# Ideal Gases

## Ideal Gases Revision

**Ideal Gases**

You may recall that the SI unit for temperature is **Kelvin**. Here we look at the conversion between **degrees** **Celsius** and **Kelvin** before looking at the ideal gas laws.

**The Kelvin Scale**

Commonly, temperature is measured in **degrees Celsius **(\degree C) which is appropriate for measuring everyday temperatures.

However, in **thermodynamics** another scale is used for measuring temperature which is known as the **Kelvin Scale**. The Kelvin Scale begins at 0 \text{ K} which is known as **absolute zero** and increases from there. There are no negative numbers in the **Kelvin Scale** as absolute zero (0 \text{ K}) is the lowest possible temperature.

**Absolute zero **(0 \text{ K}) is the point at which all particles **stop vibrating **and therefore, have **zero kinetic energy**. It is no longer possible to remove energy and therefore, temperature cannot decrease further.

0 \text{ K} is equal to -273.15 \degree C and every increase by 1\degree C corresponds to an increase in 1 \text{ K}.

To convert from degrees Celsius to Kelvin:

T\text{ (K)}= T \, (\degree C) + 273.12

- T\text{ (K)}= the
**temperature in kelvin**\text{(K)} - T \, (\degree C)= the
**temperature in degrees Celsius**(\degree C)

**Example: **The temperature at a point in space is measured to be -270 \degree C. Convert this temperature to Kelvin.

**[1 mark]**

\begin{aligned} T\text{ (K)} &= T \, (\degree C) + 273.15 \\ &= \textcolor{00d865}{-270} + 273.15 \\ &= \bold{3.15} \textbf{ K} \end{aligned}

**The Ideal Gas Laws**

There are three key gas laws which describe the relationship between **pressure, temperature **and** volume**.

**Boyle’s Law**:

**Boyle’s law**** states that the ****pressure (P) of a gas is inversely proportional to the volume (V) for a gas of constant temperature**.

P \propto \dfrac{1}{V}

- P=
**pressure**in pascals \text{(Pa)} - V=
**volume**in cubic metres \text{(m}^3\text{)}

This can also be written experimentally as:

P_1V_1=P_2V_2

- P_1= the
**initial pressure**in pascals \text{(Pa)} - V_1= the
**initial volume**in cubic metres \text{(m}^3\text{)} - P_2= the
**final pressure**in pascals \text{(Pa)} - V_2= the
**final volume**in cubic metres \text{(m}^3\text{)}

Or

PV=\text{constant}

**Example:** A 0.1 \text{m}^3 piston contains a gas under 100 \text{ Pa} of pressure. The volume of the pressure is reduced to 0.02 \text{ m}^3 piston. Calculate the new pressure assuming the temperature remains constant.

**[2 marks]**

\begin{aligned} \bold{P_1V_1} &= \bold{P_2V_2} \\ P_2 &= P_1 \dfrac{V_1}{V_2} \\ &= \textcolor{7cb447}{100} \times \dfrac{\textcolor{f95d27}{0.1}}{\textcolor{10a6f3}{0.2}} \\ &= \bold{500} \text{ Pa} \end{aligned}

**Charles’s Law:**

**Charles’s Law states that the volume of a gas is proportional to the temperature of the gas for a gas at constant pressure.**

V \propto T

- V=
**volume**in cubic metres \text{(m}^3\text{)} - T=
**temperature**in kelvin \text{(K)}

This may also be written experimentally as:

\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}

- V_1= the
**initial volume**in \text{(m}^3\text{)} - T_1= the
**initial temperature**in kelvin \text{(K)} - V_2= the
**final volume**in \text{(m}^3\text{)} - T_2= the
**final temperature**in kelvin \text{(K)}

**Example:** A container contains a gas of a fixed pressure. The initial volume of the container is 0.2 \text{ m}^3 at a temperature of 25 \degree C. The container is heated to 75 \degree C. What is the new volume of the container?

**[2 marks]**

\begin{aligned} \bold{\dfrac{V_1}{T_1}} &= \bold{\dfrac{V_2}{T_2}} \\ V_2 &= V_1 \times \dfrac{T_2}{T_1} \\ &= \textcolor{10a6f3}{0.2} \times \dfrac{\textcolor{7cb447}{75+273}}{\textcolor{ffad05}{25+273}}&= \bold{0.23} \textbf{m} \bold{^3} \end{aligned}

**Pressure Law: **

**The Pressure Law states for a fixed volume of gas, pressure and temperature are proportional. **

P \propto T

- P=
**pressure**in pascals \text{(Pa)} - T=
**temperature**in kelvin \text{(K)}

This may also be written experimentally as:

\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}

- P_1= the
**initial pressure**in pascals \text{(Pa)} - T_1= the
**initial temperature**in kelvin \text{(K)} - P_2= the
**final pressure**in pascals \text{(Pa)} - T_2= the
**final temperature**in kelvin \text{(K)}

**Example: **A fixed volume of gas is stored at 10\degree C under a pressure of 250 \text{ Pa}. The container is cooled to 5 \degree C. Calculate the new pressure inside the container.

