STATES OF MATTER - properties of gases and liquids (fluids) and solids

21. Calculations using the ideal gas equation

PV = nRT for more advanced level calculations

This page includes unit conversions and determining the relative molecular mass of a volatile organic liquid - exam practice questions.

21a. The Ideal Gas Equation of State PV = nRT AND determining molecular mass of a volatile liquid

• The most 'compact molar' form of all the P–V–T equations is known as the ideal gas equation and is the simplest possible example of an 'equation of state' for gases (see also Van der Waals equation in section 5.(e)). The explanation of the use of the word 'ideal' is explained in the Introduction to ideal gas behaviour.

  • The equation combines both Boyle's law and Charles law plus moles of gas involved.

  • The equation, known as the ideal gas equation, is given as:

  • PV = nRT  (and below is a consistent set of SI units you can use)

  • p = pressure in pascals (unit Pa)

  • V = volume in cubic metres (m³)

  • n = moles of gas (mol = mass in g / molecular mass of gas M_r)

  • R = ideal gas constant = 8.314 joules per Kelvin per mol (J K^-1 mol^-1)

  • T = temperature in Kelvin (K)

  • M_r = relative molecular mass of the gas (g/mol)

  • You must convert to these units for a correct calculation using pV = nRT

  • Make sure can do all the rearrangements!

  • V = nRT /P,    T = PV/nR,   P = nRT/V,   n = PV/RT   and   R = PV/nT

  • The latter is how R was determined experimentally from multiple measurements

• The equation is pV = nRT and requires a consistent set of units, so see below for a comparison of the two most common examples, and take care!, and SI units are pretty standard now and my calculation examples primarily use SI units (but I have left in a few examples in 'old' non-SI units).

• pressure p	volume V	n = mass g/Mr	Ideal gas constant R and its units	temperature T
  Pa Pascal 760mmHg = 1 atm = 101325 Pa Pa = kPa x 1000 Pa = MPa x 106	m3 1m3 = 106 cm3  so m3 = cm3/106 or dm3/1000 = m3	mol mol = mass (g)/Mr	8.314 J mol–1 K–1	K (Kelvin = oC + 273)
  atm atmospheres 760 mmHg = 1 atm = 101325 Pa	litre or dm3 1 litre = 1 dm3 = 103 cm3 dm3 = cm3/1000	mol mol = mass (g)/Mr	0.08206 atm dm3 mol–1 K–1	K (Kelvin = oC + 273)

• The first set are becoming the 'norm' since they are the SI units, but the mass does not have to be in kg and can be in the more 'practical unit' of g as long as M_r is in g mol^–1.

• Examples of PV = nRT calculations (all calculations assume ideal gas behaviour)

  • I'm starting this section by describing a simple school/college laboratory experiment to determine the relative molecular mass of a volatile liquid e.g. hexane or propanone.

21.b A simple method of how to determine the molecular mass of a volatile liquid

• Q1

  • (a) Describe with the aid of a diagram a simple gas syringe method for determining the molecular mass of a volatile liquid.

  •  

    • An emptied 100 cm³ gas syringe is mounted in an oven (ideally thermostated) but a humble bulb will do and the temperature is quite stable after an initial warming up period via internal convection.

    • Some of the liquid (whose M_r is toe determined), is sucked into a fine 'hypodermic' syringe (e.g. 0.2cm³) and the syringe weighed.

    • Quickly (to avoid evaporation losses), the liquid is injected into the gas syringe via a self–sealing rubber septum cap and the syringe re–weighed immediately.

    • The difference in weighings gives the mass of liquid injected.

    • When the gas volume has settled to its maximum value the volume is read (to the nearest 0.5cm³ if possible),

    • Then note the oven temperature and barometric pressure (mercury barometer for best accuracy i.e. in mmHg).

  • (b) In an experiment using the above apparatus the following data were recorded and the molecular mass of the volatile liquid calculated.

    • Mass of syringe + liquid = 10.6403 g

    • Mass of syringe after injection of liquid = 10.4227g

    • When volatilised the liquid gave 67.3 cm³ of gas.

      • (to be honest, you can really only read the gas syringe scale at best to  ±0.5cm³).

    • The temperature of the oven = 81^oC, barometric pressure 752 mmHg.

    • Using the equation PV = nRT, calculate the molecular mass of the liquid.

      • R = 8.314 J mol^–1 K^–1

    • Mass of liquid injected = 10.6405 – 10.4227 = 0.2176 g

    • p = 101325 x 752/760 =  100258 Pa, (converting pressure from mmHg to Pa)

    • V = 67.3/10⁶ = 6.73 x 10^–5 m³, T = 273 + 81 = 354 K

    • PV = nRT, substituting for moles n gives PV = m/M_rRT

    • and then rearranging gives ...

