R = 8.314 J/mol·K. Convert between atm, Pa. Worked numericals.

Ideal Gas Equation PV = nRT — Units and Applications

Question

A container holds 2 moles of an ideal gas at a temperature of 27°C and pressure of 2 atm. Calculate the volume occupied by the gas.

Given: R = 0.0821 L·atm/mol·K

Solution — Step by Step

This is the single most important step — the ideal gas equation demands absolute temperature. Kelvin, not Celsius.

T = 27 + 273 = 300 \text{ K}

List the variables before touching the formula. This prevents unit disasters.

$P = 2$ atm
$n = 2$ mol
$R = 0.0821$ L·atm/mol·K
$T = 300$ K
$V = ?$

We need $V$ , so isolate it:

V = \frac{nRT}{P}

This is algebra, but write it out explicitly — especially in board exams where steps carry marks.

V = \frac{2 \times 0.0821 \times 300}{2}

V = \frac{49.26}{2} = \textbf{24.63 L}

Why This Works

The ideal gas equation $PV = nRT$ captures how four properties of a gas — pressure, volume, amount, and temperature — are all connected through one constant $R$ .

Think of it this way: if you increase temperature, the gas molecules move faster and push harder. To keep pressure the same, the volume must increase. The equation simply quantifies this relationship.

The value of $R$ depends entirely on which units you use for pressure and volume. Use $R = 0.0821$ L·atm/mol·K when pressure is in atm and volume in litres. Use $R = 8.314$ J/mol·K (equivalently, Pa·m³/mol·K) for SI units. Both are correct — the choice is just about what units your problem gives you.

R = 8.314 \text{ J mol}^{-1} \text{K}^{-1} = 8.314 \text{ Pa·m}^3 \text{ mol}^{-1} \text{K}^{-1}

R = 0.0821 \text{ L·atm mol}^{-1} \text{K}^{-1}

R = 62.36 \text{ L·mmHg mol}^{-1} \text{K}^{-1}

Alternative Method — Using SI Units

Let’s solve the same problem with SI units to verify. Convert pressure first:

P = 2 \text{ atm} \times 101325 \text{ Pa/atm} = 202650 \text{ Pa}

Now apply $V = \frac{nRT}{P}$ with $R = 8.314$ J/mol·K:

V = \frac{2 \times 8.314 \times 300}{202650} = \frac{4988.4}{202650} \approx 0.02463 \text{ m}^3

Convert: $0.02463 \text{ m}^3 = 24.63 \text{ L}$ ✓

Same answer. This cross-check trick is worth doing in JEE numericals when you have time — a wrong unit choice will give a wildly different number.

Scoring shortcut: At STP (0°C, 1 atm), 1 mole of ideal gas occupies 22.4 L. Many NCERT problems and CBSE MCQs are just disguised versions of this fact. Recognise the pattern and you save 90 seconds.

Common Mistake

Using T in Celsius instead of Kelvin. If you plug in $T = 27$ instead of $T = 300$ , you get $V = 2.46$ L — exactly 1/10th of the correct answer. This error appears in roughly 40% of student solutions in CBSE board marking schemes. The fix is mechanical: write “K = °C + 273” as your very first line, every single time.

Related trap: using $T = 0°C = 0$ K, which would make $PV = 0$ — physically nonsensical. Zero Kelvin is absolute zero, not the freezing point of water.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using SI Units

Common Mistake

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