1 mole of H$_2$ gas is contained in a box of volume $V$ = 1.00 m$^3$ at $T$ = 300K. The gas is heated to a temperature of $T$ = 3000K and the gas gets converted to a gas of hydrogen atoms. The final pressure would be (considering all gases to be ideal) (a) same as the pressure initially. (b) 2 times the pressure initially. (c) 10 times the pressure initially. (d) 20 times the pressure initially.

> [!TLDR] > The key idea is that heating H₂ from 300 K to 3000 K dissociates each molecule into two atoms, doubling the number of moles. Combined with the tenfold temperature increase, the ideal gas law gives a final pressure 20 times the initial pressure. 1. **Start with the ideal gas law.** For an ideal gas, $PV = nRT$. Initially, we have 1 mole of H₂ gas at $T_i = 300\ \text{K}$ and volume $V = 1.00\ \text{m}^3$. The initial pressure is $P_i = \frac{n_i R T_i}{V}$. 2. **What happens when the gas is heated?** The temperature rises to $T_f = 3000\ \text{K}$. But more importantly, the H₂ molecules dissociate into hydrogen atoms: $\text{H}_2 \rightarrow 2\text{H}$. Each mole of H₂ becomes 2 moles of H atoms. So the number of moles changes from $n_i = 1$ to $n_f = 2$. 3. **Apply the ideal gas law to the final state.** The final pressure is $P_f = \frac{n_f R T_f}{V}$. Since the volume $V$ and the gas constant $R$ are unchanged, we can compare directly: \[ \frac{P_f}{P_i} = \frac{n_f T_f}{n_i T_i} = \frac{2 \times 3000}{1 \times 300} = \frac{6000}{300} = 20. \] So $P_f = 20 P_i$. …

1 mole of H₂ gas is contained in a box of volume V = 1.00 m³ at T = 300K. The gas is heated to a temperature o

NCERT Exemplar · Q6

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(a) same as the pressure initially.

(b) 2 times the pressure initially.

(d) 20 times the pressure initially.

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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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The key idea is that heating H₂ from 300 K to 3000 K dissociates each molecule into two atoms, doubling the number of moles. Combined with the tenfold temperature increase, the ideal gas law gives a final pressure 20 times the initial pressure.

Start with the ideal gas law. For an ideal gas, $PV = nRT$ . Initially, we have 1 mole of H₂ gas at $T_i = 300\ \text{K}$ and volume $V = 1.00\ \text{m}^3$ . The initial pressure is $P_i = \frac{n_i R T_i}{V}$ .
What happens when the gas is heated? The temperature rises to $T_f = 3000\ \text{K}$ . But more importantly, the H₂ molecules dissociate into hydrogen atoms: $\text{H}_2 \rightarrow 2\text{H}$ . Each mole of H₂ becomes 2 moles of H atoms. So the number of moles changes from $n_i = 1$ to $n_f = 2$ .
Apply the ideal gas law to the final state. The final pressure is $P_f = \frac{n_f R T_f}{V}$ . Since the volume $V$ and the gas constant $R$ are unchanged, we can compare directly:

\frac{P_f}{P_i} = \frac{n_f T_f}{n_i T_i} = \frac{2 \times 3000}{1 \times 300} = \frac{6000}{300} = 20.

So $P_f = 20 P_i$ . …

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

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$Q. Experimental P-V curves and theoretically predicted P-V curves are in good agreement at (A) high temperature and high pressure (B) high temperature and low pressure (C) low temperature and high pressure (D) low temperature and atmospheric pressure (E) low temperature and low pressure$
›Reveal solutionSolution

Ideal-gas behaviour holds when molecules are far apart and interactions are negligible — high temperature, low pressure.

The theoretical (ideal-gas) P-V curves assume negligible intermolecular forces and negligible molecular volume. These assumptions are best satisfied when the gas is dilute and energetic, i.e. at …
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KEAM 2026Set eng-2026-04204 marksMCQ
$2.76 \times 10^5$
›Reveal solutionSolution

Ideal gas $P=nkT\Rightarrow n=P/(kT)=5.0\times10^{25}$ per m³.

