Contents

**Introduction**

Gas pressure is a measure of the force exerted by gas molecules on the walls of their container. The more gas molecules there are in a given volume, the more collisions they will have with each other and with the container, and the higher the pressure will be. This relationship between gas pressure and the number of gas molecules in a given volume is known as the **ideal gas law**, which can be written as:

�=�(��/�)P=n(RT/V)

where P is the pressure, n is the number of moles of gas, R is the universal gas constant, T is the temperature, and V is the volume.

The ideal gas law assumes that all gases behave identically and that their behavior is independent of attractive and repulsive forces. This assumption is generally reasonable as long as the temperature of the gas is not super low (close to 0 K), and the pressure is around 1 atm.

In this article, we will explore how the pressure of a gas is related to its concentration of particles, which is defined as the number of moles of gas per unit volume (n/V). We will also look at some examples and applications of this concept in chemistry and everyday life.

**Partial Pressure and Dalton’s Law**

One important application of the ideal gas law is to calculate the pressure exerted by each component gas in a mixture of gases. We refer to this pressure as the **partial pressure** of a gas, which is the pressure that the gas would exert if it were the only one present (at the same temperature and volume).

According to **Dalton’s law of partial pressures**, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the component gases:

������=�1+�2+�3+…Ptotal=P1+P2+P3+…

where P_total is the total pressure and P_i are the partial pressures of each gas (up to n component gases).

We can also express Dalton’s law using the mole fraction of a gas, x_i, which is defined as the ratio of moles of a gas to the total moles of gas in a mixture:

��=��/������xi=ni/ntotal

Using this definition, we can write:

��=��������Pi=xiPtotal

This equation shows that the partial pressure of a gas is proportional to its mole fraction in a mixture.

**Example: Calculating Partial Pressure**

Let’s say we have a mixture of hydrogen gas, H_2(g), and oxygen gas, O_2(g). The mixture contains 6.7 mol hydrogen gas and 3.3 mol oxygen gas. The total pressure of the mixture is 1.5 atm, and the temperature and volume are constant.

We can use Dalton’s law to calculate the partial pressure of each gas in the mixture. First, we need to find the mole fraction of each gas:

��2=��2/������=6.7/(6.7+3.3)=0.67xH2=nH2/ntotal=6.7/(6.7+3.3)=0.67

��2=��2/������=3.3/(6.7+3.3)=0.33xO2=nO2/ntotal=3.3/(6.7+3.3)=0.33

Then, we can use these values to find the partial pressures:

��2=��2������=0.67×1.5=1.005 atmPH2=xH2Ptotal=0.67×1.5=1.005 atm

��2=��2������=0.33×1.5=0.495 atmPO2=xO2Ptotal=0.33×1.5=0.495 atm

We can check our answer by adding up the partial pressures and verifying that they equal the total pressure:

������=��2+��2=1.005+0.495=1.5 atmPtotal=PH2+PO2=1.005+0.495=1.5 atm

**Concentration and Collision Frequency**

Another way to understand how the pressure of a gas is related to its concentration of particles is to consider how increasing or decreasing the concentration affects the frequency of collisions between gas molecules.

The greater the frequency of successful collisions, the greater the rate of reaction. If we increase the concentration of a reacting solution or increase the pressure of a reacting gas, we are essentially bringing the reactant particles closer together. This means that:

- The frequency of collisions between reactant particles increases
- The chance of successful collisions increases
- The rate of reaction increases

Therefore, the pressure of a gas is directly related to its concentration of particles and its reaction rate.

**Example: Effect of Pressure on Reaction Rate**

Let’s look at an example of how changing the pressure of a gas affects its reaction rate. Consider the following reaction between hydrogen gas and nitrogen gas to produce ammonia gas:

�2(�)+3�2(�)→2��3(�)N2(g)+3H2(g)→2NH3(g)

This reaction is reversible, which means that it can also proceed in the opposite direction:

2��3(�)→�2(�)+3�2(�)2NH3(g)→N2(g)+3H2(g)

The equilibrium constant, K_c, for this reaction is given by:

��=[��3]2/([�2][�2]3)Kc=[NH3]2/([N2][H2]3)

where [ ] denotes the concentration of each gas in mol/L.

If we increase the pressure of the system by decreasing the volume, we are increasing the concentration of all gases involved. According to Le Chatelier’s principle^{4}, the system will try to counteract this change by shifting the equilibrium to the side with fewer moles of gas. In this case, that is the product side, since there are 4 moles of gas on the reactant side and only 2 moles of gas on the product side. Therefore, increasing the pressure will favor the forward reaction and increase the production of ammonia.

Conversely, if we decrease the pressure of the system by increasing the volume, we are decreasing the concentration of all gases involved. The system will try to counteract this change by shifting the equilibrium to the side with more moles of gas. In this case, that is the reactant side, since there are 2 moles of gas on the product side and 4 moles of gas on the reactant side. Therefore, decreasing the pressure will favor the reverse reaction and decrease the production of ammonia.

**Conclusion**

In this article, we have learned how the pressure of a gas is related to its concentration of particles. We have seen that:

- The ideal gas law shows that the pressure of a gas is directly proportional to its number of moles per unit volume (n/V), which is its concentration.
- Dalton’s law shows that the total pressure of a mixture of gases is equal to the sum of the partial pressures of each component gas, which are proportional to their mole fractions in the mixture.
- Increasing or decreasing the concentration or pressure of a reacting gas affects its collision frequency and reaction rate. According to Le Chatelier’s principle, a change in pressure will shift the equilibrium position of a reversible reaction to minimize the change.