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Import converted Year 1 Semester 2 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc

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<b>Lecture 11: Thermal Physics V: PV Planes and Transitions</b><br>
Tue 24 Feb 2026 12:00
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[[Year 1 Sem 2/LC Temperature and Matter/Lecture 10|← Previous lecture]] · [[Year 1 Sem 2/LC Temperature and Matter|Module]] · [[Year 1 Sem 2/LC Temperature and Matter/Lecture 12|Next lecture →]]

<b>Note RE: Last Lecture</b><br>To avoid confusion, the Physics convention is: <div class="pn-import-equation"><nowiki>$$W_\text{by} = \int P dV$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$W_\text{on} = - \int P dV$$</nowiki></div>

== 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:

[[File:TM-Y1S2-lec11-01.png|thumb|center|600px]]

The two transitions from <nowiki>$A \to B$</nowiki> are processes, and the nature of the process depends on the shape of the transition on the PV plot.

<ul>
<li>

Since the two possible transitions between A and B share the same start and end point, the change in internal energy <nowiki>$U_{AB}$</nowiki> is the same for both. Since we go from <nowiki>$T_A$</nowiki> to <nowiki>$T_B$</nowiki> in both cases (as <nowiki>$U$</nowiki> depends on <nowiki>$T$</nowiki>, which in turn depends on <nowiki>$P, V$</nowiki> so for the same <nowiki>$P, V$</nowiki> we have the same <nowiki>$T$</nowiki>), both paths have the same start/end internal energy, so the same change between them.

</li>
<li>

The work done by the gas in the two paths is not the same, however. <nowiki>$W_\text{by}$</nowiki> is larger for the blue path as <nowiki>$W$</nowiki> is an integral of <nowiki>$\int P dV$</nowiki>, 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.

</li>
<li>

What about heat input to the gas, <nowiki>$Q_\text{in}$</nowiki>? Since the First Law says <nowiki>$\Delta U = Q_\text{in} + W_\text{on} \implies \Delta U = Q_\text{in} - W_\text{by}$</nowiki>.

<nowiki>$\Delta U$</nowiki> is constant, and <nowiki>$-W_\text{by}$</nowiki> has a larger magnitude for blue, so must <nowiki>$Q_\text{in}$</nowiki> to keep <nowiki>$\Delta U$</nowiki> constant.

</li>
</ul>

These transitions are called a <i>PV Cycle</i> as we can start at <nowiki>$A$</nowiki>, go to <nowiki>$B$</nowiki> and end back up at <nowiki>$A$</nowiki>.

Consider a simple example with straight lines:

[[File:TM-Y1S2-lec11-02.png|thumb|center|600px]]

Where <nowiki>$1 \to 2$</nowiki> and <nowiki>$3 \to 4$</nowiki> are isochoric processes and <nowiki>$2 \to 3$</nowiki> and <nowiki>$4 \to 1$</nowiki> are isobaric. We can also look at isothermal and adiabatic processes:

[[File:TM-Y1S2-lec11-03.png|thumb|center|600px]]

A cycle must follow these rules:

<ol>
<li>

<nowiki>$\Delta U = 0$</nowiki>, for a full cycle.

</li>
<li>

Total work done is given by the area contained by the cycle.

</li>
<li>

Clockwise cycle for positive work, anticlockwise for negative work.

</li>
<li>

<nowiki>$Q_\text{in} + W_\text{on} = 0$</nowiki> as <nowiki>$\Delta U = 0$</nowiki>.

</li>
</ol>

== Otto Cycle ==

This describes a petrol internal combustion engine:

[[File:TM-Y1S2-lec11-04.png|thumb|center|600px|Blue: Isobaric, Green: Isochoric, Red: Adiabatic]]

This has the following components:

<ol>
<li>

<nowiki>$1 \to 2$</nowiki>: Intake stroke: Isobaric introduction of air and fuel to the system.

