Intro to Thermodynamics

Thermodynamics is one of the foundational forms of physics that deals with energy and heat in systems. This article will give you a comprehensive overview of thermodynamics and gives examples of using these principles.
Thermodynamics explains matter transformations and properties using energy conservation and heat transfer. It's important for devices like engines and refrigerators. Scientists/engineers use it for efficiency, energy optimisation, and environmental understanding.
There are four key laws in thermodynamics!
The First Law - Law of Energy Conservation: This law states that energy can be neither created nor destroyed; it can only change forms. It expresses the conservation of energy, often articulated as "energy is neither created nor destroyed, only transferred or converted."
The Second Law - Law of Entropy: The second law introduces the concept of entropy, the measure of the disorder or how random a system is. It asserts that in any transfer or energy transformation, an isolated system's total entropy will always increase over time.
The Third Law - Law of Absolute Zero: This law, also known as the Nernst heat theorem, describes the behaviour of matter at absolute zero temperature (0 Kelvin). It states that it is impossible to reach absolute zero in a finite number of steps.
The Zeroth Law - Law of Thermal Equilibrium: this law is a fundamental principle to measure temperature. It states that if two systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each other. This law enables the establishment of temperature scales and temperature comparisons between different systems.
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Important concepts in thermodynamicsDefinitionsHere are some important definitions to keep in mind for thermodynamics!
Adiabatic: no heat is transferred, and change in internal energy is only the result of work.
Isobaric: the process takes place under constant pressure.
Isochoric: the process takes place at the same volume.
Isothermal: the process takes place under constant temperature.
Internal Energy (U): the total kinetic energy in a system due to the motion of molecules and the potential energy in atoms.
Work (W): the transfer of energy by a force acting on an object.
For definite expansion between volumes Vi to ViWork done in an isobaric system is greatest.
Work done in an adiabatic system is the least.
Work done in an isothermal system is intermediate.
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Figure 1: Pressure vs Volume Graph for Isobaric, Adiabatic & Isothermal Processes
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WP=C>WT=C>WQ=0W_{P=C}>W_{T=C}>W_{Q=0}WP=C​>WT=C​>WQ=0​
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Wisobaric>Wisothermal>WadiabaticW_{isobaric}>W_{isothermal}>W_{adiabatic}Wisobaric​>Wisothermal​>Wadiabatic​
For a given ΔV (between Vi to Vf)Pressure change in an adiabatic system is greater than that of an isothermal one.
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Figure 2: Pressure vs Volume Graph for Adiabatic and Isothermal ProcessesIllustrating the greater pressure change in an adiabatic system (blue line), in comparison to an isothermal system (black line).
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ΔPadiabatic>ΔPisothermal{\Delta} P_{adiabatic}>{\Delta} P_{isothermal}ΔPadiabatic​>ΔPisothermal​
There is a larger increase in pressure in an adiabatic system, as there is a decrease in volume, and it would abide by Boyle's Law:
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P1V1=P2V2P_1V_1=P_2V_2P1​V1​=P2​V2​
Whereby:
P1 = pressure at the first state
V1 = volume at the first state
P2 = pressure at the second state
V2 = volume at the second state
For different processes between a given initial to final stateΔU (the change in internal energy) is the same across the three different pathways.
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ΔU1=ΔU2=ΔU3{\Delta} U_1={\Delta} U_2={\Delta} U_3ΔU1​=ΔU2​=ΔU3​
An extension of the formula for the internal energy for ideal gases:
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ΔU=cVΔT\Delta U= cV \Delta TΔU=cVΔT
Whereby:
ΔU represents the change in internal energy.
cV represents the specific heat capacity of the gas at constant pressure.
ΔT represents the change in temperature.
This is shown in Figure 3, the processes between i and f:
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ΔU=nCv(Tf−Ti){\Delta} U=nCv(T_f-T_i)ΔU=nCv(Tf​−Ti​)
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Figure 3: Pressure vs Volume Graph, Representing Three Different Processes From The Same Inital to Final State
Whereby:
Process 1 represents an isothermal process.
Process 2 represents an isochoric process.
Process 3 represents a random and general pathway from the initial state to the final state.
From this, it can be stated that the heat and work in these systems are also equal.
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ΔU=Q−W{\Delta} U=Q-WΔU=Q−W
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Q1−W1=Q2−W2=Q3−W3Q_1-W_1=Q_2-W_2=Q_3-W_3Q1​−W1​=Q2​−W2​=Q3​−W3​
The difference between heat and work (Q-W) would also be the same for all three pathways.
ExamplesWhat follows are two examples: 
The first solves using the principle that the difference between heat and work (Q-W) is the same for pathways with the same initial and final state.
