Ideal gas law

The ideal gas law is a mathematical expression that describes the behaviour of an ideal gas.

So far, all the gas laws mentioned in the previous sections work reasonably well when the experiments are done at low pressures (≤ 1 atm) where the following assumptions about gas particles are valid:

1. They are in constant random motion.
2. They are point masses with zero volume.
3. They are very small compared to the space between them, i.e. they are far apart from one another.
4. There is no intermolecular forces of attraction or repulsion between them.
5. No energy is gained or lost from collisions.

Gases that are made up of such particles are called ideal gases. We can therefore mathematically describe the properties of ideal gases by combining Boyle’s law, Charles’ law and Avogadro’s law as follows:

Boyle’s law: $V\propto\frac{1}{p}$

Charles’ law: $V\propto T$

Avogadro’s law: $V\propto n$

Combining the above laws and rearranging,

$V\propto \frac{nT}{p}\; \; \Rightarrow \; \; V=R\frac{nT}{p}\; \; \Rightarrow \; \; pV=nRT\; \; \; \; \; \; \; \; 14$

where n is defined as the number of moles of gas instead of the earlier arbitrary description of amount of gas and R is the universal gas constant with a value of 8.3145 J K^-1 mol^-1. Eq14 is called the ideal gas law. Note that the individual laws still stand, e.g. Gay-Lussac’s law of p = k₃T at constant volume for a given mass of gas, where $k_3=\frac{nR}{V}$ .

Since p₁V₁ = n₁RT₁ and p₂V₂ = n₂RT₂, Avogadro’s law can also be expressed as

$\frac{p_1V_1}{n_1T_1}=\frac{p_2V_2}{n_2T_2}\; \; \; \; \; \; \; \; 15$

Eq14 and eq15 are also known as equations of state as they interrelate a gas’ physical properties, leading to the knowledge of its physical state. Although these two equations work reasonably well at low pressures where gases behave ideally, they need to be modified to describe the physical state of gases at high pressures or low temperatures. Gases that are under such non-ideal conditions are called real gases.

Question

The pressure of a gas in a constant-volume vessel containing 4.000 moles of Pentane and 1.000 mole of Hexane at 80.00°C is 146.8 kPa. Find the pressure of the gas at 45.00°C given:

- Boiling point of pentane: 36.00°C
- Boiling point of hexane: 68.00°C
- Volume of 4 moles of liquid pentane and/or 1 mole of liquid hexane is negligible compared to the volume of the vessel

Answer

Assuming gaseous pentane and gaseous hexane behave ideally, using eq15

$\frac{146.8\times V}{(4.000+1.000)(80.00+273.15)}=\frac{p_2 V}{4.000\times(45.00+273.15)}$

$p_2=105.8\: kPa$

Question

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