In: Chemistry
A 64.0 g sample of CH3OH is condensed isothermally and reversibly to liquid at 64 C. The standard enthalpy of vaporization of CH3OH is 35.3 kJ/ mol. Find w, q, change in U, and change in H for this process.
A 64.0 g sample of CH3OH is condensed isothermally and reversibly to liquid at 64 C. The standard enthalpy of vaporization of CH3OH is 35.3 kJ/ mol. Find w, q, change in U, and change in H for this process.
Solution ;-
Lets calculate the moles of the CH3OH
Moles of CH3OH = mass / molar mass
= 64.0 g / 32.0 g per mol
= 2 mol
Lets calculate the amount of energy given by the process of condensation
1mol * 35.3 kJ/ mol = 70.6 kJ
Condensation is exothermic process therefore it given off the heat
So q = -70.6 kJ
Now lets calculate the work (w)
W= - P delta V
Lets first calculate the volume of the 2.0 mol gas at 64 C + 273 = 337 K
PV = nRT
V= nRT/P
= 2 mol * 0.08206 L atm per mol K * 337 K / 1 atm
= 55.3 L
So when this liquid condense then volume of the liquid will be geligible
So the volume change = - 55.3 L
W= - 1 atm * (-55.3 L)
W= -55.3 L atm
55.3 L atm * 101.3 J / 1 L atm = 5602 J
5602 J * 1 kJ / 1000 J = 5.602 kJ
So the w= 5.602 kJ
Now lets calculate the U
U = q+w
= -70.6 kJ + 5.602 kJ
= -65.0 kJ
So
q= -70.6 kJ
w= 5.602 kJ
U= -65.0 kJ
delta H = -70.3 kJ