The First Law of Thermodynamics
- The first law of thermodynamics is based on the principle of conservation of energy
- When energy is put into a gas by heating it or doing work on it, its internal energy must increase:
- The increase in internal energy = Energy supplied by heating + Work done on the system
- The first law of thermodynamics is therefore defined as:
$
\Delta \mathbf{U}=\mathbf{q}+\mathbf{W}
$
- Where:
- $\Delta \mathrm{U}=$ increase in internal energy $(\mathrm{J})$
- $q$ = energy supplied to the system by heating (J)
- $\mathrm{W}=$ work done on the system (J)
- The first law of thermodynamics applies to all situations, not just for gases
- There is an important sign convention used for this equation
- A positive value for internal energy $(+\Delta U)$ means:
- The internal energy $\Delta \mathrm{U}$ increases
- Heat $q$ is added to the system
- Work W is done on the system (or on a gas)
- A negative value for internal energy (- $\Delta \mathrm{U})$ means:
- The internal energy $\Delta \mathrm{U}$ decreases
- Heat q is taken away from the system
- Work W is done by the system (or by a gas) on the surroundings
- This is important when thinking about the expansion or compression of a gas
When the gas expands, it transfers some energy (does work) to its surroundings
- This decreases the overall energy of the gas
- Therefore, when the gas expands, work is done by the gas (-W)
When a gas expands, work done $W$ is negative
- When the gas is compressed, work is done on the gas (+W)
- When a gas is compressed, work done $W$ is positive
Graphs of Constant Pressure & Volume
- Graphs of pressure $\mathrm{p}$ against volume $\mathrm{V}$ can provide information about the work done and internal energy of the gas
- The work done is represented by the area under the line
- A constant pressure process is represented as a horizontal line
- If the volume is increasing (expansion), work is done by the gas and internal energy increases
- If the arrow is reversed and the volume is decreasing (compression), work is done on the gas and internal energy decreases
- A constant volume process is represented as a vertical line
- In a process with constant volume, the area under the curve is zero
- Therefore, no work is done when the volume stays the same
Worked example
The volume occupied by $1.00 \mathrm{~mol}$ of a liquid at $50^{\circ} \mathrm{C}$ is When the liquid is vaporised at an atmospheric pressure of $1.03 \times 10^5 \mathrm{~Pa}$, the vapour has a volume of $5.9 \times 10^{-2} \mathrm{~m}^3$. The latent heat required to vaporise $1.00 \mathrm{~mol}$ of this liquid at $50^{\circ} \mathrm{C}$ at atmospheric pressure is $3.48 \times 10^4 \mathrm{~J}$. Determine for this change of state the increase in internal energy $\Delta U$ of the system.
Answer/Explanation
Step 1:
Write down the first law of thermodynamics
$
\Delta \mathbf{U}=\mathbf{q}+\mathbf{W}
$
Step 2:
Write the value of heating $q$ of the system
This is the latent heat, the heat required to vaporise the liquid $=\mathbf{3 . 4 8} \times \mathbf{1 0 ^ { 4 }} \mathrm{J}$
Step 3:
Calculate the work done $W$
$
\begin{gathered}
\qquad W=p \Delta V \\
\Delta V=\text { final volume }- \text { initial volume }=5.9 \times 10^{-2}-2.4 \times 10^{-5}=0.058976 \mathrm{~m}^3 \\
p=\text { atmospheric pressure }=1.03 \times 10^5 \mathrm{~Pa} \\
\mathrm{~W}=\left(1.03 \times 10^5\right) \times 0.058976=6074.528=6.07 \times 10^3 \mathrm{~J} \\
\text { Since the gas is expanding, this work done is negative } \\
\qquad \mathrm{W}=-6.07 \times 10^3 \mathrm{~J}
\end{gathered}
$
Step 4:
Substitute the values into first law of thermodynamics
$
\Delta U=3.48 \times 10^4+\left(-6.07 \times 10^3\right)=28730=29000 \mathrm{~J}(2 \text { s.f. })
$