S1.5.4 The ideal gas equation

The Ideal Gas Law and Combined Gas Law: Understanding and Applications

The Ideal Gas Law: A Universal Equation for Gases

Definition

Ideal gas law

The ideal gas law is the equation of state of a hypothetical ideal gas which relates the pressure, volume, temperature, and amount of substance in a gas.

The ideal gas law is a mathematical relationship that connects four key variables describing a gas: pressure ( $P$ ), volume ( $V$ ), temperature ( $T$ ), and the number of moles ( $n$ ):

$P V = n R T$

Here:

$P$ : Pressure (measured in pascals, Pa),
$V$ : Volume (measured in cubic meters, m³),
$n$ : Number of moles of gas,
$R$ : Universal gas constant ( $8.31 J {mol}^{- 1} K^{- 1}$ ),
$T$ : Temperature (measured in kelvin, K).

Key Insights from the Ideal Gas Law:

Pressure and Volume Relationship:
- Compressing a gas (increasing $P$ ) reduces its volume ( $V$ ), while reducing the pressure allows the gas to expand, assuming constant temperature and number of moles.
Temperature and Volume Relationship:
- Heating a gas increases its volume because the particles move faster and exert more outward force.
Amount of Gas:
- Adding more gas molecules (increasing $n$ ) increases the pressure or volume, depending on the situation.

Tip

Always convert temperature to kelvin by adding $273.15$ to the Celsius value before using the ideal gas law. Kelvin is the absolute temperature scale required for gas law calculations.

Example question

Using the Ideal Gas Law

A 0.500 m³ tank contains $2.00 mol$ of oxygen gas at a temperature of $300 K$ . What is the pressure inside the tank?

Solution

Write the ideal gas law: $P V = n R T$
Rearrange for pressure:
$P = \frac{n R T}{V}$
Substitute the known values:
$P = \frac{(2.00) (8.31) (300)}{0.500}$
Calculate:
$P = 9, 972 Pa or approximately 10.0 kPa .$

Note

In this example, we calculated the pressure inside a tank using the ideal gas law. Notice how each unit—moles, kelvin, and cubic meters—aligns with the units of $R$ . This consistency is critical for accurate results.

The Combined Gas Law: Relating Initial and Final States of a Gas

While the ideal gas law is useful for a single set of conditions, many situations involve a gas changing state.

Example

A balloon might expand as it rises to higher altitudes where the pressure decreases.

The combined gas law relates the initial and final states of a gas:

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$

Here:

$P_{1}, V_{1}, T_{1}$ : Initial pressure, volume, and temperature,
$P_{2}, V_{2}, T_{2}$ : Final pressure, volume, and temperature.

Key Assumptions:

The amount of gas ( $n$ ) remains constant.
Temperatures must always be in kelvin.

Tip

Use the combined gas law when two or more variables (pressure, volume, temperature) change simultaneously. It’s especially useful for predicting gas behavior in dynamic conditions, such as weather changes or altitude shifts.

Example question

Using the Combined Gas Law

A weather balloon has a volume of $32.0 {dm}^{3}$ at sea level ( $P_{1} = 100.0 kPa$ , $T_{1} = 298 K$ ). At an altitude of $4500 m$ , the pressure drops to $57.7 kPa$ , and the temperature is $273 K$ . What is the balloon’s new volume?

Solution

Write the combined gas law:
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
Rearrange for $V_{2}$ :
$V_{2} = \frac{P_{1} V_{1} T_{2}}{P_{2} T_{1}}$
Substitute the known values:
$V_{2} = \frac{(100.0) (32.0) (273)}{(57.7) (298)}$
Calculate:
$V_{2} \approx 51.2 {dm}^{3}$

Note

Using the combined gas law, we predicted the balloon’s volume at a higher altitude. The balloon expands because the pressure drops more significantly than the temperature.

Applications: Determining Molar Mass from Experimental Data

The ideal gas law can also be used to calculate the molar mass of a gas.
Molar mass ( $M$ ) is defined as the mass of one mole of a substance. If you know the mass ( $m$ ) of a gas, its volume ( $V$ ), temperature ( $T$ ), and pressure ( $P$ ), you can determine its molar mass using the following formula:

$M = \frac{m R T}{P V}$

Example question

Molar Mass of a Gas

A $2.00 {dm}^{3}$ sample of an unknown gas at STP weighs $3.88 g$ . Determine its molar mass.

Solution

Recall that at STP ( $P = 100.0 kPa$ , $T = 273.15 K$ ), $R = 8.31 J {mol}^{- 1} K^{- 1}$
Write the formula:
$M = \frac{m R T}{P V}$
Substitute the known values:
$M = \frac{(4.40) (8.31) (273.15)}{(100.0 \times 10^{3}) (2.00 \times 10^{- 3})}$
Calculate:
$M \approx 44.0 g m o l^{- 1}$

Note

At STP, the molar volume of any ideal gas is $22.7 {dm}^{3} {mol}^{- 1}$ . This can simplify calculations involving volume and moles.

Units and Conversions: A Critical Step

Volume:

Use cubic meters ( $m^{3}$ ) in calculations. Convert from:
$1 {dm}^{3} = 10^{- 3} m^{3}$ ,
$1 {cm}^{3} = 10^{- 6} m^{3}$ .

Pressure:

Use pascals ( $1 Pa = 1 N / m^{2}$ ).
- Common conversions:
- $1 kPa = 10^{3} Pa$ ,
- $1 atm = 101.3 kPa$ .

Common Mistake

Many students forget to convert temperature to kelvin or use incorrect units for pressure and volume. Always double-check your units before solving gas law problems.

Reflection and Broader Implications

Self review

Can you calculate the volume of a gas at a new temperature and pressure using the combined gas law?
What steps would you follow to determine the molar mass of a gas?

Theory of Knowledge

How does the ideal gas model illustrate the role of assumptions in scientific models?
What are the limitations of applying such models to real-world phenomena?

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S1.5.4 The ideal gas equation

The Ideal Gas Law and Combined Gas Law: Understanding and Applications

The Ideal Gas Law: A Universal Equation for Gases

Key Insights from the Ideal Gas Law:

Using the Ideal Gas Law

The Combined Gas Law: Relating Initial and Final States of a Gas

Key Assumptions:

Using the Combined Gas Law

Applications: Determining Molar Mass from Experimental Data

Molar Mass of a Gas

Units and Conversions: A Critical Step

Volume:

Pressure:

Reflection and Broader Implications

You've read 2/2 free chapters this week.

Introduction to the Ideal Gas Law