<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/94bdf5b8-3f7e-42e0-a356-3263b0cce9ae/Ideal_gas.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/94bdf5b8-3f7e-42e0-a356-3263b0cce9ae/Ideal_gas.png" width="40px" /> Ideal gas: a gas with point-like non-interacting poly-atomic molecules which have internal DoF (rot, vib)
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🗒️ Note: all gases tend to an ideal gas at low enough pressures
Ideal gas law
$$ \boxed{PV=nRT \quad \text{or} \quad PV=Nk_B T} $$
where $N$ is the number of molecules, $n=N/N_A$ is the number of moles, $R=8.314 \,\mathrm{JK}^{-1}$ and $k_B=R/N_A=1.381\times 10^{-23} \,\mathrm{JK}^{-1}$
Internal energy of an ideal gas at heat capacity $C_V$ and constant volume
$$ \boxed{\text{d}E=C_V\,\text dT \quad E=C_V T} $$
Heat capacity at constant pressure and volume for ideal gases
$$ \boxed{C_P-C_V=nR} $$
For adiabatic compression and expansion of an idea gas
$$ \begin{matrix} PV^\gamma=\text{constant} & \text{where} & \gamma\equiv \frac{C_P}{C_V} \\ \\ TV^{\gamma-1}=\text{constant} & \text{and} & TP^{\gamma^{-1}-1}=\text{constant} \end{matrix} $$
where $\gamma-1=nR/C_V$
The entropy change of a $n$ mole ideal gas is
$$ \Delta S=nR \ln \left ( \left [ \frac{T_f}{T_i} \right ]^\frac{n_f}{2} \frac{V_f}{V_i} \right ) $$
since $\text dS=\frac{C_V}{T}\,\text dT$ at constant volume, it doesn't work at $T_i\to 0$
🗒️ Note: classical thermodynamics cannot predict the absolute entropy even of an ideal gas
💼 Case: consider a hill with height $h(x,y)$
The highest point is at
$$ \frac{\partial h}{\partial x}=\frac{\partial h}{\partial y}=0 $$
checking that it is a maximum
If we now constrain our highest point search to a line $y=g(x)$ or $u(x,y)=0$
We need to extremize $h(x,y)+\lambda u(x,y)$ with respect to $x$ and $y$
💃 Example: a hemisphere $h=h_0(1-x^2-y^2)$ constrained to $u(x,y)=y-mx-c=0$
$$ \begin{aligned} \frac{\partial (h+\lambda u)}{\partial x}&=0 \quad \Rightarrow -2h_0 x-\lambda m=0 \quad \Rightarrow x=-\lambda m /2h_0 \\ \frac{\partial (h+\lambda u)}{\partial y}&=0 \quad \Rightarrow -2h_0 y+\lambda=0 \quad \Rightarrow y=\lambda / 2h_0 =-x/m \end{aligned} $$
Combining the constraint with $y=x/m$ we get $x=-c(m^{-1}+m)$ so
$$ (x_m,y_m)=\frac{c}{1+m^2}(-m,1) \quad h_m=h_0\frac{1+m^2-c^2}{1+m^2} $$
$$ \begin{matrix} \text{Euler}&\cos\theta=\frac 12 (e^{i\theta}+e^{-i\theta}) & \sin\theta=\frac 1{2i} (e^{i\theta}-e^{-i\theta}) & \tan\theta=\frac{e^{i\theta}-e^{-i\theta}}{i\left [ e^{i\theta}+e^{-iv} \right ]} \\ \\ &\sec \theta \equiv \frac 1 {\cos\theta} & \cosec\theta\equiv \frac 1 {\sin \theta} & \cot \theta\equiv \frac 1 {\tan \theta} \\ \\ \text{Euler}&\cosh x=\frac 12 (e^{x}+e^{-x}) & \sinh x=\frac 12 (e^{x}-e^{-x}) & \tanh x=\frac{e^{x}-e^{-x}}{ e^{x}+e^{-x} } \\ \\ &\operatorname{sech} x \equiv \frac 1 {\cosh x} & \operatorname{cosech} x\equiv \frac 1 {\sinh x} & \operatorname{coth} x\equiv \frac 1 {\tanh x} \\ \\ &\frac{\text d \cosh x}{\text dx} =\sinh x & \frac{\text d \sinh x}{\text dx} =\cosh x & \frac{\text d \tanh x}{\text dx} =\operatorname{sech}^2 x \\ \\ \text{Taylor}& \lim_{x\to 0} \sinh x = x & \lim_{x\to 0} \cosh x = 1+\frac 12 x^2 & \lim_{x\to 0} \tanh x = x \\ \\\ \text{Taylor} & \text{large approx} & \lim_{x\to 0} \cosh x = 1 & \lim_{x\to 0} \cosh x -1 = \frac 12 x^2 \\ \\ \text{Taylor} & \text{exponential} & \lim_{x\to 0} e^x = 1 & \lim_{x\to 0} e^x-1=x \\ \\ \text{Taylor}& \lim_{x\to \infin} \sinh x = \frac 12 e^x & \lim_{x\to \infin} \cosh x =\frac 12 e^x & \lim_{x\to \infin} \tanh x = 1 \end{matrix} $$
