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 P1V1 = P2V2 V = const.´ (1/P) Charles’ Law V/T = constant V1/T1 = V2/T2 V = const.´ (T) Avogadro’s Law V/n = constant V1/n1 = V2/n2 V = const.´ (n) Ideal Gas Law PV = RnT P1V1/n1T1=P2V2/n2T2 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-m3/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 = P1V1/n1T1 = P2V2/n2T2

Since both n and T are constant (n1 = n2 & T1 = T2) we can cancel these terms out to derive Charles’ Law

V1/T1 = V2/T2

Rearranging to solve for V2 (and using Kelvins rather than Celsius)

V2 = V1T1/T2 = (8.50 L)(398 K)/(302 K) = 11.2 L

(b) Rearranging to solve for T2

T2 = V2T1/V1 = (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 SF4 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 SF4 (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 KClO3 is decomposed to produce oxygen at T = 900° C according to the following reaction:

2KClO3(s) ® 2KCl(s) + 3O2(g)

If the reactnats and products are confined to a volume of 10 cm3 what is the pressure of the products (assume ideal gas behavior, complete decomposition of KClO3, and neglect the volume of the solid product, KCl).

• P = ?
• V = 10 cm3 ´ [1 mL/1 cm3] ´ [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 KClO3 we can determine the number of moles of O2 produced.

10.0 g KClO3 ´ [1 mol KClO3/122.55 g KClO3] ´ [3 mol O2/2 mol KClO3] = 0.122 mol O2

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 ´ 103 atm