Question

The equilibrium vapor pressure of ethanol at 298.15 K is 0.078 atm. Will ethanol gas at 0.05 atm and 298.15 K spontaneously condense to liquid ehtanol? First answer this by thinking about it qualitatively. Then support it with a calculation. Use the change in entropy of the system and calculate the change in entropy of the surroundings, and the change in entropy of the universe for the condensation of one mole of ethanol gas at 0.05 atm and 298.15 K to the liquid at 0.05 atm and 298.15 K. Give all three numerical answers and state whether the second law of thermodynamics predicts that the gas will condense spontaneously. Assume that for the liquid, but not the gas, that neither S nor H change appreciably as you change the pressure of the liquid (within this pressure range) while holding the temperature constant.

Answer 1

When equilibrium vapor pressure is more than surrounding pressure, rate of evaporation is more than rate of condensation. So, ethanol condensation is not a spontaneous process.

a)

Source for data: https://en.wikipedia.org/wiki/Ethanol_(data_page)

For 1 mole of ethanol,

Entropy of liquid ethanol at 298.15 K & 1 atm = 159.9 J/K

We can assume entropy of liquid ethanol at 0.05 atm & 298 K, S_liq = 159.9 J/K

Entropy of gaseous ethanol at 298.15 K & 1 atm = 283 J/K

Entropy of gaseous ethanol at 298.15 K & 0.05 atm calculation:

At constant temperature, ΔS = - n R ln(p2/p1)

= - 1 * 8.314 J/mol-K * ln(0.05)

ΔS = 24.9 J/K

Entropy of gaseous ethanol at 298 K & 0.05 atm, S_vap = 283 + 24.9 J/K

= 307.9 J/K

ΔS_sys = S_liq – S_vap

= 159.9 – 307.9 J/K

= -148 J/K

ΔS_surr = ΔH_vap / T

ΔH_vap = 38.56 kJ/mol

ΔS_surr = 1 * 38.56 * 1000 J / 298.15 K

= 129.33 J/K

ΔS_total = ΔS_surr + ΔS_sys = 129.33 – 148 J/K

= -18.7 J/K < 0

Hence the process is not spontaneous according to second law of thermodynamics.