# Ideal gas law: definition, formula and solved exercises

The ideal gas law is a mathematical relationship that describes the behavior of gases when they are in ideal conditions.

This law establishes that, at constant temperature and quantity of gas, the pressure (P) of a gas is directly proportional to its volume (V) and inversely proportional to its absolute temperature (T).

## Ideal Gas Law Formula

Mathematically, it can be expressed as:

PV = nRT

Where:

• P is the pressure of the gas in pascals (Pa).

• V is the volume of the gas in cubic meters (m³).

• n is the amount of substance in moles (mol).

• R is the ideal gas constant, which has a value of 8.314 J/(mol·K).

• T is the absolute temperature in kelvin (K).

The ideal gas law is useful for describing and predicting the behavior of gases in various situations, such as thermodynamics, chemistry, and physics. However, it is important to note that real gases can deviate from ideal behavior under extreme conditions of temperature and pressure, which is accounted for by corrections and other more complex equations of state.

## What are ideal gases?

Ideal gases are a theoretical model used in physics and chemistry to describe the behavior of gases under ideal conditions. According to this model, ideal gases meet certain simplified assumptions:

1. Gas particles are considered to be points without volume.

2. The gas particles do not interact with each other, that is, there are no attractive or repulsive forces between them.

3. Collisions between gas particles and with the walls of the container are elastic, which means that there is no loss of kinetic energy during the collisions.

4. The amount of gas (number of moles) is large compared to the size of the container, which implies that the volume occupied by the gas particles is negligible compared to the total volume of the system.

5. Temperature and pressure are on an absolute scale, such as kelvin (K) and pascals (Pa), respectively.

Importantly, real gases can deviate from ideal behavior under extreme conditions of temperature and pressure, and more complex models are required to describe their behavior, such as modified equations of state.

## Examples of the ideal gas law from day to day

Here are some day-to-day examples that can illustrate the application of the ideal gas law:

1. Inflating a balloon: When you inflate a balloon, you are increasing the volume of air inside it. According to the ideal gas law, if you keep the pressure constant and increase the volume, the temperature of the gas inside the balloon will also increase. This is because as the gas expands, the molecules have more room to move, resulting in an increase in kinetic energy and therefore an increase in temperature.

2. Cooking with a pressure cooker: As the temperature of the water inside the pot increases, the volume remains constant and the pressure also increases due to the ideal gas law. This allows the boiling temperature of the water to rise above 100 degrees Celsius, which reduces the cooking time of the food.

3. Filling Car Tires: When you fill a car's tires with compressed air at a gas station, you are increasing the pressure of the gas inside the tires. By reducing the volume of air inside the tire, the pressure increases.

## Solved exercises

### Exercise 1

A balloon is inflated with 2 moles of gas at a pressure of 1.5 atm and a temperature of 300 K. If the balloon expands and its volume doubles, what will the new pressure be if the temperature remains constant?

Solution

We can use the ideal gas law to solve this problem. Since the temperature is constant, we can write the following equation:

P₁V₁ = P₂V₂

where P₁ and V₁ are the initial pressure and volume, respectively, and P₂ and V₂ are the final pressure and volume, respectively.

Since the volume is doubled, we have that V₂ = 2V₁. Substituting into the equation, we get:

P₁V₁ = P₂(2V₁)

P₁V₁ = 2P₂V₁

Canceling V₁ on both sides of the equation, we get:

P₁ = 2P₂

Since P₁ = 1.5 atm, we can solve the equation to find P₂:

1.5 atm = 2P₂ P₂ = 1.5 atm / 2 P₂ = 0.75 atm

Therefore, the new pressure will be 0.75 atm.

### Exercise 2

There are two containers of equal volume. In container A there is a gas at a pressure of 2 atm and a temperature of 300 K, while in container B there is the same gas at a pressure of 3 atm and a temperature of 400 K. If the two containers are are brought together and the gas is allowed to mix, what will be the final pressure and final temperature of the system?

Solution

We can use the ideal gas law to solve this problem. The ideal gas law states that the product of the pressure and volume of a gas is proportional to its absolute temperature. Therefore, we can write the following equation for each container:

P₁V = nRT₁ P₂V = nRT₂

where P₁ and P₂ are the initial pressures of containers A and B, respectively, V is the common volume, n is the amount of substance in the gas (which is the same in both containers), R is the ideal gas constant, and T₁ and T₂ are the initial temperatures of containers A and B, respectively.

If the two containers are brought together and the gas is allowed to mix, the total amount of substance and the volume remain constant. Therefore, we can add the two equations:

P₁V + P₂V = nRT₁ + nRT₂

Factoring V and nR, we have:

V(P₁ + P₂) = nR(T₁ + T₂)

Since the containers have the same volume and the amount of substance is the same, we can simplify the equation:

P₁ + P₂ = T₁ + T₂

Substituting the given values, we have:

2 atm + 3 atm = 300K + 400K 5 atm = 700K

Therefore, the final pressure of the system will be 5 atm and the final temperature will be 700 K.

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Data de publicació: July 16, 2023
Última revisió: July 16, 2023