Electric conduction in liquids (2)

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In a liquid it is ions and not electrons that lead to current conduction. In this case the resistance is given by the mobility of the ions within the liquid and the resistance must be calculated based on the concentrations of all the components.

>Model

ID:(2122, 0)



Mechanisms

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Code
Concept

Mechanisms

ID:(16035, 0)



Model

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Parameters

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$c_2$
c_2
Concentration of ions 2
mol/m^3
$\kappa_1$
kappa_1
Conductivity ions of type 1
1/Ohm m
$\kappa_2$
kappa_2
Conductivity ions of type 2
1/Ohm m
$L$
L
Conductor length
m
$c_1$
c_1
Ion concentration of type 1
mol/m^3
$m_1$
m_1
Mass of the ion 1
kg
$m_2$
m_2
Mass of the ion 2
kg
$\Lambda_1$
Lambda_1
Molar conductivity 1
m^2/Ohm mol
$\Lambda_2$
Lambda_2
Molar conductivity 2
m^2/Ohm mol
$R$
R
Resistance
Ohm
$\rho_e$
rho_e
Resistivity
Ohm m
$S$
S
Section of Conductors
m^2

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$Q_1$
Q_1
Charge of the ion 1
C
$Q_2$
Q_2
Charge of the ion 2
C
$G$
G
Conductance
S
$\kappa_e$
kappa_e
Conductivity
1/Ohm m
$\tau_1$
tau_1
Time between collisions ion 1
s
$\tau_2$
tau_2
Time between collisions ion 2
s

Calculations


First, select the equation: to , then, select the variable: to

Calculations

Symbol
Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

Variable Given Calculate Target : Equation To be used




Equations

#
Equation

$ G =\displaystyle\frac{1}{ R }$

G =1/ R


$ \kappa_e = \kappa_1 + \kappa_2 $

kappa_e = kappa_1 + kappa_2


$ \kappa_1 = \Lambda_1 c_1 $

kappa_i = Lambda_i * c_i


$ \kappa_2 = \Lambda_2 c_2 $

kappa_i = Lambda_i * c_i


$ \Lambda_1 =\displaystyle\frac{ Q_1 ^2 \tau_1 }{2 m_1 } $

Lambda_i = Q_i ^2* tau_i /(2 * m_i )


$ \Lambda_2 =\displaystyle\frac{ Q_2 ^2 \tau_2 }{2 m_2 } $

Lambda_i = Q_i ^2* tau_i /(2 * m_i )


$ R = \rho_e \displaystyle\frac{ L }{ S }$

R = rho_e * L / S


$ \rho_e =\displaystyle\frac{1}{ \kappa_e } $

rho_e = 1/ kappa_e

ID:(16024, 0)



Total conductivity (2)

Equation

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The conductivity ($\kappa_e$) of a liquid with two types of ions is calculated as the sum of the conductivity ions of type 1 ($\kappa_1$) and the conductivity ions of type 2 ($\kappa_2$). This relationship is expressed as:

$ \kappa_e = \kappa_1 + \kappa_2 $

$\kappa_e$
Conductivity
$1/Ohm m$
5487
$\kappa_1$
Conductivity ions of type 1
$1/Ohm m$
10487
$\kappa_2$
Conductivity ions of type 2
$1/Ohm m$
10488

ID:(16014, 0)



Conductivity

Equation

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The resistivity ($\rho_e$) is defined as the inverse of the conductivity ($\kappa_e$). This relationship is expressed as:

$ \rho_e =\displaystyle\frac{1}{ \kappa_e } $

$\kappa_e$
Conductivity
$1/Ohm m$
5487
$\rho_e$
Resistivity
$Ohm m$
5484

ID:(3848, 0)



Conductance

Equation

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The conductance ($G$) is defined as the inverse of the resistance ($R$). This relationship is expressed as:

$ G =\displaystyle\frac{1}{ R }$

$G$
Conductance
$1/Ohm$
5486
$R$
Resistance
$Ohm$
5485

ID:(3847, 0)



Resistance

Equation

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Using the resistivity ($\rho_e$) along with the geometric parameters the conductor length ($L$) and the section of Conductors ($S$), the resistance ($R$) can be defined through the following relationship:

$ R = \rho_e \displaystyle\frac{ L }{ S }$

$L$
Conductor length
$m$
5206
$R$
Resistance
$Ohm$
5485
$\rho_e$
Resistivity
$Ohm m$
5484
$S$
Section of Conductors
$m^2$
5475

ID:(3841, 0)



Molar conductivity (1)

Equation

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The molar conductivity ions of type i ($\Lambda_i$) is defined in terms of the charge of the ion i ($Q_i$), the time between collisions ion i ($\tau_i$), and the mass of the ion i ($m_i$), using the following relationship:

$ \Lambda_1 =\displaystyle\frac{ Q_1 ^2 \tau_1 }{2 m_1 } $

$ \Lambda_i =\displaystyle\frac{ Q_i ^2 \tau_i }{2 m_i } $

$Q_i$
$Q_1$
Charge of the ion 1
$C$
10490
$m_i$
$m_1$
Mass of the ion 1
$kg$
10496
$\Lambda_i$
$\Lambda_1$
Molar conductivity 1
$m^2/Ohm mol$
5488
$\tau_i$
$\tau_1$
Time between collisions ion 1
$s$
10493

ID:(11817, 1)



Molar conductivity (2)

Equation

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The molar conductivity ions of type i ($\Lambda_i$) is defined in terms of the charge of the ion i ($Q_i$), the time between collisions ion i ($\tau_i$), and the mass of the ion i ($m_i$), using the following relationship:

$ \Lambda_2 =\displaystyle\frac{ Q_2 ^2 \tau_2 }{2 m_2 } $

$ \Lambda_i =\displaystyle\frac{ Q_i ^2 \tau_i }{2 m_i } $

$Q_i$
$Q_2$
Charge of the ion 2
$C$
10491
$m_i$
$m_2$
Mass of the ion 2
$kg$
10497
$\Lambda_i$
$\Lambda_2$
Molar conductivity 2
$m^2/Ohm mol$
5489
$\tau_i$
$\tau_2$
Time between collisions ion 2
$s$
10494

ID:(11817, 2)



Conductivity of each ion (1)

Equation

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The conductivity ions of type i ($\kappa_i$), in terms of the molar conductivity ions of type i ($\Lambda_i$) and the concentration of ions i ($c_i$), is defined as equal to:

$ \kappa_1 = \Lambda_1 c_1 $

$ \kappa_i = \Lambda_i c_i $

$c_i$
$c_1$
Ion concentration of type 1
$mol/m^3$
5534
$\kappa_i$
$\kappa_1$
Conductivity ions of type 1
$1/Ohm m$
10487
$\Lambda_i$
$\Lambda_1$
Molar conductivity 1
$m^2/Ohm mol$
5488

ID:(11818, 1)



Conductivity of each ion (2)

Equation

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The conductivity ions of type i ($\kappa_i$), in terms of the molar conductivity ions of type i ($\Lambda_i$) and the concentration of ions i ($c_i$), is defined as equal to:

$ \kappa_2 = \Lambda_2 c_2 $

$ \kappa_i = \Lambda_i c_i $

$c_i$
$c_2$
Concentration of ions 2
$mol/m^3$
10500
$\kappa_i$
$\kappa_2$
Conductivity ions of type 2
$1/Ohm m$
10488
$\Lambda_i$
$\Lambda_2$
Molar conductivity 2
$m^2/Ohm mol$
5489

ID:(11818, 2)