Explanation:
dU=TdS-pdV (given)
To prove = 1) [tex](\frac{dS}{dV})_U=\frac{P}{T}[/tex] (at constant U)
2) [tex](\frac{dS}{dU})_v=\frac{1}{T}[/tex] (at constant V)
Solution: 1)
dU=TdS-PdV
[tex]PdV=TdS-dU[/tex]
[tex]P=\frac{(TdS)}{dV}-\frac{dU}{dV}[/tex]
Derivative of constant is zero.
Given that internal energy is ,U = constant
[tex]P=T\frac{dS}{dV}-0[/tex]
[tex]\frac{dS)}{dV}=\frac{P}{T}[/tex] (hence proved)
Solution: 2)
dU=TdS-PdV
Differentiating with respect to dU, we get:
[tex](\frac{dU}{dU})_v=T(\frac{dS}{dU})_v-P(\frac{dV}{dU})_v[/tex]
Derivative of constant is zero.
Given that volume is constant , V= constant
[tex]1=T(\frac{dS}{dU})_v[/tex]
[tex](\frac{dS}{dU})_v=\frac{1}{T}[/tex] (hence proved)