**[2 marks] **

\begin{aligned} \bold{\dfrac{P_1}{T_1}} &= \bold{\dfrac{P_2}{T_2}} \\ P_2 &= P_1 \dfrac{T_1}{T_2} \\ &= \textcolor{f95d27}{250} \times \dfrac{\textcolor{7cb447}{5 + 273}}{\textcolor{00bfa8}{10+273}} \\ &= \bold{246} \textbf{ Pa} \end{aligned}

It is important to note that **real gases do not follow these ideal gas laws**. The ideal gas laws rely upon certain **assumptions**. These assumptions (which are listed later in this page) are not met by real gases.

**Ideal Gases**

An **ideal gas** is a gas that follows all three gas laws. These combine form a new equation:

PV \propto T

Or written as an equation:

PV=kT

- P=
**pressure**in pascals \text{(Pa)} - V=
**volume**in cubic metres \text{(m}^3\text{)} - k= a
**constant of proportionality** - T=
**temperature**in kelvin \text{(K)}

An **ideal gas **consists of molecules moving around in **random directions**, at high speeds. The molecules collide with each other and the walls of the container, **exerting an outwards force **on the container. The **kinetic energy** of each particle is represented below. The molecules are moving in random directions with different amounts of kinetic energy.

As the **temperature** increases, the **kinetic energy** of each particle increases. Therefore, the molecules collide with the containers with more energy and **more frequently**, increasing the **pressure**.

Similarly, if the **volume of the container is decreased**, the molecules collide with the container more often, and therefore **increasing pressure**.

**The Ideal Gas Equation**

For a gas to be considered an **ideal gas**, we must make some assumptions about the gas:

**The volume of each molecule is considered to be negligible.****All collisions with the walls of the container are elastic.****The gas molecules do not interact with each other (no collisions, no intermolecular forces)****The gas obeys the three gas laws.**

The ideal gas equation can be written in two forms:

PV=nRT

- P=
**pressure**in pascals \text{(Pa)} - V=
**volume**in cubic metres \text{(m}^3\text{)} - n= the
**number of moles**in the gas - R= the
**molar gas constant**(=8.31 \text{ Jkg}^{-1}\text{mol}^{-1}) - T=
**temperature**in kelvin \text{(K)}

To calculate the number of moles, the equation below can be used:

n=\dfrac{m}{M}

- n=
**number of moles** - m=
**mass of the substance**in grams \text{(g)} - M=
**molar mass**of the substance in grams \text{(g)}

Alternatively, another form of the ideal gas equation can be used:

PV=NkT

- P=
**pressure**in pascals \text{(Pa)} - V=
**volume**in cubic metres \text{(m}^3\text{)} - N= the
**number of molecules**in the gas - k= the
**Boltzmann constant**(=1.38 \times 10^{-23} \text{ JK}^{-1}) - T=
**temperature**in kelvin \text{(K)}

To calculate the number of molecules in a gas we can multiply the number of moles by Avogadro’s constant (6.02 \times 10^{23} \text{ mol}^{-1}).

**Required Practical 8**

**Investigating Boyle’s Law.**

**Doing the experiment:**

- Set up the experiment as shown in the diagram.
- Ensure there are no
**leakages**when the equipment has been fully set up. If there are, these need to be resolved before starting. - Open the
**air tap**and begin pumping air into the cylinder using the**air pump**. - Keep pumping until the oil reaches the upper part of the column.
- Close the air tap once the oil stops rising and the
**pressure**reading remains**constant**. - Allow two minutes for the compressed air to
**cool down**to room temperature, then record the**pressure**and**volume**in your results table. - Repeat 10 times to collect 10 pairs of readings for pressure and volume.

**Analysis of Results:**

If completed correctly, you should find that the product of pressure and volume of each pairs of readings gives a constant:

PV=k

Alternatively, a graph of pairs of readings for P against V could be produced. A** linear graph** would be produced showing that **Boyle’s law** had been abided.

If temperature is not kept constant, the graph will not be linear and Boyle’s law cannot be applied. To keep a constant temperature it is important to keep a constant room temperature and allow the equipment and gas to cool after each trial.

## Ideal Gases Example Questions

**Question 1:** Describe the effect of increasing temperature on the pressure for a container of fixed volume of an ideal gas.

**[2 marks]**

As the temperature increases, **the kinetic energy of particles also increases**. This increases pressure as the **particles collide more often with the container**.

**Question 2:** Explain the term absolute zero?

**[2 marks]**

The temperature on the Kelvin scale that is equivalent to \bold{-273.15 \degree C}. This is the point at which particles stop vibrating and have **zero kinetic energy**.

**Question 3:** What is Boyle’s law?

**[2 marks]**

Boyle’s law states that the pressure and volume of a gas **at fixed temperature**, are **inversely proportional**.

**Question 4:** A sample 3.2 \text{ g} of oxygen has a volume of 2100 \text{ cm}^3 at a pressure of 120 \text{ kPa}. Calculate the temperature of the gas.

**[3 marks] **

\begin{aligned} \bold{PV} &= \bold{nRT} \\ T &= \dfrac{PV}{nR} \\ &= \dfrac{120 \times 10^3 \times 2.1 \times 10^{-3}}{0.05 \times 8.31} \\ &= \bold{606} \textbf{ K} \end{aligned}