STATES OF MATTER 
properties of gases and liquids (fluids) and solids
20.
The combined
gas law equation  more complicated PVT calculations P_{1}V_{1}T_{1}
= P_{2}V_{2}T_{2}
Doc Brown's
chemistry revision notes: basic school chemistry science GCSE chemistry, IGCSE chemistry, O level
and ~US grades 8, 9 and 10 school science courses or equivalent for ~1416 year old
science students for national examinations in chemistry and also helpful for UK
advanced level chemistry students aged ~1618 and US grades 1112 K12 honors.
20.
The combined
gas law equation  more complicated PVT calculations

If all the laws described
in 4b and 4c are combined, you get the following general expression

p x V/T = a constant (for a
given mass of gas).

This can be expressed in
generalised form for calculations based on an
initial set of conditions_{1} (1) changing to a new and final set of
conditions_{2} (2) for a given mass of gas, giving the
combined pressure–volume–temperature gas calculation equation ...

p_{1} x V_{1} 

p_{2} x V_{2} 
–––––––––––– 
= 
–––––––––––– 
T_{1} 

T_{2} 
 In shorthand':
p_{1}V_{1} /T_{ 1}
= p_{2}V_{2} / T_{2}
 therefore the three
permutations for problem solving involving all three variables are:

p_{2} = p_{1}V_{1}T_{2} /
V_{2}T_{1}

V_{2} =
p_{1}V_{1}T_{2} / p_{2}T_{1}

T_{2} =
p_{2}V_{2}T_{1} / p_{1}V_{1}

If one variable is constant, the
permutations for the other two variables are:

p_{2} = p_{1}V_{1} /
V_{2} (at constant
temperature)

V_{2} =
V_{1}T_{2} / T_{1}
(at constant pressure)

T_{2} =
p_{2}T_{1} / p_{1}
(at constant volume)

Note:

If the temperature is constant you get Boyle's
Law.

If p or V is constant you
get Charles's/Gay–Lussac's Law.

You can use any volume or
pressure units you like as long as both p's or both V's have the
same units.

The graphs of p or V
versus temperature become invalid once the gas has condensed into a liquid BUT
when extrapolated back all the lines seem to originate from y = 0 (for p or
V), x = –273^{o}C (for T).

This was part of the scientific evidence
that led to the belief that –273^{o}C was the lowest possible
temperature, though there is no theoretical upper limit at all.

This led to the devising of a new thermodynamic absolute
temperature scale or Kelvin scale which starts at OK.

Examples of P–V–T calculations
for you to practice (ANSWERS at end of page)

Q1

Q2

The fuel and air gases in the cylinders
of a 1200 cm^{3} car engine go from 25^{o}C before
combustion and rise to a peak temperature of 2100^{o}C after
combustion. If normal atmospheric pressure is 101 kPa, calculate the peak
pressure reached after combustion. Although the movement of the piston
changes the volume, for the sake of argument (i) assume the volume is
constant at both ends of the cycle.



(ii) To be more realistic, assume the
initial volume of fuel vapour plus air was 400 cm^{3}, now recalculate the
final pressure.

You now need to use the full PVT
expression.
 

Q3

A 5 dm^{3} (5
litre) cylinder of a gas was stored at 20^{o}C with an internal
gas pressure of 100 kPa.

At what temperature in
^{o}C would the pressure be 120 kPa if the volume of the storage
cylinder was reduced to 4 dm^{3} (4 litre).



Q4

Prior to compression, the
internal volume of one cylinder of a car engine is 250 cm^{3}.

Assume the airfuel
mixture is at a temperature of 100^{o}C and a pressure of 150
kPa.

After piston compression
and ignition the gas volume is reduced to 50 cm^{3} and the
temperature rises to 800^{o}C.

Calculate the pressure
created after compression and ignition.



Q5

A student was
investigating the speed of reaction between limestone granules and different
concentrations of hydrochloric acid. However after doing a whole series of
experiments at different acid concentrations, there was no time to do the
last planned experiment. The volume of carbon dioxide collected after 5 minutes in a
100cm^{3} gas syringe was used to determine the rate of reaction.
All the experiments were done in one lesson at a temperature of 22^{o}C
except for the last one. This was done in the next lesson, giving a carbon
dioxide volume of 47.0 cm^{3} after 5 minutes, but at a higher
temperature of 27^{o}C (when in Kelvin call this T_{1},
and the other temperature T_{2}).

To make the data
analysis fair, all the gas volumes should be ideally measured at the same
temperature, but a correction can be made for the last experiment.

(a) Calculate the volume
the of 47.0 cm^{3} of gas at 27^{o}C, would occupy at 22^{o}C.

(b) If the temperature
was ignored, what is the % error in the rate of reaction measurement?

(c) Should the
calculated value for 22^{o}C be used in the rate calculation
analysis? and are this still other sources of error? Discuss!!!

(d) Would you need to do
any correction for the volume of acid added to the limestone? Explain your
decision.
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ANSWERS
to the more complex PVT calculations

Q1

25 cm^{3} of a gas
at 1.01 atm. at 25^{o}C was compressed to 15 cm^{3} at 35^{o}C.

