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.

Meghalaya MboseMCQ· 1mImportance★★★★★est

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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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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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