Respuesta :
Explanation:
[tex]dH{}'= TdS{}'+V{}'dP+\mu dn[/tex]
For ∝ phase system,
[tex]dH{}'^{\alpha }= T^{\alpha }dS{}'^{\alpha }+V{}'^{\alpha }dP^{\alpha }+\mu^{\alpha } dn^{\alpha }[/tex]
For β phase system,
[tex]dH{}'^{\beta }= T^{\beta}dS{}'^{\beta }+V{}'^{\beta}dP^{\beta}+\mu^{\beta} dn^{\beta}[/tex]
Now we know that the total enthalpy is the sum of the enthalpy in the alpha and beta phases.
∴ [tex]H{}'_{sys}=H{}'^{\alpha }+H{}'^{\beta}[/tex]
[tex]dH{}'_{sys}=dH{}'^{\alpha }+dH{}'^{\beta}[/tex]
[tex]dH{}'_{sys}=\left (T^{\alpha }dS{}'^{\alpha }+V{}'^{\alpha }dP^{\alpha }+\mu^{\alpha } dn^{\alpha } \right )+\left (T^{\beta}dS{}'^{\beta }+V{}'^{\beta}dP^{\beta}+\mu^{\beta} dn^{\beta} \right )[/tex]
Now P, S an n are constants.
Then for isolated system, we get,
[tex]dU{}'_{sys}=d\left ( U{}'^{\alpha} + U{}'^{\beta }\right ) =0[/tex]
[tex]d U{}'^{\alpha}= - dU{}'^{\beta }[/tex]
[tex]dV{}'_{sys}=d\left ( V{}'^{\alpha} + V{}'^{\beta }\right ) =0[/tex]
[tex]d V{}'^{\alpha}= - dV{}'^{\beta }[/tex]
[tex]dn{}'_{sys}=d\left ( n{}'^{\alpha} + n{}'^{\beta }\right ) =0[/tex]
[tex]d n{}'^{\alpha}= - dn{}'^{\beta }[/tex]
[tex]dP{}'_{sys}=d\left ( P{}'^{\alpha} + P{}'^{\beta }\right ) =0[/tex]
[tex]dS{}'_{sys}=d\left ( S{}'^{\alpha} + S{}'^{\beta }\right ) =0[/tex]
[tex]dS{}'^{\alpha}= - dS{}'^{\beta }[/tex]
[tex]dH{}'_{sys} = dS{}'^{\alpha }\left ( T^{\alpha} -T^{\beta } \right )+ dP{}'^{\alpha }\left ( V^{\alpha} -V^{\beta } \right )+dn{}'^{\alpha }\left ( T^{\alpha} -T^{\beta } \right )[/tex]
For equilibrium,
[tex]dS{}'_{sys,iso}=0[/tex]
Then [tex]T^{\alpha }=T^{\beta }[/tex] ----- for thermal equilibrium
[tex]\mu ^{\alpha }=\mu ^{\beta }[/tex]----- for chemical equilibrium
[tex]P^{\alpha }=P^{\beta }[/tex] ----- for mechanical equilibrium
The above conditions are valid for one component two phase system.
