Year 1 Sem 2/LC Temperature and Matter/Lecture 11
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Note RE: Last Lecture
To avoid confusion, the Physics convention is:
PV Diagrams
We can move between points on a PV plane through processes which usually involve some kind of heat transfer and work. For example:
The two transitions from $A \to B$ are processes, and the nature of the process depends on the shape of the transition on the PV plot.
- Since the two possible transitions between A and B share the same start and end point, the change in internal energy $U_{AB}$ is the same for both. Since we go from $T_A$ to $T_B$ in both cases (as $U$ depends on $T$, which in turn depends on $P, V$ so for the same $P, V$ we have the same $T$), both paths have the same start/end internal energy, so the same change between them.
- The work done by the gas in the two paths is not the same, however. $W_\text{by}$ is larger for the blue path as $W$ is an integral of $\int P dV$, which represents the area under the curve. The blue line is further up the pressure axis, so the integral is larger and more work is done.
- What about heat input to the gas, $Q_\text{in}$? Since the First Law says $\Delta U = Q_\text{in} + W_\text{on} \implies \Delta U = Q_\text{in} - W_\text{by}$. $\Delta U$ is constant, and $-W_\text{by}$ has a larger magnitude for blue, so must $Q_\text{in}$ to keep $\Delta U$ constant.
These transitions are called a PV Cycle as we can start at $A$, go to $B$ and end back up at $A$.
Consider a simple example with straight lines:
Where $1 \to 2$ and $3 \to 4$ are isochoric processes and $2 \to 3$ and $4 \to 1$ are isobaric. We can also look at isothermal and adiabatic processes:
A cycle must follow these rules:
- $\Delta U = 0$, for a full cycle.
- Total work done is given by the area contained by the cycle.
- Clockwise cycle for positive work, anticlockwise for negative work.
- $Q_\text{in} + W_\text{on} = 0$ as $\Delta U = 0$.
Otto Cycle
This describes a petrol internal combustion engine:
This has the following components:
- $1 \to 2$: Intake stroke: Isobaric introduction of air and fuel to the system.
- $2 \to 3$: Adiabatic compression of the fluid as the piston compresses it.
- $3 \to 4$: Isochoric introduction of heat energy to the system (ignition of the fluid mixture).
- $4 \to 5$: Adiabatic expansion of the fluid mixture as the gas does work on the piston (pushing it down).
- $5 \to 2$: Isochoric expulsion of heat from the system.
- $2 \to 1$: Mass of air and exhaust gases released at a constant pressure.
Diesel Cycle
Here, the ignition and combustion process is performed at a constant pressure where work is done by the gas:
In both cases, heat is introduced (by combustion) from $3 \to 4$, denoted $Q_1$ and heat is lost from $5 \to 2$ when heat is expelled from the system. The useful work done is $Q_1 - Q_2$ and is given by the area enclosed by the loop.
Example
Consider the adiabatic gas compression stroke in a diesel engine. (This is the adiabatic stroke where pressure is increasing/volume is decreasing, i.e. from $2 \to 3$.)
$10^{-3} \mathrm{m^3}$ of nitrogen gas is initially at atmospheric pressure and $298 \mathrm{K}$ and is compressed to $1/15$th of its original volume.
You may assume that nitrogen is an ideal diatomic gas with $\gamma=1.4$ and $C_V = 20.85 J/Kmol$.
Calculate:
- The final pressure.
- The final temperature.
- The number of moles in the gas.
- The change in internal energy.
- The work done by the gas.
Part I
Atmospheric pressure is $101325 \approx 10^5\mathrm{Pa}$. As this is an adiabatic transition, $PV^\gamma = \text{const}$. Hence:
Part II
We can use either $PV = nRT$ twice, using the initial conditions to determine $n$ or we can use:
Using the same process gives:
Part III
Since we know $P_1, V_1$ or $P_2, V_2$ we can use $PV = nRT$, as mentioned above. It would be a good sanity check to do both and compare the values.
Part IV
It’s an adiabatic transition, we can related $C_V = \frac{\partial U}{\partial T}$.
Integrating both sides gives: