## The 1st law of thermodynamics

$\Delta E_{int}=Q+W$

### Internal Energy $E_{int}$

$E=E(p,V,T)$
$\Delta E_{int}=E_{int,f}-E_{int,i}$

• ideal gas
$E_{int}=\frac{i}{2}nRT$ only related to $T$

$\Delta E_{int}=\frac{i}{2}nR\Delta T$
where $i$ is degree of freedom.

|molecule|$i$|
|:-:|:-:|
|monoatomic|3|
|diatomic|5|
|triatomic|6|

• deviration omitted.
1. gas $\Rightarrow E=K=i\cdot\frac{1}{2}kT$

### Heat $Q$

$Q>0$,the system absord heat.

• thermal Conductivity $H$
$H:=\frac{Q}{\Delta t}=kA \frac{\Delta T}{\Delta x}$

$k$ is a constant related to material(W·m^-1^ ·K^-1^); $A$ is contact area;
$\Delta T$ is the temperature difference between 2 end plane; $\Delta x$ is thickness.

• heat capacity(热容)
$C:=\frac{Q}{\Delta T}$

### Work $W$

$W>0$,the system is positive worked towards.

$W=\int_{x_i}^{x_f} F_x dx=\int_{x_i}^{x_f}-pAdx=-\int_{V_i}^{V_f}pdV$
$\mathbb{NOTICE}-$

## isothermal process ($T$ is invariant)

• ideal gas
1. equation:$pV=C$, where $C$ is a constant.

$W=-\int_{V_i}^{V_f}pdV=-\int_{V_i}^{V_f}\frac{nRT}{V}dV=-nRT\ln{\frac{V_f}{V_i}}$
$E_{int}=\frac{i}{2}nRT$,$T$ is invariant, $\Delta E_{int}=0$
$\Rightarrow Q=-W=nRT\ln{\frac{V_f}{V_i}}$

## isovolumetric process ($V$ is invariant)

$V$ is invariant $\Rightarrow dV=0\Rightarrow W=0$
In this process,define constant volume specific heat $C_V:=\frac{Q}{n\Delta T}$
$\Delta E_{int}=Q=nC_V\Delta T$
$\mathbb{NOTICE}$:$C_V$is measured in isovolumetric process,
but $\Delta E_{int}=nC_V\Delta T$ holds in all process.

• ideal gas
$C_V=\frac{i}{2}R$

## isobaric process ($p$ is invariant)

$W=-\int_{V_i}^{V_f}pdV=-p(V_f-V_i)$
In this process,define constant volume specific heat $C_p:=\frac{Q}{n\Delta T}$
$\Delta E_{int}=Q+W \Rightarrow nC_V\Delta T=nC_p\Delta T - p(V_f-V_i)$

• ideal gas
$W=-p(V_f-V_i)=-nR\Delta T$

$C_p=\frac{i+2}{2}R$

## adiabatic process ($Q=0$)

• ideal gas
1. equation:$pV^\gamma=C$, where $C$ is a constant,

$\gamma=\frac{i+2}{i}>1$ is a constant related to molecule(monoatomic,diatomic,triatomic),
$\Delta E_{int}=W=-\int_{V_i}^{V_f}pdV=-\int_{V_i}^{V_f}\frac{p_iV_i^\gamma}{V^\gamma}dV=\frac{p_fV_f-p_iV_i}{\gamma-1}$

## free expansion

external force $F=0\Rightarrow W=0$, and $Q=0,\Delta E=0$
what changes?! Entropy $S$ !
to be continued.