1.5.4—Ideal gas equation

$PV = nRT$

$P₁V₁/T₁ = P₂V₂/T₂$

Worked example — amount and molar mass from gas data

The local course book gives $P=101.3\,\mathrm{kPa}$ , $V=1.91\,\mathrm{dm^3}$ , $m=3.30\,\mathrm{g}$ and $T=150\,^\circ\mathrm{C}=423.15\,\mathrm{K}$ . Because $\mathrm{kPa\,dm^3=J}$ , use $R=8.31\,\mathrm{J\,mol^{-1}\,K^{-1}}$ : $n=PV/(RT)=(101.3\times1.91)/(8.31\times423.15)=0.0550\,\mathrm{mol}$ . Then $M=m/n=3.30\,\mathrm{g}/0.0550\,\mathrm{mol}=60.0\,\mathrm{g\,mol^{-1}}$ . The final value is the mass of one mole of the vaporized compound under the stated ideal-gas model.

Convert Celsius to kelvin before substitution, and make pressure and volume units consistent with the chosen gas constant. Rearrange the equation only after the known quantities and units are identified.

Before solving, write a unit line beside P, V and T. With R = 8.31 J mol⁻¹ K⁻¹, use pressure in Pa, volume in m³ and temperature in K; using kPa with dm³ is also consistent because kPa·dm³ equals J. Judge model suitability before trusting the numerical result.

How it is assessed

Questions use mass, pressure, volume, and temperature data to determine amount or volume with the ideal gas equation.

Command terms

determine / calculate

What earns marks

Convert temperature to kelvin and volume/pressure units as required, substitute into PV=nRT, and report the amount or volume with a consistent unit and appropriate precision.

Watch for

Substituting Celsius in place of kelvin or mixing cm³ and m³ without conversion.

Representative question

[Maximum number: 2]

0.108 g of the vaporized compound was found to have a volume of $55.7 \mathrm{~cm}^{3}$ at $100^{\circ} \mathrm{C}$ and a pressure of $1.00 \times 10^{5} \mathrm{~Pa}$ .
Calculate the amount, in moles, of the compound. Use sections 1, 2 and 4 of the data booklet.

$T=373 \mathrm{~K} \text { AND } V=5.57 \times 10^{-5} \mathrm{~m}^{3} \text { OR } T=373 \mathrm{~K} \text { AND } 0.0557 \mathrm{dm}^{3} \text { and } 1 \times 10^{2} \mathrm{kPa}$ $<n=\frac{P V}{R T}=\frac{1.00 \times 10^{5} \mathrm{~Pa} \times 5.57 \times 10^{-5} \mathrm{~m}^{3}}{8.31 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 373 \mathrm{~K}}>$ $n=0.00180 \text { «mol» }$

Award [2] for correct final
answer.

Retrieve the model: ideal particles have negligible volume and forces with elastic collisions; low temperature and high pressure expose real-gas limits; molar volume and PV=nRT then connect amount, pressure, volume, and temperature.

Before calculating, check whether the question uses STP molar volume or PV=nRT, identify the fixed conditions, convert temperature to kelvin, and align pressure and volume units.

1.5.4—Ideal gas equation

Using the Ideal Gas Equation

Solving Ideal-Gas Problems

Assessment in practice

Ideal Gases Summary

Question 1