Question

10.8. Derive the third TdS equation, $\qquad TdS = C_V \left(\frac{\partial T}{\partial P}\right)_V dP + C_P \left(\frac{\partial T}{\partial V}\right)_P dV$, and show that the three TdS equations may be written as follows: (a) $TdS = C_V dT + \frac{\beta T}{\kappa} dV$. (b) $TdS = C_P dT - V\beta T dP$. (c) $TdS = \frac{C_V \kappa}{\beta} dP + \frac{C_P}{\beta V} dV$.

          10.8. Derive the third TdS equation,
$\qquad TdS = C_V \left(\frac{\partial T}{\partial P}\right)_V dP + C_P \left(\frac{\partial T}{\partial V}\right)_P dV$,
and show that the three TdS equations may be written as follows:
(a) $TdS = C_V dT + \frac{\beta T}{\kappa} dV$.
(b) $TdS = C_P dT - V\beta T dP$.
(c) $TdS = \frac{C_V \kappa}{\beta} dP + \frac{C_P}{\beta V} dV$.

10.8. Derive the third TdS equation,
TdS = CV ((∂ T)/(∂ P))V dP + CP ((∂ T)/(∂ V))P dV,
and show that the three TdS equations may be written as follows:
(a) TdS = CV dT + (β T)/(κ) dV.
(b) TdS = CP dT - Vβ T dP.
(c) TdS = (CV κ)/(β) dP + (CP)/(β V) dV.

Added by Amanda M.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Brandy Heflin

Step 1

Step 1: Start with the given third $ T dS $ equation: \[ T dS = C_V \left(\frac{\partial T}{\partial P}\right)_V dP + C_P \left(\frac{\partial T}{\partial V}\right)_P dV \] Show more…

Show all steps

Thanks for your feedback!

10.8. Derive the third T dS equation, aT dS = C dT + C dP + C dA and show that the three T dS equations may be written as follows: a) T dS = Cv dT + K 16 (0hg c T dS = 19