Using fundamental equations of thermodynamics and rules of math, derive the equa

Question

Using fundamental equations of thermodynamics and rules of math, derive the equations below: dU = [C_P - P (partial differential V/partial differential T)_P] dT - [T(partial differential V/partial differential T)_P +P (partial differential V/partial differential P)_T]dP. dH = [C_V + V(partial differential P/partial differential T)_V]dT + [T(partial differential P/partial differential T)_V + V(partial differential P/partial differential V)_T] dV b. Determine dU and dH for an ideal gas using the expression in part a. Do dU and dH depend on temperature and/or pressure? Confirm the relation between C_P and C_V for an ideal gas. c. In order to determine dU and dH for a real gas, you need to know (partial differential V/partial differential T)_P, (partial differential V/partial differential P)_T and (partial differential P/partial differential T)_V. Derive an expression for these two terms for a Clausius gas. Do dU and dH depend on temperature and/or pressure? The Clausius equation of state can be written as P = RT/V - b - a/T(V + C)^2. d. In an experiment, you can measure the coefficient of thermal expansion a and isothermal compressibility K_T. If you want to know the changes in internal energy and enthalpy with respect to changes in volume and pressure under isothermal condition, derive an expression for (partial differential U/partial differential V)_T and (partial differential H/P)_T in terms of P, T, V, alpha, and K_T. You can use any of the four Maxwell relations expressed below in your derivation. (partial differential S/partial differential P)_T = -(partial differential V/partial differential T)_P, (partial differential T/partial differential V)_S = - (partial differential P/partial differential S)_V, (partial differential T/partial differential P)_S = (partial differential B/partial differential S)_P, (partial differential S/partial differential V)_T = (partial differential P/partial differential T)_V

Explanation / Answer

a) We know that dU = Tds - Pdv -------------(1)

Cp = (dQ/dT)P   and Cp = (dQ/dT)P ------------ (2)

But we know that dQ = T ds

So, Cp = T(ds/dT)P and CV = T(ds/dT)V

Since S is a function of P, T, we can writeds = [(ds/dT)P] dT + [(ds/dP)T] dP ------------- (3)

Since V is a function of P,T, we can write dV= [(dV/dT)P] dT + [(dV/dP)T] dP -------------(4)

Substituting the values of dS and dP in equation (1), we get

dU = T{[(ds/dT)P] dT + [(ds/dP)T] dP} - P{[(dV/dT)P] dT + [(dV/dP)T] dP} -------(5)

From equation (2), Cp = T(ds/dT)P

Hence substituting the value ofCp = T(ds/dT)P in equation (5), we get

dU = T{[Cp/T] dT + [(ds/dP)T] dP} - P{[(dV/dT)P] dT + [P(dV/dP)T] dP}

By simplification and separating dT and dP terms, we get

dU = Cp dT + T[(ds/dP)T] dP - P(dV/dT)PdT - P[(dV/dP)T] dP

Hence dU = [Cp - P(dV/dT)P]dT- [T[(dV/dT)P] +P[(dV/dP)T] dP

By using Maxwell's relations, we can write (ds/dP)T = -(dV/dT)P

Thus , it is proved

Similarly for the enthalpy, we can write enthalpy H = U + PV

dH = dU + PdV + VdP

So, dH = TdS - PdV + PdV + VdP (Since dU = Tds - Pdv )

Hence dH = TdS + VdP

But

So, Cp = T(ds/dT)P and CV = T(ds/dT)V

Since S is a function of V, T, we can writeds = [(ds/dT)V] dV + [(ds/dV)T] dT ------------- (1A)

Since P is a function of V,T, we can write dP= [(dP/dV)P] dV + [(dP/dT)V] dT -------------(2A)

Hence dH = T{[(ds/dT)V] dV + [(ds/dV)T] dT} +V {[(dP/dV)T] dV + [(dP/dT)V] dT}

dH = {T(dS/dT)V)dV + V [(dP/dV)T] dV} + [T(dS/dV)T] dT + [V(dP/dT)V] dT

Hence, dH = {T(dP/dT)V + V(dP/dV)T}dV + {CP + V(dP/dT)V}dT

Using definition of Cp = T(ds/dT)P by using the Maxwell's relation (dS/dV)T= (dP/dV)T

Thus both the relations are proved.

b)

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