Advanced Thermodynamics   Exam #2

1. You know that isothermic processes occur with no change in temperature. You may even know that isochoric ones (Greek for "equal space") occur at constant volume instead. But you may not have thought about isoentropic processes, which occur at constant entropy! Let's make one.

   An ideal, monatomic gas is expanded (V₁V₂) isothermally (T₁=T₂) and reversibly entailing an increase in entropy. Next it is cooled (T₂T₃) isochorically (V₂=V₃) until that previous entropic increase is entirely cancelled out. The overall process will then be isoentropic.

   Show that VT^3/2 remains fixed through such (isoentropic) processes for an ideal, monatomic [C_V=(3/2)R, as you'll recall] gas.

S_TOT = S_A + S_B = nR ln(V₃/V₁) + nC_V ln(T₃/T₁) = 0

nR ln(V₃/V₁) = - nC_V ln(T₃/T₁) = + n (3/2) R ln(T₁/T₃) = nR ln( [T₁/T₃]^3/2 )

If we'd left it general, it would've become VT^C_V/R fixed which (when we take the C_V/R th root) becomes TV^R/C_P fixed which should ring a deja vu bell. It's one of the adiabatic ideal gas laws, arrived at through entropy! And since S = q_rev/T, it's not suprising that adiabatic, q_rev=0, would be isoentropic!

The following data on carbon is from Nash's Appendix III Thermodynamic Data at 298.15 K:

Species	Hf°	Gf°	S°	CP
	kcal/mol		cal/mol K
Graphite	0.00	0.00	1.372	2.038
Diamond	0.4553	0.6930	0.568	1.4615

from which it is clear that graphite is thermodynamically favored. Your company, eager for synthesis, asks you to find the temperature at which the reverse is true at 1 atm. Defend your answer to your supervisor.

S = S° (diamond) - S° (graphite) = 0.568 - 1.372 = - 0.804 cal/mol K

G = H - TS = 455.3 cal/mol + 0.804 T cal/mol > 0 for all T.

So there is no T for which the graphite to diamond transition is spontaneous (unless we fool around with pressure as well).

You keep diethyl ether tightly sealed for a reason. It's boiling point is only 34.5°C! Given H_vap at 25°C is 27.2 kJ/mol (and obviously doesn't change much in less than 10°C),

1. What is ether's vapor pressure at 25°C (in atm)?

ln(P_25°C/P_34.5°C) = - H_vap/R [ (1/T_25°C) - (1/T_34.5°C) ]

ln(P/1 atm) = - (27200 J/mol / 8.314 J/mol K) [ (1/298.15) - (1/307.65) ] = - 0.339

P = e^{- 0.339} atm = 0.713 atm

Since atmospheric pressure (at 25°C) falls off with altitude, A, as

P = 1 atm e^{-A / 8.4 km},

at what height (T still 25°C!) would you really not want to work with open ether cannisters? (Makes you wonder if they ship ether in the cargo holds of aircraft, doesn't it?)

But the temperature doesn't stay 25°C as you go up in altitude. Remember I told you that the temperature of the atmosphere falls off at what's called the adiabatic lapse rate? That's because the lowered pressure causes the N₂ to expand without being able to rob its neighbors of any heat; they're expanding too.

So the adiabatic formula P/T^C_P/R = constant is correct for the atmosphere. And C_V = 2.5 R for diatomic N₂ (remember why?) thus C_P = 3.5 R for the atmosphere. If it's really 25°C at sea level, that makes the temperature aloft

T = 298.15 K P ^{1 / 3.5}

which turns out to be more than enough cooling to keep the ether as a liquid as it rises in altitude! (Check it if you don't believe me.)

In fact, the way to boil ether is to lower it into a mine!! The air gets higher pressure (bad for boiling) but it also gets (adiabatically) hotter (good for boiling).

Show that if you lower the ether to a depth where the atmospheric pressure is 1.1786 atm, it will, in fact, boil due to the atmosphere's adiabatic heating!

ln(P/1 atm) = - H_vap/R [ (1/T) - (1/T_34.5°C) ]

1/T = 1/T_34.5°C - R ln(P) / H_vap

1/T = (1/307.65) - 8.314 J/mol K × ln(1.1786) / 27200 J/mol = 3.200×10^-3 K^-1

T = 312.5 K required to boil ether there.

But what's available from the adiabatic lapse rate?

Totally weird. (And less than a mile underground.)