__Gas Laws__

The physical condition or state of a gas is defined by the following variables:

- Temperature, T
- Pressure, P
- Volume, V
- # of Moles, n

The relationships between these variables have been determined over the centuries through experimentation. They are given in the book as the following:

Boyle’s Law |
PV = constant |
P |
V = const.´ (1/P) |

Charles’ Law |
V/T = constant |
V |
V = const.´ (T) |

Avogadro’s Law |
V/n = constant |
V |
V = const.´ (n) |

Ideal Gas Law |
PV = RnT |
P |
V = const.´ (nRT/P) |

The ideal gas law is a combination of the first three gas laws, it is the only one you need to remember. The constant R is the ideal gas constant. The value of R depends upon the units you use.

R = 8.314 Pa-m^{3}/mol-K
= 0.08206 L-atm/mol-K = 62.36 L-torr/mol-K

One of the consequences of
the ideal gas law is that the state of a gas does not depend upon
its identity. Therefore, __1 mole of any gas at Standard
temperature and pressure (STP) occupies 22.4 L__.

STP ® T = 273 K and P = 1 atm

Its very
important to remember to __always write the temperature in
Kelvins__ rather than Celsius when working gas law problems.

To illustrate the power of the ideal gas law I will end this section by working out several example problems.

__Example__

A fixed quantity of gas, at constant pressure, occupies a volume of 8.50 L and has a temperature of 29.0° C. (a) What volume will the gas occupy if the temperature is increased to 125° C? (b) At what temperature will the volume be 5.00 L?

(a) Rearranging the ideal gas law to represent the state of the gas both before and after the temperature change:

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

Since both
n and T are constant (n_{1} = n_{2} & T_{1}
= T_{2}) we can cancel these terms out to derive
Charles’ Law

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

Rearranging
to solve for V_{2} (and using Kelvins rather than
Celsius)

V_{2}
= V_{1}T_{1}/T_{2} = (8.50 L)(398 K)/(302
K) = 11.2 L

(b)
Rearranging to solve for T_{2}

T_{2}
= V_{2}T_{1}/V_{1} = (5.00 L)(302
K)/(8.50 L) = 178 K

__Example__

What is the number of moles of an ideal gas with V = 2.50 L, T = 37° C and P = 725 torr?

Before using the ideal gas equation I like to convert each quantity to the appropriate units, so I don’t get confused down the line.

- P = 725 torr ´ [1 atm/760 torr] = 0.954 atm
- T = 37 + 273 = 310 K
- V = 2.50 L
- n = ?

Now I take the ideal gas law and rearrange to solve for n:

n = PV/RT = [(0.954 atm)(2.50 L)]/[(0.08206 L-atm/mol-K)(310 K)]

n = 0.0938 moles

Now lets move onto some more advanced applications of the ideal gas law, such as calculating the density of a gas or evaluating a gas phase reaction.

__Example
3__

What is
the density of SF_{4} vapour at 650 torr and 100° C?

- P = 650 torr ´ [1 atm/760 torr] = 0.855 atm
- T = 100 + 273 = 373 K
- n = ?
- V = ?

We aren’t given either the number of moles or the volume so at first glance it would seem that we are stuck. However, if we were to calculate the number of moles per unit volume we could use the molecular weight to convert into mass per unit volume, which is density.

Rearranging the ideal gas law:

n/V = P/RT = (0.855 atm)/[(0.08206 L-atm/mol-K)(373 K)] = 0.0279 mol/L

Multiplying
by the molecular weight of SF_{4} (108.1 g/mol) will give
the density:

r = (n/V)´ MW = (0.0279 mol/L)(108.1 g/mol) = 3.02 g/L

__Example
__

10.0 g of
KClO_{3} is decomposed to produce oxygen at T = 900° C according to the
following reaction:

2KClO_{3}(s)
® 2KCl(s) + 3O_{2}(g)

If the
reactnats and products are confined to a volume of 10 cm^{3}
what is the pressure of the products (assume ideal gas behavior,
complete decomposition of KClO_{3}, and neglect the
volume of the solid product, KCl).

- P = ?
- V =
10 cm
^{3}´ [1 mL/1 cm^{3}] ´ [1 L/1000 mL] = 1 ´ 10^{-2}L - T = 900 + 273 = 1173 K
- n = ?

Once again
it would appear that we are stuck, because we don’t know 2
quantities. However, if we calculate the theoretical yield of
KClO_{3} we can determine the number of moles of O_{2}
produced.

10.0 g
KClO_{3} ´ [1 mol KClO_{3}/122.55 g KClO_{3}]
´ [3 mol O_{2}/2
mol KClO_{3}] = 0.122 mol O_{2}

Now I rearrange the ideal gas law to solve for pressure.

P = nRT/V
= [(0.122 mol)(0.08206 L-atm/mol-K)(1173 K)]/(1 ´ 10^{-2} L)
= 1.2 ´ 10^{3} atm