A hot air balloon is a sphere of radius 8 m. The air inside is at a temperature of 60°C. How large a mass can the balloon lift when the outside temperature is 20°C? (Assume air is an ideal gas, $R = 8.314$ J mole$^{-1}$K$^{-1}$, 1 atm. $= 1.013\times 10^{5}$ Pa; the membrane tension is 5 N m$^{-1}$.)

A hot air balloon is a sphere of radius 8 m. The air inside is at a temperature of 60°C. How large a mass can

NCERT Exemplar · Q23

R = 8.314

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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 balloon floats because the hot air trapped inside is less dense than the cooler air outside at the same pressure. The membrane's tension raises the internal pressure by a negligible amount (about 1.25 Pa, against roughly $10^5$ Pa atmospheric), so it can essentially be ignored. Comparing the mass of air displaced (cool, outside) to the mass of air enclosed (hot, inside) shows the balloon can lift about $3.1\times10^2$ kg (about 310 kg).

Why the membrane tension barely matters

For a spherical balloon of radius $r=8$ m, surface tension $S=5$ N m $^{-1}$ produces an excess internal pressure of

\Delta P = \frac{2S}{r} = \frac{2\times5}{8} = 1.25\text{ Pa}

Compared with atmospheric pressure ( $1.013\times10^5$ Pa), this is utterly negligible — about one part in $10^5$ . So to excellent approximation, the pressure of the air inside the balloon equals the outside atmospheric pressure, $P\approx1.013\times10^5$ Pa.

Setting up the buoyancy calculation

The balloon's volume is

V = \frac43\pi r^3 = \frac43\pi(8)^3 \approx 2.14\times10^3\text{ m}^3

Using the ideal gas law $PV=nRT$ at essentially the same pressure $P$ for both the hot air inside and the cool air outside (same volume $V$ ), the number of moles of each is

n = \frac{PV}{RT}

with $T_{out}=20°\text{C}=293$ K and $T_{in}=60°\text{C}=333$ K:

n_{out} = \frac{(1.013\times10^5)(2.14\times10^3)}{8.314\times293} \approx 8.92\times10^4\text{ mol}

n_{in} = \frac{(1.013\times10^5)(2.14\times10^3)}{8.314\times333} \approx 7.85\times10^4\text{ mol}

Finding the liftable mass …

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Why the membrane tension barely matters

Setting up the buoyancy calculation

Finding the liftable mass …

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