entropy difference $\Delta S$

equilibrium(reversible) process

$\Delta S:=\int_i^f\frac{dQ}{T}$
ちょと待ってください！What is equilibrium process?
equilibrium process $\Rightarrow$ the process can be drawn in the figure $p-V$

$\Delta E_{int}=Q+W$
$nC_VdT=dQ-pdV$
$\frac{dQ}{T}=p\frac{dV}{T}+nC_V\frac{dT}{T}$

  • ideal gas
    $\frac{dQ}{T}=nR\frac{dV}{V}+nC_V\frac{dT}{T}$

$\Delta S = nR\ln{\frac{V_f}{V_i}}+nC_V\ln\frac{T_f}{T_i}$
using $pV=nRT$, we can also obtain:
$\Delta S = nC_p\ln{\frac{T_f}{T_i}}-nR\ln\frac{p_f}{p_i}$
$\Delta S = nC_p\ln{\frac{V_f}{V_i}}+nC_V\ln\frac{p_f}{p_i}$

irreversible process

Entropy $S$ is a state function.
We utilize reversible process to calculate the entropy difference of irreversible process.

  • ideal gas
    We only need to know initial state $(p_i,V_i,T_i)$ and the final state$(p_f,V_f,T_f)$ and use the formula mentioned above to calculate it.

Carnot cycle(卡诺循环)

  • ideal gas

State:A->B->C->D->A
A->B:$T=T_H,\Delta E_{int,AB}=0, Q_{AB} = Q_H=T_H\Delta S_H>0, W_{AB}=-Q_H<0$, gas works towards enrivonment.
B->C:$T=T_H\rightarrow T_L,\Delta E_{int,BC}=nR\Delta T$
C->D:$T=T_L,\Delta E_{int,CD}=0, Q_{CD} = Q_L=T_L\Delta S_L<0, W_{CD}=-Q_L>0$
D->A:$T=T_L\rightarrow T_H,\Delta E_{int,DA}=nR\Delta T$

efficiency of a Carnot engine $\varepsilon$

$\varepsilon=\frac{E_{out}}{E_{in}}=\frac{|W|}{|Q_H|}=\frac{|Q_H|-|Q_L|}{|Q_H|}=1-\frac{|Q_L|}{|Q_H|}$

for refrigeration

$\varepsilon=\frac{E_{out}}{E_{in}}=\frac{Q_H}{|W|}=\frac{|Q_H|}{|Q_H|-|Q_L|}$

a statistical view of entropy

We denote the number of micro states by $\omega$
$S=k\ln\omega$, where $k$ is Boltzmann constant.

the 2nd law of thermodynamics

  1. For a closed system, in any process, $\Delta S\geq0$
  2. No process is possible whose sole result is the transfer of heat from a reservoir at one temperature to another reservoir at a higher temperature
    That is, there is no perfect engine!