An ideal gas is taken from state 1 to state 2 along a curve on a $P$-$V$ diagram for which $P=\dfrac{\text{Constant}}{V}$ (that is, $PV=$ constant, a rectangular hyperbola). State 1 lies at the upper-left (high pressure, small volume) and state 2 at the lower-right (low pressure, large volume), so the pressure decreases from 1 to 2. Which of the following $T$-$P$ diagrams (temperature $T$ vertical, pressure $P$ horizontal) correctly represents this process? The candidate diagrams are: (i) a straight sloping line rising from point 1 (lower-left) to point 2 (upper-right); (ii) a vertical line at a fixed pressure with point 1 below point 2; (iii) a horizontal line at a fixed temperature with point 2 on the left (lower pressure) and point 1 on the right (higher pressure); (iv) a horizontal line at a fixed temperature with point 1 on the left (lower pressure) and point 2 on the right (higher pressure). (a) (iv) (b) (ii) (c) (iii) (d) (i)

> [!TLDR] > The given curve $PV=$ constant is an isotherm, so $T$ does not change. On a $T$-$P$ graph a constant-$T$ process is a horizontal line, and because the pressure drops from state 1 to state 2, state 1 is the right end (high $P$) and state 2 the left end (low $P$). That is diagram (iii). ### Concept For an ideal gas, $PV=nRT$. The stated relation $PV=\text{Constant}$ therefore forces $T=\text{Constant}$: the process is **isothermal**. ### Building the $T$-$P$ graph - $T$ is constant throughout, so the locus on a $T$ (vertical) vs $P$ (horizontal) plot must be a **horizontal straight line**. This immediately rules out (i) (sloping) and (ii) (vertical). - From state 1 to state 2 the pressure decreases ($P_1>P_2$). On the horizontal line, the higher-pressure state 1 is to the right and the lower-pressure state 2 is to the left. - Diagram (iii) shows exactly this: point 2 on the left, point 1 on the right, at constant $T$. Diagram (iv) reverses them (1 left, 2 right) and so implies pressure increases from 1 to 2 — contradicting the given process. > [!ANSWER] > Option (C) — the $T$-$P$ diagram (iii) (a horizontal isotherm with 2 at lower pressure on the left and 1 at higher pressure on the right).

An ideal gas is taken from state 1 to state 2 along a curve on a P-V diagram for which P=dfracConstantV (that

NCERT Exemplar · Q3

P

(i) a straight sloping line rising from point 1 (lower-left) to point 2 (upper-right);

(ii) a vertical line at a fixed pressure with point 1 below point 2;

(iii) a horizontal line at a fixed temperature with point 2 on the left (lower pressure) and point 1 on the right (higher pressure);

(iv) a horizontal line at a fixed temperature with point 1 on the left (lower pressure) and point 2 on the right (higher pressure).

(a)

(iv)

(b)

(ii)

(c)

(iii)

(d) (i)

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Concept understanding — Ideal Gas Law

The Ideal Gas Law: From Intuition to Equation

Imagine you're blowing up a balloon. You feel the resistance as you push more air in. The balloon gets tighter, harder to squeeze. Now imagine leaving that balloon in a hot car — it might even pop. Or take it to the top of a mountain, and it suddenly looks half-deflated.

These everyday experiences are telling you something deep about gases: their pressure, volume, temperature, and the amount of gas inside are all connected. The Ideal Gas Law is the single equation that captures that connection.

The Four Players

Every gas has four measurable properties:

Pressure ( $P$ ) — how hard the gas pushes on its container (like the tightness of the balloon)
Volume ( $V$ ) — how much space the gas occupies (the size of the balloon)
Temperature ( $T$ ) — how hot the gas is (measured in Kelvin, not Celsius)
Amount ( $n$ ) — how many gas particles are present (measured in moles)

The Ideal Gas Law says: if you know any three of these, you can calculate the fourth. It's the master relationship.

The Precise Statement

PV = nRT

Where $R$ is the universal gas constant. Its value depends on the units you use, but the most common one for exams is:

R = 0.0821 \ \frac{\text{L·atm}}{\text{mol·K}}

This means: if pressure is in atmospheres (atm), volume in litres (L), amount in moles (mol), and temperature in Kelvin (K), then $R = 0.0821$ .

Watch out

Temperature must be in Kelvin. Never plug Celsius into this equation. To convert: $K = °C + 273.15$ . For most exam problems, using $K = °C + 273$ is fine.

Why It Makes Physical Sense

The equation $PV = nRT$ isn't just a random formula — it's a compact summary of three simpler laws that were discovered earlier:

Boyle's Law (pressure-volume relationship): At constant $n$ and $T$ , $P \propto 1/V$ . Squeeze a gas into half the volume, pressure doubles.
Charles's Law (volume-temperature relationship): At constant $n$ and $P$ , $V \propto T$ . Heat a gas, it expands.
Avogadro's Law (amount-volume relationship): At constant $P$ and $T$ , $V \propto n$ . More gas particles need more space.

The Ideal Gas Law combines all three into one clean statement.

What "Ideal" Means

Real gases don't always follow this law perfectly. At very high pressures or very low temperatures, gas particles start interacting with each other and taking up significant space themselves. The "ideal" gas is a simplified model where:

Particles have negligible volume
No forces act between particles (except during collisions)
Collisions are perfectly elastic

For most exam problems at normal conditions (room temperature, atmospheric pressure), real gases behave close enough to ideal that the law works beautifully.

A Quick Example

A 2.0 L container holds 0.50 mol of gas at 300 K. What's the pressure?

P = \frac{nRT}{V} = \frac{(0.50)(0.0821)(300)}{2.0}

P = \frac{12.315}{2.0} = 6.16 \ \text{atm}

Tip

Always write the equation, plug in numbers with units, then calculate. This catches unit mistakes and shows your work for partial credit.

The Big Picture

The Ideal Gas Law is your go-to tool whenever a gas changes conditions or you need to find one property from the others. It's the foundation for understanding how gases behave in everything from car engines to weather balloons to your own breathing.

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

The given curve $PV=$ constant is an isotherm, so $T$ does not change. On a $T$ - $P$ graph a constant- $T$ process is a horizontal line, and because the pressure drops from state 1 to state 2, state 1 is the right end (high $P$ ) and state 2 the left end (low $P$ ). That is diagram (iii).

Concept

For an ideal gas, $PV=nRT$ . The stated relation $PV=\text{Constant}$ therefore forces $T=\text{Constant}$ : the process is isothermal.

Building the $T$ - $P$ graph

$T$ is constant throughout, so the locus on a $T$ (vertical) vs $P$ (horizontal) plot must be a horizontal straight line. This immediately rules out (i) (sloping) and (ii) (vertical).
From state 1 to state 2 the pressure decreases ( $P_1>P_2$ ). On the horizontal line, the higher-pressure state 1 is to the right and the lower-pressure state 2 is to the left.
Diagram (iii) shows exactly this: point 2 on the left, point 1 on the right, at constant $T$ . Diagram (iv) reverses them (1 left, 2 right) and so implies pressure increases from 1 to 2 — contradicting the given process.

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Concept

Building the TTT-PPP graph

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Building the $T$ - $P$ graph

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