Ideal gas equation PV = nRT — numerical problems and deviations from ideal behavior

Question

(a) Calculate the volume occupied by 2 moles of an ideal gas at STP ( $T = 273$ K, $P = 1$ atm). (b) Why do real gases deviate from ideal behaviour at high pressure and low temperature?

(NEET 2023, similar pattern)

Solution — Step by Step

PV = nRT

where $P$ = pressure, $V$ = volume, $n$ = number of moles, $R$ = gas constant, $T$ = temperature (in Kelvin).

$R = 0.0821$ L atm mol⁻¹ K⁻¹ (when $P$ is in atm and $V$ in litres).

Given: $n = 2$ mol, $T = 273$ K, $P = 1$ atm

V = \frac{nRT}{P} = \frac{2 \times 0.0821 \times 273}{1}

V = \frac{44.83}{1}

\boxed{V = 44.83 \text{ L} \approx 44.8 \text{ L}}

Note: 1 mole of ideal gas at STP occupies 22.4 L, so 2 moles occupy $2 \times 22.4 = 44.8$ L. The formula and the shortcut give the same answer.

The ideal gas equation assumes:

Gas molecules have zero volume (they’re point particles)
There are no intermolecular forces between molecules

Real gases deviate because:

At high pressure, molecules are squeezed close together. Their actual volume becomes significant compared to the container volume — assumption (1) fails.
At low temperature, molecules move slowly. Intermolecular attractions become significant because molecules spend more time near each other — assumption (2) fails.

The corrected equation is:

\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT

$\frac{an^2}{V^2}$ corrects for intermolecular attractions (pressure correction)
$nb$ corrects for finite molecular volume (volume correction)
$a$ and $b$ are Van der Waals constants specific to each gas

Why This Works

The ideal gas equation works well under conditions where both assumptions hold: low pressure (molecules far apart, so their volume is negligible) and high temperature (molecules move fast, so attractions don’t matter). These are the conditions under which real gases behave most like ideal gases.

The deviations are captured by the compressibility factor $Z = PV/(nRT)$ . For an ideal gas, $Z = 1$ always. For real gases, $Z < 1$ at moderate pressures (attractions dominate) and $Z > 1$ at very high pressures (volume effect dominates).

Alternative Method — Using R in SI units

$R = 8.314$ J mol⁻¹ K⁻¹, $P = 1.013 \times 10^5$ Pa

V = \frac{nRT}{P} = \frac{2 \times 8.314 \times 273}{1.013 \times 10^5} = \frac{4539.4}{1.013 \times 10^5} = 0.0448 \text{ m}^3 = 44.8 \text{ L}

For NEET, memorise: 1 mole of any ideal gas at STP = 22.4 L. This shortcut handles most gas volume questions in seconds. But if the question changes temperature or pressure, use $PV = nRT$ with proper units.

Common Mistake

The biggest unit error: mixing $R = 8.314$ J/(mol K) with pressure in atm and volume in litres. If you use $R = 8.314$ , then $P$ must be in Pa and $V$ in m³. If you use $R = 0.0821$ , then $P$ must be in atm and $V$ in litres. Mixing units is the single most common error in gas law numericals and leads to answers that are off by orders of magnitude.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using R in SI units

Common Mistake

Try These Next