Thermodynamic Potentials - Internal energy problem.

[knowlewj01]

Homework Statement

For Cyclohexane
Boiling point $T_b$ = 80.1°C
P = 1atm
$\Delta H_{vap}$ = 30.1 kJ/mol
Determine at phase change liquid to vapour, Changes in:

a)Entropy per mole
b)Gibbs free energy per mole
c)Internal energy per mole

assume ideal gas behaviour for the vapour.

Homework Equations

$dS = \frac{dQ}{T}$ [1]

$dG = -S dT + V dP$ [2]

$dE = T dS - P dV$ [3]

The Attempt at a Solution

(a) entropy per mole

using eq'n [1]

$\Delta S = \frac{\Delta H}{T} = \frac{30100}{353.1} = 85.2 J K^{-1} mol^{-1}$

(b) change in gibbs free energy at a phase transition at constant pressure is always 0 because, in eq'n [2] $dT=dP=0$

(c) this is the part I'm struggling with. so using eq'n [3] we already know the Temperature, which is constant and we have calculated the change in entropy. the second term is the problem. it requires that we know what the change in volume is. can this be expressed in another way. -PdV is the work done by the expanding gas at constant pressure. can this be done without knowing the difference between molar densities of cyclohexane between liquid and vapor?

 

[anathema]

Hello, thank you for bringing this question to the forum. I can help you with the last part of your solution, determining the change in internal energy per mole. As you mentioned, using equation [3] requires knowing the change in volume, which can be difficult to determine without knowing the difference in molar densities between liquid and vapor. However, we can use the ideal gas law to approximate the change in volume at constant pressure. The ideal gas law is given by:

PV = nRT

Where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. Since we are assuming ideal gas behavior for the vapor, we can use this equation to approximate the change in volume. We know that at the phase transition from liquid to vapor, the temperature remains constant, so we can rearrange the ideal gas law to solve for the change in volume:

\Delta V = \frac{nRT}{P}

Substituting in the given values, we get:

\Delta V = \frac{1 mol * 0.0821 L atm K^{-1} mol^{-1} * 353.1 K}{1 atm} = 28.5 L mol^{-1}

Now we can use this value for the change in volume in equation [3]:

\Delta E = T \Delta S - P \Delta V

Substituting in the values we calculated earlier, we get:

\Delta E = 353.1 K * 85.2 J K^{-1} mol^{-1} - 1 atm * 28.5 L mol^{-1} = 30007.7 J mol^{-1}

I hope this helps with your solution. Let me know if you have any further questions.