A2 · Practice questions · The equation of state

pV = nRT = NkT, with T always in kelvin.

Six original Cambridge-style questions on the equation of state of an ideal gas in its molar and molecular forms, the Boltzmann constant, and fixed-mass changes of pressure, volume and temperature.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

Two forms, one constant, kelvin every time.

Per mole: pV = nRT with n in mol and R = 8.31 J K⁻¹ mol⁻¹; p in Pa, V in m³, T in K.
Per molecule: pV = NkT with N the number of molecules and k = R / N_A = 1.38 × 10⁻²³ J K⁻¹; never pair n with k or N with R.
Fixed mass: pV / T is constant, so p₁V₁ / T₁ = p₂V₂ / T₂; the shortcuts p ∝ T and V ∝ T hold only with T in kelvin.

Analysis

[2 marks]

The equation of state of an ideal gas may be written as pV = nRT or as pV = NkT. Show that the two forms are consistent, and hence express the Boltzmann constant k in terms of R and the Avogadro constant N_A.

The number of molecules is N = nN_A, so nRT can be written as (N / N_A)RT = N(R / N_A)T, which has the form NkT. ✓
Comparing, k = R / N_A = 8.31 / 6.02 × 10²³ = 1.38 × 10⁻²³ J K⁻¹, the gas constant per molecule. ✓

Calculation

[3 marks]

The air in a car tyre occupies 0.025 m³ at a pressure of 2.4 × 10⁵ Pa and a temperature of 17 °C. Calculate the amount of air in the tyre. After a long journey the air temperature is 45 °C and the volume is unchanged. Calculate the new pressure.

T = 17 + 273 = 290 K, so n = pV / RT = (2.4 × 10⁵ × 0.025) / (8.31 × 290) = 2.5 mol. ✓
At constant volume and fixed mass p / T is constant: p₂ = p₁T₂ / T₁ = 2.4 × 10⁵ × (318 / 290). ✓
p₂ = 2.6 × 10⁵ Pa; using 45 / 17 in place of the kelvin ratio would wrongly give more than double the pressure. ✓

Calculation

[3 marks]

Air in a laboratory is at a pressure of 1.01 × 10⁵ Pa and a temperature of 293 K. Treating it as an ideal gas, calculate the number of molecules per cubic metre, and hence estimate the average distance between neighbouring molecules.

pV = NkT gives N / V = p / kT = 1.01 × 10⁵ / (1.38 × 10⁻²³ × 293) = 2.5 × 10²⁵ m⁻³. ✓
Volume per molecule = 1 / (2.5 × 10²⁵) = 4.0 × 10⁻²⁶ m³. ✓
Treating each as a small cube, the spacing is the cube root, (4.0 × 10⁻²⁶)^1/3 = 3.4 × 10⁻⁹ m, about ten molecular diameters, which is why the ideal-gas model works well for air. ✓

Analysis

[3 marks]

A student heats a fixed mass of gas at constant volume and plots pressure p against Celsius temperature θ. The graph is a straight line that does not pass through the origin. Explain why the line misses the origin, state what the intercept on the θ axis represents, and explain how the graph shows that p is proportional to the thermodynamic temperature.

Pressure is proportional to the thermodynamic temperature T, not to θ; since T = θ + 273.15 the gas still has pressure at 0 °C, so the line cuts the p axis above the origin. ✓
Extrapolating the line back to p = 0 gives an intercept at θ = −273 °C, which is absolute zero, 0 K. ✓
Measured from this intercept the graph is a straight line through the origin, so p ∝ T on the kelvin scale, which is the shortcut pV = nRT gives for constant V and n. ✓

Calculation

[3 marks]

A diver's cylinder of internal volume 0.012 m³ is filled with air to a pressure of 2.0 × 10⁷ Pa at 293 K. Calculate the volume this air would occupy at the surface pressure of 1.0 × 10⁵ Pa and the same temperature, and the mass of air in the cylinder. The molar mass of air is 29 g mol⁻¹.

At constant temperature pV is constant: V = p₁V₁ / p₂ = (2.0 × 10⁷ × 0.012) / 1.0 × 10⁵ = 2.4 m³. ✓
n = pV / RT = (2.0 × 10⁷ × 0.012) / (8.31 × 293) = 99 mol (98.6). ✓
m = nM = 98.6 × 0.029 = 2.9 kg. ✓

Calculation

[4 marks]

A sealed rigid flask of volume 5.0 × 10⁻³ m³ contains helium at a pressure of 1.2 × 10⁵ Pa and a temperature of 300 K.

(a) Calculate the amount of helium in the flask. (b) Calculate the number of helium atoms. (c) The flask is cooled to 150 K. Calculate the new pressure. (d) Instead, at 300 K, half of the helium is allowed to escape. State the pressure that remains and explain, using the equation of state, why it equals the answer to (c).

(a) n = pV / RT = (1.2 × 10⁵ × 5.0 × 10⁻³) / (8.31 × 300) = 0.24 mol. ✓
(b) N = nN_A = 0.241 × 6.02 × 10²³ = 1.4 × 10²³ atoms. ✓
(c) Fixed mass at constant volume: p ∝ T, so p = 1.2 × 10⁵ × (150 / 300) = 6.0 × 10⁴ Pa. ✓
(d) 6.0 × 10⁴ Pa also: p = nRT / V, so halving n at fixed T and V halves p exactly as halving T at fixed n and V does; the gas cannot tell which product, nT, was halved. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.