STATES OF MATTER 
properties of gases and liquids (fluids) and solids
19.
Charles Law
and GayLussac's Law
Calculations
involving V_{1}/V_{2}
= T_{1}/T_{2}
and
P_{1}T_{1}
=P_{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.
19.
Charles's Law and GayLussac's Law for pressure/volume and
temperature calculations

The particle theory of gas pressure
was explained in Part 1 so this section concentrates on the gas law
calculations involving pressure and volume and their variation with
temperature.

(i)
Charles's Law
states for a fixed mass of gas at constant pressure:

The volume of a gas is directly
proportional to the absolute temperature (K) at constant pressure

V = constant x T (right graph), or

V/T = constant,
or

V_{1}/V_{2}
= T_{1}/T_{2} for conditions changing from 1 (initial)
to 2 (final)

or V_{1}/T_{1}
= V_{2}/T_{2} for constant pressure

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

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

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

Kinetic particle
model reasoning  increasing
the temperature increases the kinetic energy of the molecules giving
more forceful collisions which 'push out' to expand the gas at constant
pressure (major factor) and the chance of particle collision with the
sides of the container is also increased (minor factor).

So both effects contribute to an
increase in volume at constant pressure with increase in temperature
(above right graph.

Note that the graphs
extrapolate
back to 0K (absolute zero, Kelvin scale) or 273^{o}C
(Celsius scale).

(ii) GayLussac's Law states that for a fixed mass of
gas at constant volume:

The pressure of a gas is directly
proportional to the absolute temperature (K) at constant volume,

p = constant x T (right graph),
or

p/T = constant,
or

p_{1}/p_{2}
= T_{1}/T2 for conditions changing from 1 (initial)
to 2 (final),

or p_{1}/T_{1}
= p_{2}/T_{2} for constant volume

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

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

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

Kinetic particle
model reasoning  increasing
the temperature increases the kinetic energy of the molecules giving
more forceful collisions (major factor) and greater chance of collision
(minor factor), both of which contribute to an increase in the pressure if the volume is
constrained (kept constant).

Note again that the graphs
extrapolate back to 0K (absolute zero, Kelvin scale) or 273^{o}C
(Celsius scale).
Some practice
questions based on Charles's Law and Gay Lussac's Law

Q1

Q2

Q3

Q4

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 assuming the volume is
constant at both ends of the cycle.


ANSWERS
Learning objectives for calculations based
on
Charles's Law and GayLussac's Law
calculations
Know that for a fixed mass
of gas the volume is directly proportional to the absolute temperature
on the Kelvin scale (Charles's Law).
Know that for a fixed mass
of gas at constant volume the pressure is directly proportional to the
absolute temperature (K) of the gas (Gay Lussac's Law).
Be able explain Charles's
Law and Gay Lussac's Law in terms of the kinetic particle theory i.e.
the frequency, and more importantly, the kinetic energy of the particles
and their force of impact.
Be able to do calculations
based on Charles's Law
i.e. V = constant x T and V1 x T2 = V2
x T1 for a fixed mass of gas
at constant pressure:
Be able to do calculations
based on Gay Lussac's Law
i.e.
p = constant x T and p1/p2
= T1/T2 for a
fixed mass of gas at constant volume:
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ANSWERS
to Charles's Law and Gay Lussac's Law calculations

Q1

The pressure exerted by
a gas in sealed container is 100kPa at 17^{o}C. It was found that
the container might leak if the internal pressure exceeds 120kPa. Assuming
constant volume, at what temperature in ^{o}C will the container
start to leak?

17^{o}C + 273 =
290K

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

rearranging to scale up to the higher
temperature

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

T_{2} = 290 x 120/100 = 348 K or 348
– 273
= 75^{o}C
when the container might leak

Q2

A cylinder of propane
gas at 20^{o}C exerted a pressure of 8.5 atmospheres. When exposed
to sunlight it warmed up to 28^{o}C. What pressure does the
container side now experience?

20^{o}C = 273 +
20 = 293K, 28^{o}C = 273 + 28 = 301K

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

p_{2} =
8.5 x 301/293 = 8.73 atm

Q3

12.0 dm^{3} of gas in a cylinder
and piston system is heated from 290 K to 340 K. If the pressure remains
constant, calculate the final volume of gas in the cylinder.

V/T = constant

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

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

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

V_{2} = 12 x 340/290 =
14.1
dm^{3}

Q4

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 assuming 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