</li>
<li>

<nowiki>$2 \to 3$</nowiki>: Adiabatic compression of the fluid as the piston compresses it.

</li>
<li>

<nowiki>$3 \to 4$</nowiki>: Isochoric introduction of heat energy to the system (ignition of the fluid mixture).

</li>
<li>

<nowiki>$4 \to 5$</nowiki>: Adiabatic expansion of the fluid mixture as the gas does work on the piston (pushing it down).

</li>
<li>

<nowiki>$5 \to 2$</nowiki>: Isochoric expulsion of heat from the system.

</li>
<li>

<nowiki>$2 \to 1$</nowiki>: Mass of air and exhaust gases released at a constant pressure.

</li>
</ol>

== Diesel Cycle ==

Here, the ignition and combustion process is performed at a constant pressure where work is done by the gas:

[[File:TM-Y1S2-lec11-05.png|thumb|center|600px|Blue: Isobaric, Green: Isochoric, Red: Adiabatic]]

In both cases, heat is introduced (by combustion) from <nowiki>$3 \to 4$</nowiki>, denoted <nowiki>$Q_1$</nowiki> and heat is lost from <nowiki>$5 \to 2$</nowiki> when heat is expelled from the system. The useful work done is <nowiki>$Q_1 - Q_2$</nowiki> 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 <nowiki>$2 \to 3$</nowiki>.)

<nowiki>$10^{-3} \mathrm{m^3}$</nowiki> of nitrogen gas is initially at atmospheric pressure and <nowiki>$298 \mathrm{K}$</nowiki> and is compressed to <nowiki>$1/15$</nowiki>th of its original volume.

You may assume that nitrogen is an ideal diatomic gas with <nowiki>$\gamma=1.4$</nowiki> and <nowiki>$C_V = 20.85 J/Kmol$</nowiki>.

Calculate:

<ol>
<li>

The final pressure.

</li>
<li>

The final temperature.

</li>
<li>

The number of moles in the gas.

</li>
<li>

The change in internal energy.

</li>
<li>

The work done by the gas.

</li>
</ol>

<b>Part I</b><br>Atmospheric pressure is <nowiki>$101325 \approx 10^5\mathrm{Pa}$</nowiki>. As this is an adiabatic transition, <nowiki>$PV^\gamma = \text{const}$</nowiki>. Hence: <div class="pn-import-equation"><nowiki>$$P_1 V_1^\gamma = P_2 V_2^\gamma$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\implies P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\implies P_2 = P_1 \left(\frac{V_1}{(V_1 / 15)}\right)^\gamma$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\implies P_2 = P_1 \left(15\right)^\gamma = 4.5 \mathrm{M\,Pa}$$</nowiki></div>

<b>Part II</b><br>We can use either <nowiki>$PV = nRT$</nowiki> twice, using the initial conditions to determine <nowiki>$n$</nowiki> or we can use: <div class="pn-import-equation"><nowiki>$$T_1 V_1^{(\gamma - 1)} = T_2 V_2^{(\gamma - 1)}$$</nowiki></div>

Using the same process gives: <div class="pn-import-equation"><nowiki>$$T_2 = T_1 (15)^{(\gamma - 1)} = 880 \mathrm{K}$$</nowiki></div>

<b>Part III</b><br>Since we know <nowiki>$P_1, V_1$</nowiki> or <nowiki>$P_2, V_2$</nowiki> we can use <nowiki>$PV = nRT$</nowiki>, as mentioned above. It would be a good sanity check to do both and compare the values.

<b>Part IV</b><br>It’s an adiabatic transition, we can related <nowiki>$C_V = \frac{\partial U}{\partial T}$</nowiki>.

Integrating both sides gives: <div class="pn-import-equation"><nowiki>$$\int dT = \int \frac{\partial U}{\partial T}$$</nowiki></div>

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Ash: Import converted Year 1 Semester 2 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc