Volume and pressure calculations

In a nutshell

When gas pressure is constant, volume is directly proportional to temperature. When temperature is constant, pressure is inversely proportional to volume.

Equations

Word Equation	Symbol Equation
$pressure\space \times \space volume = constant$	$p\times V=C$
$pressure \space object \space 1 \times volume \space object \space 1=pressure \space object \space 2 \times volume \space object \space 2$	$p_1\times V_1=p_2\times V_2$

Variable definitions

Quantity Name	Symbol	Unit Name	Unit
$pressure$	$p$	$Pascal$	$Pa$
$volume$	$V$	$cubic \space metre$	$m^3$
$constant$	$C$	$Pascal \space cubic \space metre$	$Pa\,m^3$

Volume and pressure

Consider a container of gas with a movable wall:

Physics; Matter; KS4 Year 10; Volume and pressure calculations

Initially, the gas inside the container has the same pressure as the air outside, so the wall remains stationary.

Moving the wall inwards, decreases the volume available for the gas. This means the gas particles hit the walls of the container more often. Assuming that temperature remains constant, the pressure increases.

The formula relating the pressure and volume of a gas at a constant temperature is:

$pressure\, \times \, volume = constant \\ \ \\ p\times V=C$

where the variables are defined above.

The product of an object's pressure and volume at constant temperature is equal to a constant, $C$ . This means for two different objects, the products of their pressure and volumes at the same temperature will equal the same constant.

This means that we can equate the two objects as they are equal to each other:

$pressure \space object \space 1 \times volume \space object \space 1=pressure \space object \space 2 \times volume \space object \space 2 \\ \ \\ p_1\times V_1=p_2\times V_2$

Note: Object $1$ and Object $2$ can either be the same object but in two different points in time or they could be two completely independent objects as long as the temperature remains constant.

Example

The volume of a gas in a sealed cylinder starts at $5.0 \times 10^{-5} \, m^3$ with a pressure of $1.2\times 10^6 \, Pa$ . A piston is pushed slowly into the cylinder, increasing the gas pressure to $2.0 \times 10^6 \, Pa$ . What is the new volume of the gas assuming the temperature remains constant?

Write down the relevant information provided in the question (converting to base units where appropriate):

$p_1=1.2\times 10^6\,Pa \\ \ \\ V_1=5.0\times 10^{-5}\, m^3 \\ \ \\p_2=2.0\times 10^6 \, Pa.$

Write down the relevant formula:

$p\times V=constant \\ \ \\ p_1\times V_1=p_2\times V_2$ .

Rearrange the formula:

$V_2=\frac{p_1\times V_1}{p_2}$ .

Substitute the values into the correct formula:

$V_2=\frac{1.2\times 10^6\,\times \, 5.0\times10^{-5}}{2.0\times10^6} \\ \ \\ V_2= \underline{3.0\times10^{-5}\,m^3}.$

So the new volume of the gas is $\underline{3.0\times 10^{-5} \, m^3}$ .