where the system is rigid. Adiabatic processes can—and often do—involve substantial work. The adiabatic expansion of steam in a turbine, for example, does enormous amounts of work. This is incorrect.

\Delta U = -w

Watch out

Students often confuse adiabatic with isothermal because both sound like "nothing changes." Remember: adiabatic = no heat transfer ( $q=0$ ), isothermal = no temperature change ( $\Delta T=0$ ). They are fundamentally different!

✓Final answer

q = 0

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Q.For the process to occur under adiabatic conditions, the correct condition is:

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