Question

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.

Answer 1

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