Question

314 J/mol K 78.4+ 273,15 All work must be show to receive credit. Remember, significant figures are important! P_TAH P, -AH / T,+1, IT R 0,0821 L atm/mol K = 8.314 J/mol K "P, RT TI? 1. (4 points) Ethanol has a heat of vaporization of 38 56 kJ/mol and a normal boiling point of 78.4°C. a. What is the vapor pressure of ethanol at 250°C? In P2 - Nroup CT-T2 ) 25.0+ 273. P2L n =-A 38.56 kJ last 78.4°C -25.06 ) 0.0821 atm b. What is the boiling temperature of ethanol on Pikes Peak, where the atmo- spheric pressure iş 540 torr? 540 torr tum = 0.7105 am 760 torr o.1109 atm (4 points) A newly formulated substance has a normal boiling point of 314°C, a nor- mal freezing point of 16°C, a triple point at 460 torr and 29°C, and a critical point at 7302 torr and 603°C. Draw a phase diagram for this substance, labeling the liquid, gas, and solid phases, the triple point, the critical point, and the supercritical fluid. Which is more dense for this substance, the liq- uid or the solid state? Explain how you ar- rived at this answer.

Answer 1

Question 1.

Part a.

Here, P1 = 1 atm, T1 = 78.4+273 = 351.4 K

T2 = 25+273 = 298 K, $\Delta$H_vap = 38.56 kJ/mol = 38560 J/mol

Now, ln(P2/1 atm) = (-38560 J/mol)/(8.314 J/mol.K) * {(351.4-298)/(351.4*298)} K^-1

i.e. ln(P2/1 atm) = -2.365

i.e. P2/1 atm = e^-2.365 = 0.09394

Therefore, the vapor pressure of ethanol at 25 ^oC = 0.09394 atm

Part b.

Here, P1 = 1 atm, T1 = 351.4 K

$\Delta$H_vap = 38560 J/mol, P2 = 540 torr

Now, ln(540 torr/760 torr) = (-38560 J/mol)/(8.314 J/mol.K) * {(351.4-T2)/(351.4*T2)} K^-1

i.e. -0.342 = -4637.96 * (1/T2 - 0.002846)

Therefore, the boiling temperature of ethanol at 540 torr (T2) = 342.53 K = 69.53 ^oC