%%need to add a small addendum.%% # Intro: The Adiabatic System? The isolated system. Where cars come to push, and bad thermodynamics textbooks come to introduce. Imagine this - you're a gas piston in an (idealised) car. In other words, you're your own system (go you!), and conservation of energy applies within your system. A spark forces your hand, starting the process - you push against gas, gas inevitably pushes back, creating thrust. Notice this - car manufacturers typically don't want their engines to get *too* hot - it'll lead to mechanical failures down the line. But at the same time, they need the gas in the pistons to have enough heat such that there's enough outward pressure when the heat energy is converted back to kinetic energy. So what do they do? They stop the transfer of heat between each piston so that the amount of heat in the system remains the same. In other words, there is **no net heat transfer across the system.** This is an ***adiabatic*** system, where the grass is green, the sky is blue, and the net change in the heat energy of the system is 0. In this feature, we'll unpack the equation for the expansion of each adiabatic gas, so stay tuned! ![[gas-piston-mishap-comic.png]] *A 'small' misunderstanding!* # Understanding the Expansion Law The formula for adiabatic expansion is: $pV^{\gamma} = \mathrm{constan t}$ Where: - $p$ **is the pressure on the system** - $V$ **is the volume of the system** - $\gamma$ **is a constant that changes depending on the nature of the gas (e.g if it's diatomic, etc)** So, before we begin, let's lay down an important aspect of adiabatic gases. As there's no change in the heat energy within the sample, it's possible for us to create a special case for the First Law, namely: $dU = -dW$ Where $dU$ and $dQ$ are the changes in the internal energy and heat energy within the system respectively. This is similar to an isovolumetric process, where no work is being done by the system as there is no change in the volume. Therefore, if we rewrite the equation with [[Kinetic Theory & Ideal Gases (Thermodynamics)#Internal Energy|the formula for internal energy]] and [[Specific and Molar Heat Capacity (Thermodynamics)#EXTENSION Molar Specific Heats|the formula for the specific molar heat energy]], we get: $\frac{3}{2}nR dT = -pdV$ $\frac{3}{2} nC_{V} dT = -pdV$ $\frac{3}{2} n dT = -\frac{p}{C_{V}}dV$ There are two variables in this equation but none of the variables they correspond to. Let's try to create an equation so that we can solve this differential equation. The first obvious thing is the substitution with the ideal gas equation; note that the equation for the internal energy is similar to the ideal gas equation. If we take a derivative against the temperature $T$ across the whole expression, we get: $Vdp + pdV = nRdT$ As per [[Specific and Molar Heat Capacity (Thermodynamics)|molar heat capacity under constant pressure]], we can make a substitution for $R$ in terms of $C_P$ and $C_V$: $\frac{V dp + p dV}{C_{P}-C_{V}} = n dT$ Substitute the $ndT$ term from the internal energy equivalence equation we first got to get: $\frac{V dp + p dV}{C_{P}-C_{V}} = - \frac{p}{C_{V}} dV$ $V dp + p dV = -\frac{C_{P}}{C_{V}}pdV + pdV$ $Vdp = -\frac{Cp}{C_{V}} pdV$ $\frac{dp}{p} + \frac{C_{P}}{C_{V}} \frac{dV}{V} = 0$ Now all we have to do is solve the equations! Remember, it might not look like it, but these quantities are all functions of the temperature. See heat contour maps for more - we're simply differentiating along a 'gradient' of expansion in the temperature function. Our solution is: $\ln |p| + \frac{C_{P}}{C_{V}}\ln|V| = c$ Where $c$ is the integration constant which will be referred to as $\mathrm{constant}$ from now on. If we remove the $\ln$ terms, using log rules we get: $pV^{\frac{C_{P}}{C_{V}}} = \mathrm{cons\tan