Respuesta :
Answer:
In order to understand the Maxwell equations and relations derived from Maxwell relations, you will have to know Euler's Reciprocity.
It states that for any state thermodynamic quantity (let x and y) having a state function phi, it must satisfy the following condition.
[tex](\frac{ {∂}^{2}\phi }{∂x \cdot∂y}) = (\frac{ {∂}^{2}\phi }{∂y \cdot∂x})[/tex]
Maxwell equations: These are a set of equations derived from the application of Euler's Reciprocity. The four Maxwell equations are as follows:
[tex]dH = TdS + VdP[/tex]
[tex]dG = VdP - SdT[/tex]
[tex]dA = -PdV - SdT[/tex]
[tex]dU = TdS - PdV[/tex]
Let's derive each Maxwell relations step by step,
1) dH = TdS + VdP
Enthalpy is a function of Entropy & Pressure
[tex] \sf \qquad \qquad H = f(S,P)[/tex]
[tex]dH = TdS + VdP[/tex]
[tex]dH = (\frac{∂H}{∂S})_{_P} dS + (\frac{∂H}{∂P})_{_S}dP[/tex]
Comparing both the above equation,
[tex]T = (\frac{∂H}{∂S})_{_P}[/tex]
[tex]V =(\frac{∂H}{∂P})_{_S}[/tex]
now we know that Enthalpy is a state function hence applying cross reciprocality,
[tex](\frac{∂V}{∂S})_{_P}= (\frac{∂T}{∂P})_{_S}[/tex]
This is called the first Maxwell relation,
Similarly,
2) dG = -SdT + VdP
Free energy is a function of Temperature & Pressure,
[tex] \sf \qquad \qquad G = f(T,P)[/tex]
[tex]dG = -SdT + VdP[/tex]
[tex]dG = (\frac{∂G}{∂P})_{_T} dP + (\frac{∂G}{∂T})_{_P}dT[/tex]
Comparing both the above equation,
[tex](\frac{∂G}{∂P})_{_T} = V [/tex]
[tex] (\frac{∂G}{∂T})_{_P}= -S[/tex]
now we know that Free energy is a state function hence applying cross reciprocality,
[tex]-(\frac{∂S}{∂T})_{_P}= (\frac{∂V}{∂P})_{_T}[/tex]
This is called the second Maxwell relation,
3) dA = -PdV - SdT
helmholtz free energy is a function of Temperature & Volume,
[tex] \sf \qquad \qquad A = f(T,V)[/tex]
[tex]dA = -PdV - SdT[/tex]
[tex]dA = (\frac{∂A}{∂V})_{_T} dV + (\frac{∂A}{∂T})_{_V}dT[/tex]
Comparing both the above equation,
[tex](\frac{∂A}{∂V})_{_T} = -P [/tex]
[tex] (\frac{∂A}{∂T})_{_V}= -S[/tex]
now we know that helmholtz free energy is a state function hence applying cross reciprocality,
[tex]-(\frac{∂S}{∂V})_{_T}= -(\frac{∂P}{∂T})_{_V}[/tex]
[tex](\frac{∂S}{∂V})_{_T}= (\frac{∂P}{∂T})_{_V}[/tex]
This is called the third Maxwell relation,
4) dU = TdS - PdV
Internal energy is a function of Entropy & Volume,
[tex] \sf \qquad \qquad A = f(S,V)[/tex]
[tex]dU = TdS - PdV [/tex]
[tex]dU = (\frac{∂U}{∂S})_{_V} dS + (\frac{∂U}{∂V})_{_S}dV[/tex]
Comparing both the above equation,
[tex](\frac{∂U}{∂V})_{_S} = -P [/tex]
[tex] (\frac{∂U}{∂S})_{_V}= T[/tex]
now we know that Internal energy is a state function hence applying cross reciprocality,
[tex](\frac{∂T}{∂V})_{_S}= -(\frac{∂P}{∂S})_{_V}[/tex]
This is called the fourth Maxwell relation,
The main significance of the Maxwell relation is that those thermodynamic quantities which are unmeasurable can be replaced with measurable quantities with the help of the Maxwell relation.
The derivative of the extensive asset in relation to the extensive asset gives the intensive asset; with respect to the intensive asset, the derivative of the extensive asset gives the extensive asset. This is the result of the overall Maxwell relations.
The coefficient of expansion and compression of a gas in thermodynamics is the application of the Maxwell relations.
There are 3 coefficients introduced,
Coefficient of thermal expansion (expansivity) α
[tex]α= \frac{1}{V} (\frac{∂V}{∂T} )_{_P}[/tex]
Coefficient of isothermal compressibility β
[tex]β = -\frac{1}{V} (\frac{∂V}{∂P} )_{_T}[/tex]
Isochoric thermal expansion coefficient γ
[tex] γ= \frac{1}{P} (\frac{∂P}{∂T} )_{_V}[/tex]
For ideal gases,
PV = nRT
For one mole ideal gas (n=1),
PV = RT
Taking derivative with respect to T at constant pressure,
[tex]V(\frac{∂P}{∂T} )_{_P}+ P(\frac{∂V}{∂T} )_{_P}= R(\frac{∂T}{∂T} )_{_P}[/tex]
At constant pressure ∂P = 0, & R.H.S = 1, hence
[tex](\frac{∂V}{∂T} )_{_P}= \frac{R}{P}[/tex]
[tex]α= \frac{1}{V} (\frac{∂V}{∂T} )_{_P}[/tex]
[tex]α= \frac{1}{V} \cdot \frac{R}{P}[/tex]
[tex]\sf Also, PV=RT\\ \frac{R}{PV} = \frac{1}{T}[/tex]
[tex]\boxed{α= \frac{1}{T}}[/tex]
Following the same procedure, by taking derivating w.r.t. pressure at constant temperature.
[tex]V(\frac{∂P}{∂P} )_{_T}+ P(\frac{∂V}{∂P} )_{_T}= R(\frac{∂T}{∂P} )_{_T}[/tex]
[tex]V+ P(\frac{∂V}{∂P} )_{_T}= 0[/tex]
[tex](\frac{∂V}{∂P} )_{_T}= \frac{-V}{P}[/tex]
Substituting the above value in,
[tex]β = -\frac{1}{V} (\frac{∂V}{∂P} )_{_T}[/tex]
[tex]β = -\frac{1}{V} \cdot \frac{-V}{P}[/tex]
[tex] \boxed{β = \frac{1}{P}}[/tex]
Repeating the same procedure again, i.e. derivative w.r.t. T at constant volume
[tex]V(\frac{∂P}{∂T} )_{_V}+ P(\frac{∂V}{∂T} )_{_V}= R(\frac{∂T}{∂T} )_{_V}[/tex]
At constant Volume ∂V = 0, and R.H.S = 1, hence overall equation becomes,
[tex]V(\frac{∂P}{∂T} )_{_V} = R \\ (\frac{∂P}{∂T} )_{_V} = \frac{R}{V}[/tex]
Substituting above value in,
[tex]γ= \frac{1}{P} (\frac{∂P}{∂T} )_{_V}[/tex]
[tex]γ= \frac{1}{P} \cdot \frac{R}{V}[/tex]
[tex] \boxed{γ= \frac{1}{T}}[/tex]
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