Compute.
$n=\frac{P}{kT}=\frac{2.76\times10^5}{(1.38\times10^{-23})(400)}.$
Denominator $=1.38\times400\times10^{-23}=552\times10^{-23}=5.52\times10^{-21}$ . …
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KEAM 2026Set eng-2026-04224 marksMCQ
$f$
›Reveal solutionSolution

Real gases deviate because, when compressed and cooled, both attractive forces and the finite molecular volume matter.

At high pressure molecules are close together, so their own finite volume $V$ can no longer be neglected relative to the container.

At low temperature the intermolecular attractive force $f$ becomes appreciable. …
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KEAM 2026Set eng-2026-04224 marksMCQ
$300\ K$
›Reveal solutionSolution

Use $PV=nRT$ at constant volume: $P/(nT)$ is constant.

Initial: $n_1=18$ , $P_1=15$ atm, $T_1=300$ K. Final: $n_2=12$ , $P_2=9$ atm. …
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$Q. A cylindrical vessel contains 16 kg of gas at a pressure of 1 atmosphere. A certain amount of gas is taken out and the pressure of gas in the vessel becomes 0.75 atmosphere. The amount of gas taken out is (A) 2.5 kg (B) 4 kg (C) 7.5 kg (D) 8.25 kg (E) 10 kg$
›Reveal solutionSolution
The amount of gas taken out is 4 kg.

Concept and Intuition

At constant temperature and volume, the mass of gas in the vessel is proportional to its pressure (from $PV = nRT$ , with $n \propto$ mass). Reducing the pressure to 0.75 atm reduces the mass proportionally.

Step-by-Step Solution

Mass $\propto$ pressure at constant $T,V$ .

Remaining mass $= 16 \times \frac{0.75}{1} = 12\,kg$ . …
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KEAM 2025Set eng-2025-04254 marksMCQ
$E$
›Reveal solutionSolution

From kinetic theory the translational internal energy is $E = \tfrac{3}{2}nRT = \tfrac{3}{2}PV$ , hence $PV = \tfrac{2E}{3}$ .

Translational (kinetic) internal energy of an ideal gas:
$E = \frac{3}{2}nRT.$
Ideal gas equation:

$PV = nRT.$ …
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KEAM 2025Set eng-2025-04264 marksMCQ
$M$
›Reveal solutionSolution

The number of moles is mass divided by molar mass, and multiplying by Avogadro's number gives the number of molecules: $N = \left(\dfrac{M}{M_o}\right)N_A$ .

Step 1 — number of moles:

$n = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{M}{M_o}$ …
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›Reveal solutionSolution

Total pressure $=$ sum of partial pressures is Dalton's law.

Dalton's law of partial pressures states that for a mixture of non-reacting ideal gases, the total pressure equals the sum of the partial pressures each gas would exert alone:

$P=P_1+P_2+\cdots$ …
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KEAM 2025Set eng-2025-04284 marksMCQ
$T$
›Reveal solutionSolution

Kinetic-theory form of the ideal gas law: $PV=Nk_BT$ .

Writing the ideal gas law with the total number of molecules $N$ and the Boltzmann constant $k_B$ (where $k_B=R/N_A$ ):

$PV=Nk_BT.$ …
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›Reveal solutionSolution

Ideal behaviour requires negligible intermolecular forces and volume: low pressure and high temperature.

Why low pressure. At low pressure the gas is dilute, so the volume of the molecules is negligible compared to the container and intermolecular interactions are minimal. …
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›Reveal solutionSolution

Real gases approach ideal behaviour at low pressure and high temperature.

The ideal-gas model assumes negligible molecular volume and negligible intermolecular forces. At low pressure the molecules are far apart so their own volume is unimportant, and at high temperature the kinetic energy dominates over intermolecular attractions. Under these conditions the devi …
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Q.1 mole of H2_22​ gas is contained in a box of volume VVV = 1.00 m3^33 at TTT = 300K. The gas is heated to a temperature of TTT = 3000K and the gas gets converted to a gas of hydrogen atoms. The final pressure would be (considering all gases to be ideal)

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Q.1 mole of H $_2$ gas is contained in a box of volume $V$ = 1.00 m $^3$ at $T$ = 300K. The gas is heated to a temperature of $T$ = 3000K and the gas gets converted to a gas of hydrogen atoms. The final pressure would be (considering all gases to be ideal)

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