STATES OF MATTER 
properties of gases and liquids (fluids) and solids
21.
Calculations using the ideal gas equation PV =
nRT
Helpful for UK
advanced level chemistry students aged ~1618, IB courses and US grades 1112 K12 honors.
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 (m^{3})

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)

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' nonSI units).

pressure p 
volume V 
n = mass g/M_{r} 
Ideal gas constant R
and its units 
temperature T 
Pa Pascal
760mmHg = 1 atm =
101325 Pa
Pa = kPa x 1000
Pa = MPa x 10^{6} 
m^{3}
1m^{3} = 10^{6} cm^{3}
so m^{3} = cm^{3}/10^{6}
or dm^{3}/1000 = m^{3} 
mol
mol = mass (g)/M_{r} 
8.314
J mol^{–1} K^{–1} 
K
(Kelvin =
^{o}C + 273) 
atm atmospheres
760
mmHg = 1 atm
=
101325 Pa 
litre or dm^{3}
1 litre = 1 dm^{3}
= 10^{3} cm^{3}
dm^{3} = cm^{3}/1000 
mol
mol = mass (g)/M_{r} 
0.08206
atm dm^{3} mol^{–1} K^{–1} 
K
(Kelvin =
^{o}C + 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)
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 cm^{3} 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. 0.2cm^{3}) 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 0.5cm^{3} 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 cm^{3}
of gas.

The temperature of the oven = 81^{o}C,
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 = 67.3/10^{6} =
6.73 x 10^{–5}
m^{3}, T = 273 + 81 = 354 K

PV = nRT, substituting for moles
n
gives PV = m/M_{r}RT

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.

(d) State very briefly, a method of
determining the molecular mass of ANY compound that can be vapourised intact.
21c.
PRACTICE EXAM QUESTIONS
More examples of
the PV = nRT ideal gas equation calculations
Worked out
ANSWERS to the PV=nRT based questions (at the end of the
page)

Q2

Q3

Q4

Q5

A cylinder of ethane gas has a volume of 1600 dm^{3}.

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 ^{o}C?



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 cm^{3} at a pressure of 101 kPa.

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

(c)
What important information can theses analyses NOT
tell you?
Worked out
ANSWERS to the PV=nRT based questions (at the end of the
page)
Learning objectives on using the PV=nRT
equation and determining molecular mass
Be able to perform
calculations with the equation PV=nRT, including rearrangements and
selecting appropriate sets of units.
Be able to describe and
explain a simple method for determining the molecular mass of a volatile
organic liquid, calculating the molecular mass from the observation
measurements.
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

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

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

Law of Conservation of Mass and
simple reacting mass calculations

Composition by
percentage mass of elements in a compound

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

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

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

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

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

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

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

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

Electrolysis
products calculations (negative cathode and positive anode products)

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

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

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

Radioactivity & half–life calculations including dating materials
All my
UK GCSE level (~US grade 810) school chemistry revision
notes
All my UK advanced level (~US grades 1112)
preuniversity chemistry revision notes
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Worked out
ANSWERS to the PV=nRT based questions

Q2

Q3

(a) A 5 litre container
contained 0.5kg of butane gas (C_{4}H_{10}).

Assuming
ideal gas behaviour calculate the pressure of the gas if the cylinder is
stored at 25^{o}C.

M_{r}(C_{4}H_{10})
= (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

PV = nRT, P = nRT/V, 5
litre = 5 dm^{3} = 5 x 10^{–3} m^{3}

P = 8.621 x 8.314 x 298/(5
x 10^{–3})

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

Q4

A 100 dm^{3} (100 litre) cylinder of
oxygen gas exerts a pressure of 900 kPa at 20^{o}C.

Calculate the mass of oxygen in the cylinder in
kg.

R = 8.314 J K^{1} mol^{1},
V = 100/1000 = 0.1 m^{3}, 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}(O_{2})
= 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 dm^{3}.

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 ^{o}C?

pV = nRT, T = pV/nR, M_{r}(C_{2}H_{6})
= (2 x 12) + 6 = 30

pressure p = 1.2 x 10^{6} Pa

volume V = 1600/1000 = 1.6
m^{3}

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

temperature T = (1.2 x 10^{6} x 1.6)/(666.7 x
8.314) = 346 K (3 sf)

T = 346  273 =
73^{o}C.
(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 cm^{3} 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} m^{3}

m = 0.275 g (OK to work in g, because Mr is then
computed in g mol1  1 mole definition)

T = 60 + 273 = 333 K

M_{r} = (0.275 x 8.314 x 333) / (101000 x 7.5 x
105)

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
C_{7}H_{16}

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 2,2,3trimethylbutane!

For lots on the
structure and naming of C_{7}H_{16} isomers see
Alkane structure and
naming
