For a fixed mass of an ideal gas, its volume $V$ (vertical axis, in litres) is plotted against its absolute temperature $T$ (horizontal axis, in kelvin) at two different constant pressures $P_1$ and $P_2$. Both plots are straight lines passing through the origin. The line labelled $P_2$ is steeper (has the larger slope) than the line labelled $P_1$. What can be inferred about the relation between $P_1$ and $P_2$? (a) $P_1 > P_2$ (b) $P_1 = P_2$ (c) $P_1 < P_2$ (d) data is insufficient.

For a fixed mass of an ideal gas, its volume V (vertical axis, in litres) is plotted against its absolute temp

NCERT Exemplar · Q5

V

(a)

P_1 > P_2

(b)

P_1 = P_2

(c)

P_1 < P_2

(d) data is insufficient.

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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 …

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On a $V$ versus $T$ plot at constant pressure, the ideal-gas law gives a straight line through the origin whose slope is $nR/P$ . The larger the slope, the smaller the pressure. Since the $P_2$ line is the steeper one, $P_2$ is the lower pressure, so $P_1 > P_2$ .

Concept

For a fixed amount of ideal gas at constant pressure, volume is directly proportional to absolute temperature (Charles' law), which plots as a straight line through the origin.

Why this formula

From $PV = nRT$ at constant $P$ :

V = \left(\frac{nR}{P}\right) T,

so the slope of the $V$ – $T$ line is

\text{slope} = \frac{nR}{P}.

With $n$ and $R$ fixed, slope is inversely proportional to $P$ : a steeper line corresponds to a smaller pressure.

Steps

Both lines pass through the origin, confirming $V \propto T$ for each fixed pressure.
Slope $= nR/P \Rightarrow P = nR/\text{slope}$ . …

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

COMEDK 2026Set 2026-A1 markMCQ
$27 \times 10^4 \mathrm{cc}$
›Reveal solutionSolution
At STP, one mole of any ideal gas occupies 22.4 L. The total volume gives the total moles of gas; subtracting the moles of hydrogen (from its given mass) yields the moles of oxygen, whose mass is then found using its molar mass. The result is 129.7 g, option (B).

Concept & Intuition

Standard Temperature and Pressure (STP) is a fixed reference state: 0 °C and 1 atm. A cornerstone of ideal‑gas behaviour is that one mole of any gas occupies 22.4 L at STP. This is independent of the gas’s identity — a mole of hydrogen and a mole of oxygen each take up the same volume.

Here we have a mixture of two gases in a known total volume. If we can find the total number of moles in the vessel, and we already know the number of moles of hydrogen from its mass, then the difference gives the moles of oxygen. Converting those moles back to mass yields the answer.

Convert the vessel’s volume to litres The volume is given as $27 \times 10^4 \ \mathrm{cc}$ . Since $1 \ \mathrm{L} = 1000 \ \mathrm{cc}$ ,

$V = \frac{27 \times 10^4}{1000} = 270 \ \mathrm{L}.$

Find the total number of moles of gas in the vessel At STP, 1 mol occupies 22.4 L. Therefore,

$n_{\text{total}} = \frac{270}{22.4} \ \text{mol}.$
Compute:
$n_{\text{total}} = \frac{270}{22.4} = 12.0536\ \text{mol} \quad (\text{approximately}).$

Find the number of moles of hydrogen Mass of hydrogen = 16 g, molar mass = 2 g mol⁻¹.

$n_{\mathrm{H_2}} = \frac{16}{2} = 8\ \text{mol}.$

Determine the moles of oxygen The mixture contains only hydrogen and oxygen, so

$n_{\mathrm{O_2}} = n_{\text{total}} - n_{\mathrm{H_2}} = 12.0536 - 8 = 4.0536\ \text{mol}.$

Convert moles of oxygen to mass Molar mass of oxygen ( $\mathrm{O_2}$ ) = 32 g mol⁻¹. …
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COMEDK 2026Set 2026-M1 markMCQ
$B$
›Reveal solutionSolution
Using the ideal gas law and the fact that equal masses of different gases contain different numbers of moles, the pressure ratio is found to be 1:8, so the correct option is (A).

Concept & Intuition

The problem gives equal masses of oxygen (O₂) and hydrogen (H₂) at the same temperature, but different volumes. Pressure depends on the number of gas molecules (moles), not the mass directly. Since oxygen has a much larger molar mass than hydrogen, the same mass of oxygen contains far fewer moles. Combined with the volume difference, this drastically affects the pressure ratio. The ideal gas law $PV = nRT$ is the perfect tool here.

Step-by-step solution

Write the ideal gas law for each vessel.

For vessel A (oxygen): $P_A V_A = n_A R T$

For vessel B (hydrogen): $P_B V_B = n_B R T$

Temperature $T$ and gas constant $R$ are the same for both.

Relate the number of moles to the given masses.

