The General Gas Equation Revision Notes for Grade 11 NSC Matric Physical Sciences

The General Gas Equation

Understanding the general gas equation

The general gas equation is a powerful tool that combines multiple gas law relationships into one comprehensive equation. This equation allows us to solve problems where pressure, volume, and temperature all change simultaneously for a fixed amount of gas.

The general gas equation builds upon the individual gas laws you have already learned. When we combine Boyle's law and the pressure-temperature relationship, we create a single equation that can handle complex gas behaviour scenarios.

infoNote

The beauty of the general gas equation lies in its versatility - it can handle situations where all three gas properties (pressure, volume, and temperature) change at once, making it incredibly useful for real-world applications.

Development from individual gas laws

Boyle's law foundation

Boyle's law tells us that pressure is inversely proportional to volume when temperature remains constant:

$p \propto \frac{1}{V} \text{ (at constant temperature)}$

This means that as volume increases, pressure decreases, and vice versa.

Pressure-temperature relationship

Similarly, we know that pressure is directly proportional to temperature when volume remains constant:

$p \propto T \text{ (at constant volume)}$

This tells us that as temperature increases, pressure increases proportionally.

Combining the relationships

When we combine these two relationships, we get a more comprehensive picture. If both volume and temperature can vary, we find that:

$p \propto \frac{T}{V}$

This means pressure is directly proportional to temperature and inversely proportional to volume.

Mathematical formulation

Introducing the proportionality constant

To convert our proportionality into an equation, we introduce a proportionality constant, $k$ :

$p = \frac{kT}{V}$

We can rearrange this equation to give us:

$pV = kT$

Alternative forms

The equation can also be written as:

$\frac{pV}{T} = k$

This form clearly shows that for a fixed mass of gas, the ratio of pressure times volume to temperature remains constant.

The general gas equation

Since the ratio $\frac{pV}{T}$ remains constant for a fixed mass of gas, we can compare two different sets of conditions:

$\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}$

This is the general gas equation. The subscripts 1 and 2 refer to the initial and final conditions of the same mass of gas.

Key conditions and requirements

chatImportant

Critical Requirements for Using the General Gas Equation:

Temperature units: Temperature must always be expressed in Kelvin (K) when using the general gas equation. To convert from Celsius to Kelvin, simply add 273:

$T(K) = T(°C) + 273$

Unit consistency: The units used for pressure and volume must be the same on both sides of the equation. You can use any units (kPa, atm, dm³, cm³, etc.) as long as you are consistent.

Mass requirement: The general gas equation only applies when the mass (or number of moles) of gas remains constant. If gas escapes or is added to the system, you cannot use this equation.

Worked examples

lightbulbExample

Worked Example 1: Truck Tyre Pressure

Problem: At the beginning of a journey, a truck tyre has a volume of 30 dm³ and an internal pressure of 170 kPa. The temperature of the tyre is 16°C. By the end of the trip, the volume of the tyre has increased to 32 dm³ and the temperature of the air inside the tyre is 40°C. What is the tyre pressure at the end of the journey?

Solution:

Step 1: Write down all known information

$T_1 = 16°C$
$T_2 = 40°C$
$V_1 = 30 \text{ dm}^3$
$V_2 = 32 \text{ dm}^3$
$p_1 = 170 \text{ kPa}$
$p_2 = ?$

Step 2: Convert temperatures to Kelvin

$T_1 = 16 + 273 = 289 \text{ K}$
$T_2 = 40 + 273 = 313 \text{ K}$

Step 3: Choose the appropriate equation Since temperature, pressure, and volume are all changing, we use the general gas equation:

$\frac{p_2V_2}{T_2} = \frac{p_1V_1}{T_1}$

Step 4: Substitute and calculate

$\frac{(32)p_2}{313} = \frac{(170)(30)}{289}$

$\frac{(32)p_2}{313} = 17.647$

$(32)p_2 = 5523.529$

$p_2 = 172.6 \text{ kPa}$

The pressure will be 172.6 kPa.

lightbulbExample

Worked Example 2: Gas Sample Volume Change

Problem: A sample of gas exerts a pressure of 100 kPa at 15°C. The volume under these conditions is 10 dm³. The pressure increases to 130 kPa and the temperature increases to 32°C. What is the new volume of the gas?

Solution:

Step 1: Write down all known information

$T_1 = 15°C$
$T_2 = 32°C$
$p_1 = 100 \text{ kPa}$
$p_2 = 130 \text{ kPa}$
$V_1 = 10 \text{ dm}^3$
$V_2 = ?$

Step 2: Convert temperatures to Kelvin

$T_1 = 15 + 273 = 288 \text{ K}$
$T_2 = 32 + 273 = 305 \text{ K}$

Step 3: Use the general gas equation

$\frac{p_2V_2}{T_2} = \frac{p_1V_1}{T_1}$

Step 4: Substitute and calculate

$\frac{(130)V_2}{305} = \frac{(100)(10)}{288}$

$\frac{(130)V_2}{305} = 3.47$

$(130)V_2 = 1059.027$

$V_2 = 8.15 \text{ dm}^3$

The volume will be 8.15 dm³.

lightbulbExample

Worked Example 3: Propane Gas Cylinder

Problem: A cylinder of propane gas is kept at a temperature of 298 K. The gas exerts a pressure of 5 atm and the cylinder holds 4 dm³ of gas. If the pressure of the cylinder increases to 5.2 atm and 0.3 dm³ of gas leaks out, what temperature is the gas now at?

Solution:

Step 1: Write down all known information

$T_1 = 298 \text{ K}$
$T_2 = ?$
$V_1 = 4 \text{ dm}^3$
$V_2 = 4 - 0.3 = 3.7 \text{ dm}^3$
$p_1 = 5 \text{ atm}$
$p_2 = 5.2 \text{ atm}$

Step 2: Convert units if necessary Temperature is already in Kelvin. All other values use consistent units.

Step 3: Use the general gas equation

$\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}$

Step 4: Substitute and calculate

$\frac{(5)(4)}{298} = \frac{(5.2)(3.7)}{T_2}$

$0.067 = \frac{19.24}{T_2}$

$(0.067)T_2 = 19.24$

$T_2 = 286.7 \text{ K}$

The temperature will be 286.7 K.

Problem-solving strategy

infoNote

Systematic Approach to Gas Equation Problems

When tackling general gas equation problems, always follow these four steps:

Write down all known information - list all given values clearly
Convert units if necessary - ensure temperature is in Kelvin and other units are consistent
Choose the appropriate equation - use the general gas equation when multiple variables change
Substitute and calculate - carefully substitute values and solve for the unknown

This systematic approach will help you avoid common mistakes and ensure accurate solutions every time.

bookmarkSummary

Key Points to Remember:

The general gas equation is $\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}$ and applies when pressure, volume, and temperature all change for a fixed mass of gas
Temperature must always be in Kelvin - add 273 to convert from Celsius
Units must be consistent on both sides of the equation, but you can use any appropriate units
The equation only works when the mass of gas stays constant - if gas escapes or is added, this equation cannot be used
Follow the four-step problem-solving approach for systematic and accurate solutions

The General Gas Equation (Grade 11 NSC Matric Physical Sciences): Revision Notes