1000 Solved Problems in Modern Physics

4.3 Solutions 267

Therefore,

dU=Tds−PdV (6)

whereUis the internal energy,Qthe heat absorbed,Wthe work done by the

system,Sthe entropy,Pthe pressure andTthe Kelvin temperature.

Let the independent variables be calledxandy. Then

U=U(x,y);V=V(x,y);S=S(x,y)(7)

Now,

df=

(

∂f

∂x

)

y

dx+

(

∂f

∂y

)

x

dy (8)

Therefore

dU=

(

∂U

∂x

)

y

dx+

(

∂U

∂y

)

x

dy (9)

dV=

(

∂V

∂x

)

y

dx+

(

∂V

∂y

)

x

dy (10)

dS=

(

∂S

∂x

)

y

dx+

(

∂S

∂y

)

x

dy (11)

Eliminating internal energyUand substituting (9), (10) and (11) in (6)

(

∂U

∂x

)

y

dx+

(

∂U

∂y

)

x

dy=T

[(

∂S

∂x

)

y

dx+

(

∂S

∂y

)

x

dy

]

−P

[(

∂V

∂x

)

y

dx+

(

∂V

∂y

)

x

dy

]

(12)

Equating the coefficients of dxand dy

(

∂U

∂x

)

y

=T

(

∂S

∂x

)

y

−P

(

∂V

∂x

)

y

(13)

(

∂U

∂y

)

x

=T

(

∂S

∂y

)

x

−P

(

∂V

∂y

)

x

(14)

Differentiating (13) with respect toywithxfixed, and differentiating (14)

with respect toxwithyfixed