I need answers for only problems (d), (e), (f). Phase changes Explain what the C

Question

I need answers for only problems (d), (e), (f).

Phase changes Explain what the Clapeyron equation describes Explain what the Clausius-Clapeyron equation describes Calculate the boiling temperature of water at an elevation of 1700 m (approximately the height of Boulder, CO) given that atmospheric pressure there is about 0.83 atm. You may refer to the CHEM 452 units table (on Canvas under 'Files miscellaneous', for information on water properties. Calculate the (equilibrium) sublimation temperature of CO_2 at a pressure of 2 atm assuming a constant enthalpy of sublimation of AH-26 kJ/mol. The sublimation temperature of CO_2 at 1 atm is -78.5 degree C. e. Gold has an equilibrium melting temperature of 1337 K at 1 atm and an enthalpy of fusion of 12, 550 J/mol. Find the pressure at which the equilibrium melting temperature is 1400 K. The densities of solid gold and liquid gold are 19.3 g/mL and 17.3 g/mL and are assumed to be approximately constant under all conditions. (Gold, continued.) Using your answer from part e), calculate the entropy change for melting 1 mol of gold.

Explanation / Answer

Along a phase transition line , temperature and pressure are not independent of each other since the system is univarient, that is , only one intensive parameter can be varied independently. The dependence of pressure on temperature and vice versa in a two phase equilibrium is expressed by Clapeyron equation.

dP/dT = dH/tdV

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Clapeyron eqn allipes to all the equilibriums whereas clausius -clapeyron eqn is allpied only to gaseous equilibrium where we can assume that dV = Vvap since volume of vapor is much greater than volume of solid or gas.

So, Clapeyron eqn takes the form:

dlnP/dT = dHvap/RT^2

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Atmospheric pressure at 177m = 0.819 atm

ln (P2/P1) = dHvap/R (1/T1-1/T2)

or, ln (0.819/0.83) = 40657 Jmol-1 / 8.314 J/K/mol (1/373-1/T2)

or, T2 = 372.62K = 99.62oC

Boiling point of water at this altitude is 99.62oC.

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ln (P2/P1) = dHvap/R (1/T1-1/T2)

or, ln (2/1) = -26*1000J/mol/8.314 J/K/mol (1/194.5-1/T2)

or, T2 =203.26 K = -69.74oC

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