• The pH-scale. Learn about the basics of the pH-scale. Videos included. Read More
  • pKa & Ka and examples on how they are used in calculations. Read More
  • Ionic Strength - see how I is used to adjust pKa & Ka values to their real values based on I. Read More
  • Acids & Bases - basic definitions and models. Learn how an acid functions. Read More
  • Gases - the focus is on the ideal gas law and computations using pV=nRT. Read More
  • Equilibriums - focus is on the H2CO3 <=> HCO3- <=> CO32- buffer and calculations. Read More


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Gases - the ideal gas law explained

Chemical compounds in aqueous solutions are fairly easy to handle as their quantities can either be expressed in weight such as grams or kilos, moles per liter (molarity) or moles per kg (molality).

Topic: pKa & Ka
The basics

For gases it is a bit more complicated. In order to determine how much gas is present in some air volume, both temperature and pressure must be known. In some cases, not considered here, a factor such as buoyancy also has to be considered. The ideal gas law is frequently used equation (1) relating pressure (p), volume (V), number of moles (n), the universal gas constant (R) and the temperature in Kelvin (T) to one another.

(1) p . V = n . R . T

Gases that behave according to eq. 1 are called ideal gases. So if (1) is true for a gas, the gas is said to behave ideally. However, gases are not behaving ideally; they are behaving ideally to a large extent, so in most cases the ideal gas law can be used for routine computations involving gases. In high-pressure situations and at very low temperatures, the ideal gas law should not be used.

To better understand how the ideal gas law is used, some examples have been constructed for consideration:

Topic: Equilibriums
The pH scale

Example 1.
In a 1.5 Liter volume, the pressure of Argon(g) is 18000 ppm at 42o Fahrenheit. How many moles of Argon are there in the 1.5 L?

Preliminary consideration: 18000 ppm is 18000/1000000 or 0.018 Atm; R is 0.08206 L.atm.K-1.mol-1 and the temperature, 42 degrees Fahrenheit is 5/9*(42+459,67) or 279o Kelvin.
This information is entered into eq.1:
0.018 . 1.5 = n . 0.08206 . 279
To isolate n, the number of moles, we divide this equation by 0.08206 . 279 on both sides of the equation. This yields that the number of moles is 0.00179 or 1.179 mmol.

Topic: Ideal Gas Law
Ionic Strength

Example 2.
0.01 gram of toluene(l) is injected into a 2 L tube at 293o Kelvin. The toluene immediately evaporates. How many atmosphere of toluene is now in the tube?

Preliminary consideration: The molecular weight of toluene is 92.13 g/mol, so 0.01 g toluene is 0.000109 moles.
This is simple: we divide by V on both sides of eq. 1. The answer is 0.001304 atm.

Web references

An online calculater for the ideal gas law
A short but very precise explanation

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