Defining Internal Energy
- Energy can generally be classified into two forms: kinetic or potential energy
- The molecules of all substances contain both kinetic and potential energies
- The amount of kinetic and potential energy a substance contains depends on the phases of matter (solid, liquid or gas), this is known as the internal energy
- The internal energy of a substance is defined as:
- The sum of the random distribution of kinetic and potential energies within a system of molecules
- The symbol for internal energy is $U$, with units of Joules (J)
- The internal energy of a system is determined by:
- Temperature
- The random motion of molecules
- The phase of matter: gases have the highest internal energy, solids have the lowest
- The internal energy of a system can increase by:
- Doing work on it
- Adding heat to it
- The internal energy of a system can decrease by:
- Losing heat to its surroundings
Exam Tip
When an exam question asks you to define “internal energy”, you can lose a mark for not mentioning the “random motion” of the particles or the “random distribution” of the energies, so make sure you include one of these in your definition!
Internal Energy & Temperature
- The internal energy of an object is intrinsically related to its temperature
- When a container containing gas molecules is heated up, the molecules begin to move around faster, increasing their kinetic energy
- If the object is a solid, where the molecules are tightly packed, when heated the molecules begin to vibrate more
- Molecules in liquids and solids have both kinetic and potential energy because they are close together and bound by intermolecular forces
- However, ideal gas molecules are assumed to have no intermolecular forces
- This means there have no potential energy, only kinetic energy
- The (change in) internal energy of an ideal gas is equal to:
$
\Delta \mathrm{U}=\frac{3}{2} \mathrm{k} \Delta \mathrm{T}
$
- Therefore, the change in internal energy is proportional to the change in temperature:
$
\Delta \mathbf{U} \propto \mathbf{T}
$
- Where:
- $\Delta \mathrm{U}=$ change in internal energy $(\mathrm{J})$
- $\Delta \mathrm{T}=$ change in temperature $(\mathrm{K})$
Worked example: Internal energy & temperature
A student suggests that when an ideal gas is heated from $50^{\circ} \mathrm{C}$ to $150^{\circ} \mathrm{C}$, the internal energy of the gas is trebled.
State and explain whether the student’s suggestion is correct.
Answer/Explanation
Step 1:
Write down the relationship between internal energy and temperature
The internal energy of an ideal gas is directly proportional to its temperature
$
\Delta \mathbf{U} \propto \mathbf{T}
$
Step 2:
Determine whether the change in temperature (in $\mathrm{K}$ ) increases by three times
The temperature change is the thermodynamic temperature ie. Kelvin
The temperature change in degrees from $50^{\circ} \mathrm{C}$ to $150^{\circ} \mathrm{C}$ increases by three times
The temperature change in Kelvin is:
$
\begin{aligned}
& 50{ }^{\circ} \mathrm{C}+273.15=323.15 \mathrm{~K} \\
& 150^{\circ} \mathrm{C}+273.15=423.15 \mathrm{~K} \\
&
\end{aligned}
$
$
\frac{423.15}{323.15}=1.3
$
Therefore, the temperature change, in Kelvin, does not increase by three times
Step 3:
Write a concluding statement relating the temperature change to the internal energy The internal energy is directly proportional to the temperature The thermodynamic temperature has not trebled, therefore, neither has the internal energy Therefore, the student is incorrect
Exam Tip
If an exam question about an ideal gas asks for the total internal energy, remember that this is equal to the total kinetic energy since an ideal gas has zero potential energy
Work Done by a Gas
- When a gas expands, it does work on its surroundings by exerting pressure on the walls of the container it’s in
- This is important, for example, in a steam engine where expanding steam pushes a piston to turn the engine
- The work done when a volume of gas changes at constant pressure is defined as:
$
\mathbf{W}=\mathbf{p} \Delta \mathbf{V}
$
- Where:
- $\mathrm{W}=$ work done $(\mathrm{J})$
- $p=$ external pressure $(\mathrm{Pa})$
- $\mathrm{V}=$ volume of gas $\left(\mathrm{m}^3\right)$
- For a gas inside a cylinder enclosed by a moveable piston, the force exerted by the gas pushes the piston outwards
- Therefore, the gas does work on the piston
- The volume of gas is at constant pressure. This means the force $F$ exerted by the gas on the piston is equal to :
$
\mathbf{F}=\mathbf{p} \times \mathbf{A}
$
- Where:
- $\mathrm{p}=$ pressure of the gas $(\mathrm{Pa})$
- $A=$ cross sectional area of the cylinder $\left(\mathrm{m}^2\right)$
- The definition of work done is:
$
\mathbf{W}=\mathbf{F} \times \mathbf{d}
$
- Where:
- $\mathrm{F}=$ force $(\mathrm{N})$
- $d$ = perpendicular displacement to the force $(m)$
- The displacement of the gas $d$ multiplied by the cross-sectional area $A$ is the increase in volume $\Delta \mathrm{V}$ of the gas:
$
\mathbf{W}=\mathbf{p} \times \mathbf{A} \times \mathbf{d}
$
- This gives the equation for the work done when the volume of a gas changes at constant pressure:
$
\mathbf{W}=\mathbf{p} \Delta \mathbf{V}
$
- Where:
- $\Delta \mathrm{V}=$ increase in volume of the gas in the piston when expanding $\left(\mathrm{m}^3\right)$
- This is assuming that the surrounding pressure $p$ does not change as the gas expands
- This will be true if the gas is expanding against the pressure of the atmosphere, which changes very slowly
- When the gas expands ( $V$ increases), work is done by the gas
- When the gas is compressed ( $V$ decreases), work is done on the gas
Worked example
When a balloon is inflated, its rubber walls push against the air around it.
Calculate the work done when the balloon is blown up from $0.015 \mathrm{~m}^3$ to $0.030 \mathrm{~m}^3$.
Atmospheric pressure $=1.0 \times 10^5 \mathrm{~Pa}$.
Answer/Explanation
Step 1: Write down the equation for the work done by a gas
$
\mathbf{W}=\mathbf{p} \Delta \mathbf{V}
$
Step 2:
Substitute in values
$
\begin{aligned}
& \Delta V=\text { final volume – initial volume }=0.030-0.015=0.015 \mathrm{~m}^3 \\
& \qquad W=\left(1.0 \times 10^5\right) \times 0.015=1500 \mathrm{~J}
\end{aligned}
$
Exam Tip
The pressure $p$ in the work done by a gas equation is not the pressure of the gas but the pressure of the surroundings. This because when a gas expands, it does work on the surroundings.