An inflated rubber balloon contains one mole of an ideal gas, has a pressure $p$, volume $V$ and temperature $T$. If the temperature rises to 1.1 $T$, and the volume is increaset to 1.05 $V$, the final pressure will be (a) 1.1 $p$ (b) $p$ (c) less than $p$ (d) between $p$ and 1.1.

An inflated rubber balloon contains one mole of an ideal gas, has a pressure p, volume V and temperature T. If

NCERT Exemplar · Q8

p

(a) 1.1

p

(b)

p

p

(d) between

p

and 1.1.

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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 ideal gas law $pV = nRT$ governs the relationship between state variables. When temperature increases by 10% but volume increases by only 5%, the pressure must rise to approximately 1.048 $p$ , placing it between $p$ and 1.1 $p$ .

Why the ideal gas law determines the outcome

An ideal gas obeys the equation of state $pV = nRT$ , where $n$ is the number of moles and $R$ is the universal gas constant. For a fixed quantity of gas (one mole in this case), any change in temperature and volume must produce a corresponding change in pressure to maintain the equality. The balloon's rubber membrane adjusts to accommodate these changes, but the gas inside still follows the ideal gas law.

The key insight is that pressure responds to the ratio of temperature to volume. If temperature grows faster than volume, pressure must increase; if volume grows faster, pressure must decrease.

Step-by-step calculation

Write the initial state equation The balloon initially contains one mole at pressure $p$ , volume $V$ , and temperature $T$ :

p V = RT

Write the final state equation After heating, the final state has pressure $p_f$ , volume $1.05V$ , and temperature $1.1T$ :

p_f (1.05V) = R(1.1T)

Take the ratio of final to initial states Dividing the second equation by the first eliminates $R$ :

\frac{p_f (1.05V)}{pV} = \frac{1.1T}{T}

Simplifying:

\frac{p_f \cdot 1.05}{p} = 1.1

Solve for the final pressure

p_f = \frac{1.1}{1.05} \, p = \frac{11}{10.5} \, p = \frac{22}{21} \, p \approx 1.048 \, p

Interpret the result …

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Q.An inflated rubber balloon contains one mole of an ideal gas, has a pressure ppp, volume VVV and temperature TTT. If the temperature rises to 1.1 TTT, and the volume is increaset to 1.05 VVV, the final pressure will be

Why the ideal gas law determines the outcome

Step-by-step calculation

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Q.An inflated rubber balloon contains one mole of an ideal gas, has a pressure $p$ , volume $V$ and temperature $T$ . If the temperature rises to 1.1 $T$ , and the volume is increaset to 1.05 $V$ , the final pressure will be

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