The second example uses a P-V (pressure-volume) diagram to calculate the efficiency of a cycle.
Example 1If: ﻿Q(iaf)
:80.00cal​
, ﻿W(iaf)
:50.00cal​
, and for path f(i), ﻿W(a)
:-30.00cal​
.
The value of Q for path f(i) will be which one of the following?
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(1) 60 cal(2) 30 cal(3) −30 cal(4) −60 cal(1)\ 60\ cal\\(2)\ 30\ cal\\(3)\ -30\ cal\\(4)\ -60\ cal(1) 60 cal(2) 30 cal(3) −30 cal(4) −60 cal
Given:
﻿﻿W (fi)
:-30.00​
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﻿﻿W (if)
:30.00​
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Figure 4: Pressure vs Volume Graph for Example 1
This graph depicts two pathways: one with an isochoric and isobaric process, and another depicting the reversal (of the final to the initial state).
For processes i(af) and i(f), same ΔU and Q-W:
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Qiaf−Wiaf=Qif−WifQ_{iaf}-W_{iaf}=Q_{if}-W_{if}Qiaf​−Wiaf​=Qif​−Wif​
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∴Qif=Qiaf−Wiaf+Wif\therefore Q_{if}=Q_{iaf}-W_{iaf}+W_{if}∴Qif​=Qiaf​−Wiaf​+Wif​
﻿﻿Q(if)
:60.00​
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﻿﻿Q(fi)
:-60.00​
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∴ Option 4 is correct. 
Example 2As shown in the P-V diagram, one mole of Helium gas undergoes a cyclic process ABCA. Find the efficiency of the cycle.
Total work done = area bounded by the cycle
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Total Work=12×4γ0× P0= 4P0V0Total\ Work=\frac{1}{2}\times4\gamma_0\times\ P_0\\ =\ 4P_0V_0Total Work=21​×4γ0​× P0​= 4P0​V0​
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Figure 5: Pressure vs Volume Graph for Example 2
Process AB:
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WAB=12(P0+3P0)(4V0)=8P0V0W_{AB}=\frac{1}{2}(P_0+3P_0)(4V_0) =8P_0V_0WAB​=21​(P0​+3P0​)(4V0​)=8P0​V0​
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From: T=PVnRFrom:\ T=\frac{PV}{nR}From: T=nRPV​
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Temp at A:TA=P0V0nRTemp\ at\ A: T_A =\frac{P_0V_0}{nR}Temp at A:TA​=nRP0​V0​​
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Temp at B:TB=(3P0)(5V0)nR=15P0V0nRTemp\ at\ B: T_B =\frac{(3P_0)(5V_0)}{nR}=\frac{15P_0V_0}{nR}Temp at B:TB​=nR(3P0​)(5V0​)​=nR15P0​V0​​
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Figure 6: Pressure vs Volume Graph for Example 2
Change in internal energy:
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ΔUAB=nCV(TB−TA) = n3R2(15−1)P0V0nR=21P0V0{\Delta} U_{AB}=nCV (T_B-T_A)\ =\ n\frac{3R}{2}(15-1) \frac{P_0V_0}{nR}=21P_0V_0ΔUAB​=nCV(TB​−TA​) = n23R​(15−1)nRP0​V0​​=21P0​V0​
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QAB=WAB+ΔUAB=8P0V0+21P0V0=29P0V0Q_{AB}=W_{AB}+{\Delta}U_{AB}=8P_0V_0+21P_0V_0=29P_0V_0QAB​=WAB​+ΔUAB​=8P0​V0​+21P0​V0​=29P0​V0​
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∴ Efficiency =4P0V029P0V0=429\therefore\ Efficiency\ =\frac{4P_0V_0}{29P_0V_0}=\frac{4}{29}∴ Efficiency =29P0​V0​4P0​V0​​=294​
﻿﻿Efficiency (η)
:0.14​
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*Note: Both BC and CA are cooling processes; hence, heat is being supplied in only the AB process.
Additional ResourcesIf you liked this, check out our other articles and resources!
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ReferencesEncyclopaedia Britannica. 2023. Thermodynamics. [ONLINE] Available at: https://www.britannica.com/science/thermodynamics. [Accessed 24 October 2023].
LinkedIn. 2023. Important Thermodynamic Concepts. [ONLINE] Available at: https://www.linkedin.com/feed/update/urn:li:activity:7095576421120548864?utm_source=share&utm_medium=member_desktop. [Accessed 24 October 2023].
Wikipedia. 2023. Internal Energy. [ONLINE] Available at: https://en.wikipedia.org/wiki/Internal_energy. [Accessed 24 October 2023].
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