Let the common mass be $m$ .

Molar mass of O₂ = 32 g/mol, of H₂ = 2 g/mol.

So

$n_A = \frac{m}{32}, \quad n_B = \frac{m}{2}.$

Use the volume relation.

Given $V_B = 2 V_A$ . Let $V_A = V$ , then $V_B = 2V$ .

Find the pressure ratio.

From the ideal gas law:

$P_A = \frac{n_A R T}{V}, \quad P_B = \frac{n_B R T}{2V}.$
Take the ratio:

$\frac{P_A}{P_B} = \frac{n_A R T / V}{n_B R T / (2V)} = \frac{n_A}{n_B} \cdot 2.$ …
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KCET 2026Set C21 markMCQ
$Q. The graph of pressure P and volume V of 1 mole of an ideal gas at constant temperature is [FIGURE: Four separate P-vs-V graphs labelled Graph-I (a horizontal line, constant P), Graph-II (a straight line through the origin with positive slope), Graph-III (a straight line decreasing linearly from a high P to a lower P as V increases), and Graph-IV (a curve starting high on the P-axis and decreasing smoothly/non-linearly toward the V-axis, i.e. a hyperbola-like shape).] (A) Graph-I (B) Graph-II (C) Graph-III (D) Graph-IV$
›Reveal solutionSolution

The ideal gas law at constant temperature gives $PV = \text{constant}$ , which traces a rectangular hyperbola on a $P$ - $V$ diagram.

Step 1 — Apply the ideal gas law at constant T

For $n=1$ mole of ideal gas held at constant temperature $T$ (an isothermal process):
$PV = nRT = \text{constant} \;\Rightarrow\; P = \frac{nRT}{V} \propto \frac{1}{V}$
Step 2 — Identify the shape of the curve

An inverse relationship $P \propto 1/V$ is a rectangular hyperbola: $P$ starts high at small $V$ and decreases smoothly and non-linearly as $V$ increases, never touching either axis. This matches Graph-IV, described as a curve starting high on the $P$ -axis and decreasing smoothly toward the $V$ -axis. …
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COMEDK 2025Set 2025-A1 markMCQ
$P T^2=$
›Reveal solutionSolution

The coefficient of volume expansion for an ideal gas under the condition $PT^2 = \text{constant}$ is $\frac{3}{T}$ , which corresponds to option (D).

The coefficient of volume expansion, usually denoted $\gamma$ or $\beta$ , is defined as the fractional change in volume per unit change in temperature at constant pressure:
$\beta = \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_P$
But here the gas is not expanding at constant pressure — instead, pressure and temperature are linked by $P T^2 = \text{constant}$ . So we cannot simply use the ideal gas law with $P$ fixed. Instead, we must find how $V$ depends on $T$ along the given path, then differentiate.

1. Write the given condition and the ideal gas law.

The ideal gas law:
$PV = nRT$
Given: $P T^2 = k$ (a constant). So $P = \frac{k}{T^2}$ .

2. Substitute $P$ into the ideal gas law to eliminate $P$ .
$\frac{k}{T^2} \cdot V = nRT$
Multiply both sides by $T^2$ :
$k V = nR T^3$
So
$V = \frac{nR}{k} T^3$
This shows that along this expansion, volume is proportional to $T^3$ .

3. Differentiate $V$ with respect to $T$ .
$\frac{dV}{dT} = \frac{nR}{k} \cdot 3 T^2 = 3 \frac{nR}{k} T^2$
But from step 2, $\frac{nR}{k} = \frac{V}{T^3}$ . Substitute:
$\frac{dV}{dT} = 3 \cdot \frac{V}{T^3} \cdot T^2 = \frac{3V}{T}$
4. Compute the coefficient of volume expansion.

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COMEDK 2024Set 2024-M1 markMCQ
$300 \mathrm{~cm}^3$
›Reveal solutionSolution
Using $N=\dfrac{PV}{RT}N_A$ with SI values gives $\approx 9.65\times10^{18}$ molecules.

Convert the data to SI units:

$V = 300\ \text{cm}^3 = 3.0\times10^{-4}\ \text{m}^3$

$P = 1\ \text{mm Hg} = 133.3\ \text{Pa}$

$T = 27^\circ\text{C} = 300\ \text{K}$

Number of moles: …
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COMEDK 2021Set 20211 markMCQ
$\frac{1}{p}$
›Reveal solutionSolution

(This is why the coefficient of volume expansion of an ideal gas at 0 degrees C, i.e. T = 273 K, is 1/273 per K.)

Concept: the coefficient of volume expansion is defined as

gamma = (1/V) * (dV/dT) at constant pressure.

For an ideal gas at constant pressure, PV = nRT => V = nRT/P, so

(dV/dT) at constant P = nR/P = V/T.

Hence gamma = (1/V) * (V/T) = 1/T. …
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