## Table of Contents

## Ideal Gas

An ideal gas is a theoretical construct that allows scientists and engineers to simplify the complex behaviors of gases under various conditions. The concept is rooted in several assumptions that help in modeling gas behavior with a high degree of accuracy under certain conditions. The primary characteristics and equations associated with an ideal gas can be outlined as follows:

## Characteristics of an Ideal Gas

**Obeying the Ideal Gas Law:**The behavior of an ideal gas is most accurately described by the ideal gas law, which is expressed as $PV = nRT$. In this equation, $P$ represents the pressure of the gas, $V$ its volume, $n$ the number of moles of gas, $R$ the ideal gas constant, and $T$ the temperature of the gas in Kelvin.**No Long-Range Forces:**Particles of an ideal gas are assumed not to exert long-range forces on each other. This means that there are no attractive or repulsive forces affecting the gas particles except during collisions.**Negligible Volume:**The individual particles of an ideal gas occupy a negligible volume compared to the total volume of their container. This assumption implies that the actual space taken up by the gas molecules is so small that it can be considered zero for the purposes of calculations involving the ideal gas law.

## Behavior Under Different Conditions

- Ideal gases tend to exhibit behavior that closely aligns with the ideal gas law at
**low pressures**(or low concentration of air molecules) and**high temperatures**(significantly higher than their boiling points). Under these conditions, the lack of attractive forces between particles becomes a reasonable approximation, as the particles are moving rapidly and are far apart from each other.

## Internal Energy of an Ideal Gas

- For a monatomic ideal gas (a gas consisting of individual atoms, not molecules), the internal energy ($U$) is directly related to its kinetic energy ($E_k$). This relationship indicates that the internal energy of an ideal gas depends solely on its temperature, with $U = E_k$. As the temperature increases, so does the kinetic energy of the gas particles, which in turn increases the internal energy of the gas.

## Useful Equations

- A useful equation for linking pressure, volume, and temperature changes in processes involving ideal gases is derived from the combined gas law, expressed as:

$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$

This equation demonstrates how changes in pressure, volume, and temperature are related. For example, if the temperature of a gas increases, either the volume must increase or the pressure must decrease if the gas is to remain in an ideal state, assuming the amount of gas (in moles) remains constant.

The ideal gas model serves as a fundamental tool in thermodynamics and fluid mechanics, providing a simple yet powerful way to predict and understand the behavior of gases under a variety of conditions. While real gases deviate from ideal behavior under high pressures and low temperatures (due to the effects of intermolecular forces and the finite volume of gas molecules), the ideal gas law remains a crucial starting point for many calculations and theoretical analyses in physics and engineering.

## Worked Examples on Ideal Gas Law

### Example 1: Calculating Moles of Gas in a Balloon

A helium balloon has a volume of 2.0 liters at a pressure of 1.0 atm and a temperature of 300 K. How many moles of helium gas are in the balloon? Use $R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}$.

**Click here to show/hide answer**

To find the number of moles ($n$) of helium gas in the balloon, we use the ideal gas law:

$$PV = nRT$$

Rearranging the formula to solve for $n$ and substituting given values:

$$\begin{aligned} n &= \frac{PV}{RT} \\ &= \frac{(1.0 \, \text{atm})(2.0 \, \text{L})}{(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1})(300 \, \text{K})} \\ &= \frac{2.0}{24.63} \\ &\approx 0.081 \, \text{mol} \end{aligned}$$

**Conclusion:** The balloon contains approximately 0.081 moles of helium gas.

### Example 2: Pressure Change with Temperature in a Fixed Volume

A sealed container with a fixed volume of 1.5 liters contains nitrogen gas at 2.0 atm pressure and a temperature of 250 K. If the temperature is raised to 500 K, what will be the new pressure inside the container?

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We can use the combined gas law, keeping volume constant ($V_1 = V_2$), thus it simplifies to:

$$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$

Rearranging to solve for $P_2$:

$$\begin{aligned} P_2 &= P_1 \frac{T_2}{T_1} \\ &= 2.0 \, \text{atm} \times \frac{500 \, \text{K}}{250 \, \text{K}} \\ &= 2.0 \, \text{atm} \times 2 \\ &= 4.0 \, \text{atm} \end{aligned}$$

**Conclusion:** When the temperature is raised to 500 K, the pressure inside the container increases to 4.0 atm.

### Example 3: Determining Final Temperature after Compression

A gas occupies a volume of 3.0 liters at a temperature of 350 K. If the gas is compressed to a volume of 1.0 liter and the initial pressure and final pressure are the same, what is the final temperature of the gas?

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Since the initial and final pressures are the same and cancel out, we can use the relationship between volume and temperature for a constant amount of gas at constant pressure, known as Charles’s law:

$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$

Rearranging to solve for $T_2$:

$$\begin{aligned} T_2 &= T_1 \frac{V_2}{V_1} \\ &= 350 \, \text{K} \times \frac{1.0 \, \text{L}}{3.0 \, \text{L}} \\ &= \frac{350}{3} \\ &\approx 116.7 \, \text{K} \end{aligned}$$

**Conclusion:** After compression to a volume of 1.0 liter, the final temperature of the gas is approximately 116.7 K.