#Thermodynamics *For a brief overview of some of the concepts here make sure you take a look at* [[Heat Transfers (Thermo)]]! # Internal Energy - A Brief Introduction Most of the time, thermodynamic systems are going to be difficult to model. Them often being fluids doesn't really help either - heck, even modelling temperature becomes a differential equation! # Work Done - Thermodynamic Systems *Take a look at p-V diagrams - they're virtually inseparable from thermodynamics! Learn about them* [[Kinetic Theory & Ideal Gases (Thermodynamics)#Kinematic Graphs - The pV Diagram|here.]] Let's think of a piston. In between the piston and the walls of a chamber is an ideal gas. You know what, let's draw it out! ![[Pasted image 20231208204843.png]] This work done allows the piston to push itself into the gas - amazing! Now we can consider conditions when the thermal reservoir is turned off - there'll be less pressure on the piston, meaning that it'll move. Displacement doesn't have to be constant, so to be as accurate as possible we model displacement and work as [[Calculus Basics (Maths)#The Derivative|derivatives]] - changes in time. Let's write down the equation for $dW$ with this knowledge! $dW = \overrightarrow{F} ds = (pA)(ds) = pdV$ To find the instantaneous value of the cumulative work done, we can use the integral: $W = \int \, dW = \int_{V_{i}}^{V_{f}} p \, dV $ So we just take the initial and final volumes to get an expression for the work done. Amazing! This is an example of a *p-V diagram*, or a pressure-volume diagram. Below is a visual representation of how these two variables link up with the work done on the system! ![[Pasted image 20231208205629.png]] *Learn about path dependance and more* [[Kinetic Theory & Ideal Gases (Thermodynamics)#Kinematic Graphs - The pV Diagram|here.]] >[!Tip]- Work Done Directions >Remember, we're taking the work done *by the gas on the piston*, so negative work done means that we're doing the opposite, where the piston needs to expend energy in order to reverse the change to the system. # The First Law - Not so Intuitive! Specifically speaking, the first law of thermodynamics is a differential equation relating internal energy, heat and work together - giving us: $\def \dbar{\mkern 5 mu \rlap{\bar{}}} \dbar U = \dbar Q - \dbar W$ Notice that these differentials have a little bar to them - they're actually [[Stokes' and Green's Theorem - More Multivariable Laws (Maths)#Inexact Derivatives|Inexact Derivatives]], dependant on a certain dimension to integrate them on. Thermodynamics isn't bound by the second dimension, after all. *Once you learn about* [[The Haymaker - The Heat Equation (Thermodynamics)|the heat equation]] *this is going to seem coherent - I swear!* Unfortunately, there's no algebraic derivation for this - all we know is that empirically, $Q-W$ is the only quantity not [[Kinetic Theory & Ideal Gases (Thermodynamics)#Kinematic Graphs - The pV Diagram|path dependant]]. Instead, let's try to think of the different thermodynamic systems and how this relates to them! ## Case 1: The Adiabatic System *Main Article:* [[Adiabatic Processes (Thermodynamics)]] The Adiabatic Process is a fun one - a product of idealisation, much like Sadi Carnot's [[Entropy - The Second (Third) Law (Thermodynamics)#Carnot Engines and Perfect Refrigerators|Carnot Engine.]] Essentially, a process happens so fast that heat transfers cannot occur - so $Q$ is 0: $U_{int} = -W$ %%diagram for this. what's a fast process? brainstorm!%% ## Case 2: The Isovolumetric System When volume *must* stay the same, we can't let the gas expand - meaning that it will not do work on its surroundings. In a graphical sense, one of the dimensions for the work done area will be 0. Therefore, we get an expression for the internal energy: $U_{int} = Q$ %%brainstorm examples%% ## Case 3: Cyclical Expansions Some systems are cyclic, meaning that they always revert to the same point after a while of mucking about. Since this system must revert to a certain point, the total internal energy change will never change. Therefore, as $U = Q-W$, for a cyclic process the heat energy is equal to the work done: $Q = W$ %%brainstorm examples%% ## Case 4: Free Expansions *More on these in* [[The Maxwell-Boltzmann Distribution (Thermodynamics)]]! Free expansions occur when nothing acts on a gas - hence the name 'free', the tourist of all 4 cases! If nothing acts on this gas, there won't be any internal energy within the gas to speak of, so there won't be any heat transfers or work being done by the gas either: $Q = W = 0$ %%diagram for this.%% # # Example Questions