    • M_r = mRT/PV

    • = (0.2176 x 8.314 x 354)/(100258 x 6.73 x 10^–5) = 94.9 (95 to 2sf)

  • (c) If the compound was formed from the reaction of bromine and a hydrocarbon, suggest a possible molecular formula for the compound.

    • bromomethane, CH₃Br, M_r = 12 + 3 + 80 = 95

      • If it was bromomethane, what is the percentage difference (percentage error)?

      • % difference = (experimental value - true value) / true value

      • % difference = (94.9 - 95.0) x 100 / 95.0 = 0.11 %

      • I should say that the gas syringe method is NOT this accurate at all, but this is how you do the calculation!

  • (d) State very briefly, a method of determining the molecular mass of ANY compound that can be vapourised intact.

    • Mass spectrometer, from the molecular ion peak (see mass spectrometry).

21c. PRACTICE EXAM QUESTIONS

More examples of the PV = nRT ideal gas equation calculations

• For other gas calculations see

• Mole definition and Avogadro Constant

• molar gas volume

• reacting gas volume ratios.

• Q2

  • (a) What is the volume of 6g of chlorine at 27^oC and 101kPa (approx. 1 atm)?

    • -

  • (b) What is the volume of the chlorine in dm³ and cm³?

    • -

• Q3

  • (a) A 5 litre container contained 0.5kg  of butane gas (C₄H₁₀).

    • Assuming ideal gas behaviour calculate the pressure of the gas if the cylinder is stored at 25^oC.

  • -

• Q4

  • A 100 dm³ (100 litre) cylinder of oxygen gas exerts a pressure of 900 kPa at 20^oC.

  • Calculate the mass of oxygen in the cylinder in kg.

  • -

• Q5

  • A cylinder of ethane gas has a volume of 1600 dm³.

  • The safe limit of storage pressure for the cylinder is 1.2 MPa.

  • If the cylinder contains 20 kg of ethane gas, what is the highest safest storage temperature in ^oC?

  • -

• Q6 Determining the relative molecular mass of a volatile organic compound using heated gas syringe method.

• (a) On volatilisation, 0.275 g of a liquid hydrocarbon produced a vapour volume of 75 cm³ at a pressure of 101 kPa.

  • Calculate the molecular mass of the hydrocarbon.

  • -

• (b) On analysis the hydrocarbon consisted of 83.9% carbon and 16.1% hydrogen and does not decolourise bromine water.

  • Calculate the molecular formula of the hydrocarbon.

  • Atomic masses: C = 12.01  and  H = 1.01

  • -

• (c) What important information can theses analyses NOT tell you?

  • -

OTHER USEFUL PAGES

See also for gas calculations

Moles and the molar volume of a gas, Avogadro's Law

Reacting gas volume ratios, Avogadro's Law & Gay–Lussac's Law Calculations

All other calculation pages

1. What is relative atomic mass?, relative isotopic mass and calculating relative atomic mass

2. Calculating relative formula/molecular mass of a compound or element molecule

3. Law of Conservation of Mass and simple reacting mass calculations

4. Composition by percentage mass of elements in a compound

5. Empirical formula and formula mass of a compound from reacting masses (easy start, not using moles)

6. Reacting mass ratio calculations of reactants and products from equations (NOT using moles) and brief mention of actual percent % yield and theoretical yield, atom economy and formula mass determination

   • Reacting masses, concentration of solution and volumetric titration calculations (NOT using moles)

7. Introducing moles: The connection between moles, mass and formula mass – the basis of reacting mole ratio calculations (relating reacting masses and formula mass)

8. Using moles to calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition)

9. Moles and the molar volume of a gas, Avogadro's Law

10. Reacting gas volume ratios, Avogadro's Law and Gay–Lussac's Law (ratio of gaseous reactants–products)

11. Molarity, volumes and solution concentrations (and diagrams of apparatus)

12. How to do volumetric titration calculations e.g. acid–alkali titrations (and diagrams of apparatus)

13. Electrolysis products calculations (negative cathode and positive anode products)

14. Other calculations e.g. % purity, % percentage & theoretical yield, volumetric titration apparatus, dilution of solutions (and diagrams of apparatus), water of crystallisation, quantity of reactants required, atom economy

15. Energy transfers in physical/chemical changes, exothermic/endothermic reactions

16. Gas calculations involving PVT relationships, Boyle's and Charles Laws (this page)

17. Radioactivity & half–life calculations including dating materials

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Worked out ANSWERS to the PV=nRT based questions

• Q2

  • (a) What is the volume of 6g of chlorine at 27^oC and 101kPa (approx. 1 atm)?

