Question

Please answer parts A-D and show work/solution

For A and D DONT INTEGRATE

For B and C INTEGRATE

ANSWER A-D

A B C D

ANSWER PARTS A, B, C, AND D.........

8. We have two "named" equations which allow us to quantify the pressure- temperature relationship of a two-phase equilibrium, the Clausius equation, A, and the Clausius-Clapeyron equation, B din P ΔΗ dP AS A: dT RT2 The Clausius equation is an exact result for a one-component system, derived with no approximations, while the Clausius-Clapeyron equation has some constraints. We will use both equations to examine water whose enthalpy of vaporization, AH is 40.66 kJ mol at the normal boiling point of 373.15 K. a) Start with the Clausius equation and assume that the entropy change for the phase transition is inde- pendent of temperature and that V,) V,m(g). Derive an expression for the temperature variation of the vapor pressure of a liquid. (Hint: it will not have AH vap in it) b) Use your relation in part a) to determine the vapor pressure of water at 3 c) Use your relation in part a) to determine the vapor pressure of water at d) The Clausius-C Clapeyron equation can be derived from the Clausius equation under certain approxi- mations. Perform this derivation, give all the approximations, and tell which systems the equation is applicable to on equation and assume that the enthalpy of vaporization is inde- ndent of temperature. Derive an expression for the temperature variation of the vapor pressure of a liq- uid 0 Use your relation in part e) to determine the vapor pressure of water at 300 K g) Use your relation in part e) to determine the vapor pressure of water at 500 K h) Which equation accords best with observation? Suggest a reason(s) for why this may be so

Answer 1

a) Clausius equation:

dP/dT = $\Delta$S_m/V_m(g) - V_m(l)

Here, V_m(g) >> V_m(l), i.e. V_m(l) can be neglected when you are subtracting it from V_m(g).

i.e. dP/dT = $\Delta$S_m/V_m(g)

According to Ideal gas equation:

V_m(g) = RT/P

Therefore, dP/dT = $\Delta$S_m*P/RT

b) By integrating the final equation in part a), you will get

ln(P2/P1) = $\Delta$H_vap/R (1/T1 - 1/T2)

i.e. ln(P2/1 atm) = (40660 J/mol)/(8.314 J.mol^-1.K^-1) * (1/373.15 - 1/300) K^-1

i.e. P₂ = 0.0409 atm

c) ln(P2/1 atm) = (40660 J/mol)/(8.314 J.mol^-1.K^-1) * (1/373.15 - 1/500) K^-1

i.e. P₂ = 27.8 atm

d) The Clausius-Clapeyron equation is applicable to liquid-vapour equilibrium only.

Here, V_m(g) >> V_m(l), i.e. V_m(l) can be neglected when you are subtracting it from V_m(g).

i.e. $\Delta$V_m(g) = V_m(g) >> V_m(l) ~ V_m(g)

Therefore, from Clausius equation: dP/dT = $\Delta$S_m/V_m(g)

From ideal gas equation: P.V_m(g) = RT, then V_m(g) = RT/P

i.e. dP/dT = $\Delta$S_m*P/RT

i.e. 1/P dP/dT = $\Delta$S_m/RT

i.e. d(ln P)/dT = $\Delta$S_m/RT

Here, the liquid-vapor system is at equilibrium, i.e. $\Delta$G = 0

i.e. $\Delta$H_vap - T.$\Delta$S_m = 0

i.e. $\Delta$S_m = $\Delta$H_vap/T

Hence, d(ln P)/dT = $\Delta$H_vap/RT²