t}$ This constant $\frac{C_P}{C_V}$ is the $\gamma$ we get initially. In other words, an adiabatically-expanding gas will expand so that $pV^{\gamma}$ will ALWAYS equal to the same constant. So; $p_i V_i^{\gamma} = p_f V_f^{\gamma}$Where a subscript of $i$ denotes the initial state of the gas whilst the $f$ subscript denotes the final state of the gas. The constant $\gamma$ varies depending on the nature of the gas; for example, a diatomic gas (so oxygen, which consists of $\mathrm{O_{2}}$ molecules) has a $\gamma$ of $\frac{7}{5}$ whilst a monoatomic gas has a $\gamma$ of $\frac{5}{3}$. Can you find a similar expression using the temperature instead of the pressure? *TEMPORARY NOTE TO SELF: REMEMBER THE DIFFERENT TYPES OF EXPANSIONS AND HOW THEY'RE REPRESENTED ON THE DIAGRAM! YOU'RE OUT OF PRACTICE, BUT NOT OUT OF TIME. GOOD LUCK!* *ISOTHERMAL - DOWN AN ISOTHERM (PERFECT 1/T PROPORTIONALITY)* *ISOBARIC - STRAIGHT HORIZONTAL LINE* *ISOVOLUMETRIC - STRAIGHT VERTICAL LINE* *ADIABATIC - SIMILAR TO ISOTHERMAL BUT HOPS BETWEEN TWO ISOTHERMS (BECAUSE THERE IS NO HEAT INTO THE SYSTEM!)* *REMEMBER THAT NO WORK - ISOVOLUMETRIC BECAUSE OF W = PdV!* ## Free Expansions # Example Problems >[!Example]- Chapter 19.9 Problem 54 >We know that for an adiabatic process $p V^{\gamma}$ is a constant. Evaluate “a constant” for an adiabatic process involving exactly 2.0 mol of an ideal gas passing through the state having exactly $p =  1.0 \mathrm{atm}$ and $T =  300 \mathrm{K}$. Assume a diatomic gas whose molecules rotate but do not oscillate. > >**SOLUTION:** > >This question is mostly fluff that doesn't matter. One might be fooled into believing that the "diatomic" nature of the gas will force us to do some funky stuff, but that's not true! >First, let's find the volume using the [[Kinetic Theory & Ideal Gases (Thermodynamics)#The Ideal Gas Law|Ideal Gas Law:]] >$V = \frac{nRT}{p} = \frac{2.0 * 8.3145 * 300}{1.0 \times 10^5} = 0.049 m^3$ >So: >$(1.0 \times 10^5) (0.049)^{\gamma} = \mathrm{constant}$ >As the gas is diatomic, we can just bring up our revision booklets to get $\gamma = \frac{7}{5}$ for a diatomic gas, which will give us our constant: >$\mathrm{constant} = (1.0 \times 10^5)(0.049)^\frac{7}{5} = 1.5 * 10^3 \mathrm{N m^{2.2}}$ >[!Example]- Chapter 19.9 Problem 56 > Suppose $1.00 \ \mathrm{L}$ of a gas with $g = 1.30$, initially at $273 \ \mathrm{K}$ and $1.00 \ \mathrm{atm}$, is suddenly compressed adiabatically to half its initial volume. Find its final (a) pressure and (b) temperature. (c) If the gas is then cooled to 273 K at constant pressure, what is its final volume? > > **SOLUTION** > This is another ideal gas law question! > > (a) Remember that the adiabatic gas expansion equation must **always** be equal to the same constant. So since the volume of this sample is changing, the pressure must also change: > $p_{i}V^1.3 = p_{f}(0.5V)^1.3$ > $p_{f} = p_{i} \left( \frac{1}{0.5^1.3} \right)$ > $p_{f} = 2.46 \ \mathrm{atm}$ > > (b) Use the ideal gas law to find the temperature! > $T = \frac{pV}{nR}$ > Since only the volume and pressure changes, the temperature can be written as: > $T = 2.46 \times 0.5 \times 273= 336 \ \mathrm{K}$ > > (c) We'll use [[Kinetic Theory & Ideal Gases (Thermodynamics)#Boyle, Guy-Lussac, Charles - The Prelude|Charles' law]] to find an expression: > $\frac{V_{1}}{V_{2}} = \frac{T_{1}}{T_{2}}$ > So since the volume $V_2$ is now half of the original we must take the ratio to find our target volume $V_1$: > $\frac{273}{336} * 0.5 = 0.406 \ \mathrm{L}$ >[!Example]- Chapter 19.9 Problem 62 >An ideal diatomic gas, with rotation but no oscillation, undergoes an adiabatic compression. Its initial pressure and volume are $1.20 \ \mathrm{atm}$ and $0.200\ \mathrm{m^3}$. Its final pressure is $2.40 \ \mathrm{atm}$. How much work is done by the gas? $$ $$