    • pV = nRT, V = nRT/p

    • T = 273 + 27 = 300K,

    • n = 6/71 = 0.08451 mol chlorine, Mr(Cl₂) = 2 x 35.5 = 71

    • and p = 101 x 1000 = 101000 Pa.

    • V = 0.08451 x 8.314 x 300/101000 = 0.00209 m³ (3 sf)

  • (b) What is the volume of the chlorine in dm³ and cm³?

    • 1 m³ = 1000 dm³ = 10⁶ cm³

    • V = 0.002087 x 1000 = 2.087 dm³

    • V = 0.002087 x 10⁶ = 2087 cm³ (2090 3sf)

• Q3

  • (a) A 5 litre container contained 0.5kg  of butane gas (C₄H₁₀).

    • Assuming ideal gas behaviour calculate the pressure of the gas if the cylinder is stored at 25^oC.

    • M_r(C₄H₁₀) = (4 x 12) + 10 = 58, 0.5kg = 500g

    • moles of gas n = 500/58 = 8.621, T = 273 + 25 = 298K

    • R = 8.314 J mol^–1 K^–1 or 0.08206 atm litre mol^–1 K^–1

  • (i) Using SI units to calculate the gas pressure

    • (1 m³ = 1000 dm³ = 10⁶ cm³, so dm³/10³ = m³)

  • PV = nRT, P = nRT/V, 5 litre = 5 dm³ = 5 x 10^–3 m³

  • P = 8.621 x 8.314 x 298/(5 x 10^–3)

  • P = 4271830 Pa = 4272 kPa = 4.27 MPa (3 sf)

  • (ii) 'old units'

    • P = 8.621 x 0.08206 x 298/5 = 42.16 atm

• Q4

  • A 100 dm³ (100 litre) cylinder of oxygen gas exerts a pressure of 900 kPa at 20^oC.

  • Calculate the mass of oxygen in the cylinder in kg.

  • R = 8.314 J K^-1 mol^-1,  V = 100/1000 = 0.1 m³,  p = 900 000 Pa,  T = 20 + 273 = 293 K

  • pV = nRT, n = pV/RT

  • n mol = pV/RT = (900 000 x 0.1)/(8.314 x 293) = 36.95 mol

  • mol = mass/M_r,    M_r(O₂) = 2 x 16 = 32

  • mass = mol x M_r = 36.95 x 32 = 1182 g = 1.18 kg (3sf)

• Q5

  • A cylinder of ethane gas has a volume of 1600 dm³.

  • The safe limit of storage pressure for the cylinder is 1.2 MPa.

  • If the cylinder contains 20 kg of ethane gas, what is the highest safest storage temperature in ^oC?

  • pV = nRT,  T = pV/nR, M_r(C₂H₆) = (2 x 12) + 6 = 30

  • pressure p = 1.2 x 10⁶ Pa

  • volume V = 1600/1000 = 1.6 m³

  • moles n = 20 000/30 = 666.7, R = 8.314 J K^-1 mol^-1

  • temperature T =  (1.2 x 10⁶ x 1.6)/(666.7 x 8.314) = 346 K (3 sf)

  • T = 346 - 273 = 73^oC. (3 sf)

• Q6 Determining the relative molecular mass of a volatile organic compound using heated gas syringe method.

• (a) On volatilisation, 0.275 g of a liquid hydrocarbon produced a vapour volume of 75 cm³ at a pressure of 101 kPa.

  • Calculate the molecular mass of the hydrocarbon.

  • PV = nRT = (m/M_r) RT

  • Mr = mRT / PV

  • P = 101000 Pa

  • V = 75 x 10^-6 = 7.5 x 10^-5 m³

  • m = 0.275 g (OK to work in g, because Mr is then computed in g mol-1 - 1 mole definition)

  • T = 60 + 273 = 333 K

  • M_r = (0.275 x 8.314 x 333) / (101000 x 7.5 x 10-5)

  • M_r = 100.5

• (b) On analysis the hydrocarbon consisted of 83.9% carbon and 16.1% hydrogen and does not decolourise bromine water.

  • Calculate the molecular formula of the hydrocarbon.

  • Atomic masses: C = 12.01  and  H = 1.01

  • mass ratio: 83.9 : 17.1

  • mole ratio: 83.9/12.01 : 16.1 / 1.01

  • mole ratio: 6.986 : 15.94

  • mole ratio: 7 : 16 (the closest integer values)

  • Therefore the hydrocarbon formula is C₇H₁₆

  • This is a saturated alkane and would not decolourise water,

• (c) What important information can theses analyses NOT tell you?

  • This analysis tells you nothing about the molecular structure, of which there are several carbon chain structural isomers from linear heptane to highly branched 2,2,3-trimethylbutane!

  • For lots on the structure and naming of C₇H₁₆ isomers see Alkane structure and naming