Calculate the final
pressure of the gas.

p_{1} = 1.01 atm,
p_{2} = ?, V_{1} = 25 cm^{3}, V_{2} = 15 cm^{3},

T1 = 25 + 273 = 298
K, T2 = 35 + 273 = 308 K

(p_{1} x V_{1})/T_{1}
= (p_{2} x V_{2})/T_{2}

p_{2} = (p_{1} x V_{1}
x T_{2})/(V_{2} x T_{1})

p_{2} =
(1.01 x 25 x 308)/(15 x 298) = 1.74 atm



Q2

The fuel and air gases in the cylinders
of a 1200 cm^{3} car engine go from 25^{o}C before
combustion and rise to a peak temperature of 2100^{o}C after
combustion. If normal atmospheric pressure is 101 kPa, calculate the peak
pressure reached after combustion. Although the movement of the piston
changes the volume, for the sake of argument (i) assume the volume is
constant at both ends of the cycle.

T_{1} = 25 + 273 = 298 K, T_{2}
= 2100 + 273 = 2373 K, P1 = 101 KPa

p/T = constant

p_{1}/p_{2}
= T_{1}/T2

p_{2} = p_{1}
x T_{2}/T_{1}

p_{2} = 101 x 2373/298 =
804 kPa

(ii) To be more realistic, assume the
initial volume of fuel vapour plus air was 400 cm^{3}, now recalculate the
final pressure.

You now need to use the full PVT
expression.
 (p_{1} x V_{1})/T_{1}
= (p_{2} x V_{2})/T_{2}
 p_{2} = (p_{1} x V_{1}
x T_{2})/(V_{2} x T_{1})
 p_{2} = (101 x 400 x 2373)/(1200 x 298) =
268 kPa

Q3

A 5 dm^{3} (5
litre) cylinder of a gas was stored at 20^{o}C with an internal
gas pressure of 100 kPa.

At what temperature in
^{o}C would the pressure be 120 kPa if the volume of the storage
cylinder was reduced to 4 dm^{3} (4 litre).

T_{2} = P_{2}V_{2}T_{1}
/ P_{1}V_{1}

T_{2} = (120 x 4
x(20 + 273)) / (100 x 5)

T_{2} = 14060 /
500 = 281.3 K

T_{2} = 281 K (3
sf)

T_{2} = 281  273
= 8^{o}C

Q4

Prior to compression, the
internal volume of one cylinder of a car engine is 250 cm^{3}.

Assume the airfuel
mixture is at a temperature of 100^{o}C and a pressure of 150
kPa.

After piston compression
and ignition the gas volume is reduced to 50 cm^{3} and the
temperature rises to 800^{o}C.

Calculate the pressure
created after compression and ignition.

P_{2} = P_{1}V_{1}T_{2}
/ V_{2}T_{1}

P_{2} = (150 x 250
x (800 + 273)) / (50 x (100 + 273))

P_{2} = 40237500 /
18650 = 2157.5

P_{2} =
2160 kPa
(3 sf)

Q5

A student was
investigating the speed of reaction between limestone granules and different
concentrations of hydrochloric acid. However after doing a whole series of
experiments at different acid concentrations, there was no time to do the
last planned experiment. The volume of carbon dioxide collected after 5 minutes in a
100cm^{3} gas syringe was used to determine the rate of reaction.
All the experiments were done in one lesson at a temperature of 22^{o}C
except for the last one. This was done in the next lesson, giving a carbon
dioxide volume of 47.0 cm^{3} after 5 minutes, but at a higher
temperature of 27^{o}C (when in Kelvin call this T_{1},
and the other temperature T_{2}).

To make the data
analysis fair, all the gas volumes should be ideally measured at the same
temperature, but a correction can be made for the last experiment.

(a) Calculate the volume
the of 47.0 cm^{3} of gas at 27^{o}C, would occupy at 22^{o}C.

V_{1}/V_{2}
= T_{1}/T_{2}
so V_{2} = V_{1}
x T_{2}/T_{1}

V_{1}
= 47.0 cm^{3}, T_{1} = 273 + 27 = 300K, T_{2} = 273 +
22 = 295K

V_{2} = 47.0 x 295/300 =
46.2 cm^{3}

(b) If the temperature
was ignored, what is the % error in the rate of reaction measurement?

Volume error = 47.0
– 46.2 = +0.8 cm^{3}, therefore ....

% error = 0.8 x 100/47
= +1.7%
(so you would over calculate the reaction rate without this correction)

The % error in the
volume would be the same as calculated for the rate e.g. in cm^{3}/min.

(c) Should the
calculated value for 22^{o}C be used in the rate calculation
analysis? and are this still other sources of error?

The
theoretical–calculated gas volume for 22^{o}C should be used for
calculating the rate, it will improve the accuracy a little, BUT there is
another problem!

If the reaction was
unfortunately carried out at a higher temperature (i.e. 27^{o}C)
there is a second source of error. At a higher temperature the reaction is
faster, so you are bound to get a higher volume of gas formed in five
minutes. Therefore you will calculate a faster rate of reaction e.g. in cm^{3}
gas/minute at 27^{o}C, that would have occurred/been measured at 22^{o}C
and so an unfair comparison with all the other results from the previous
lesson.

So, although you can
correct reasonably well the volume error due to an 'expanded' gas volume at
the higher temperature, the gas volume will still be too high because of
the faster rate of reaction at 27^{o}C and there isn't much you
can do about that error except repeat the experiment at 22^{o}C,
which is the best thing to do anyway!

Note that if the
temperature of a rates experiment was too low compared to all the other
experiments, the 'double error' would occur again, but this time the
measured gas volume and the calculated speed/rate of reaction would be lower
than expected.

(d) Would you need to do
any correction for the volume of acid added to the limestone? Explain your
decision.

No correction needed
for this at all. Although liquids expand/contract on heating/cooling, the volume
changes are far less compared to gas volume changes for the same temperature
change. This is because of the relatively strong intermolecular forces between liquid
molecules, which are almost absent in gases.
