1.- One mole of Nitrogen (N2) has been compressed at T0=273 K to the volume
V0=1liter. The gas goes through the free expansion process (Q = 0, W = 0), in which
the pressure drops down to the atmospheric pressure Patm=1 bar. Assume that the
gas obeys the van der Waals equation of state in the compressed state, and that it
behaves as an ideal gas at the atmospheric pressure.
1.- Does the temperature increase or decrease?
For 1 mole of the “vdW” N2 gas:
For 1 mole of ideal N2 gas (after expansion, Tf is T after expansion)
As it is a free expansion delta W=0, delta U =0 (Q=0). So The internal energy needs to
be conserved and we obtain Tf making UvdW=Uideal.
So T decreases
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2.- Why?
T decreases because in the vdW U we have kinetic (temperature) and
potential energy terms. In the ideal U we only have the kinetic term.
This means that the potential energy region was attractive on expanding
the gas molecules work against attraction forces and this work comes
at the expense of kinetic energy.
3.- Will the temperature always change in the same direction as you
answered in 1 for any free expansion process like this where the
compressed state is a vdW gas and the expanded state is an ideal gas?
Yes, according to the equation any free expansion should always
decrease the T.
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2.- The pressure of the saturated water vapor at T = 0.010C is 0.006 bar (these T and P
correspond to the triple point of water). The latent heat of the solid-liquid transformation at
0.010C is 335 kJ/kg, the latent heat of the liquid-gas transformation is 2500 kJ/kg. Find the
pressure of the saturated water vapor at T = -10C.
HINT: The vapor gas behaves like an ideal gas.
HINT2: The volume of the solid is negligible as compared to the gas
To solve this problem we just need to use Clausius-Clapeyron.
Latent heat of sublimation at the triple point satisfies:
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Above I neglected the volume of solid as compared to the gas and expressed
the gas volume using the ideal gas law.
n is the number of moles in 1 kg.
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3.- 1 kg of water at 200C is converted into ice at -100C (all this happens at P = 1
bar). The latent heat of ice melting Lmelt = 334 kJ/kg, the heat capacity of water at
constant pressure is 4.2 kJ/(kg·K) and that of ice 2.1 kJ/(kg·K), the heat of fusion
of ice at 00C is 336 kJ/kg.
(a) What is the total change in entropy of the water-ice system?
(b) If the density of water at 00C is taken as 10% greater than that of ice, what is
the slope of the melting curve of ice at this temperature? Give both sign and size.
(c)Is this sign normal?
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(a) This is almost identical to the problem we did on tuesday, but now we are
asked for the change in entropy. We need to calculate 3 entropy changes ((1)
water from 20 oC to 0 oC; (2) freezing of water;(3) cooling of ice from 0 oC to
-10 oC).
(1)
(2)
(3)
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(b) We use Clausius-Clapeyron
(b) The sign is not normal, for an ideal substance the solid should be
more dense than the liquid.
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4.- The phase diagram of a solution of B in A, at a pressure of 1 bar, is shown
in the Figure. The upper bounding curve (the dew-point curve) of the twophase region can be represented by
The lower bounding curve (the bubble-point curve) can be represented by
A beaker containing equal mole numbers of A and B is brought to its boiling
temperature at the bubble-point curve.
What is the composition of the vapor as it first begins to boil off?
Does boiling tend to increase or decrease the mole fraction of B in the remaining
liquid?
T0
T*
T1
A
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B
We are asked to compute what is XB in the dew point curve at T=T*.
We first need to compute T* and then replace this value in the dew point
curve to solve for XB.
- boiling tends to decrease the mole fraction of B in the remaining liquid
(because there is much more B in the gas).
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P
liquid
solid
gas
Ttr
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T
5.- The schematic phase diagram shown is for
hydrogen H2. Here are the things that we know
about the system (all values near the triple
point):
1.- The triple point T is Ttr=14 K
2.- The molar weight of H2=2g/mole
3.- Latent heat of vaporization Lv=1.01 kJ/mol
4.- Molar density of liquid H2 is rho=71 kg/m3
5.- 4.- Molar density of solid H2 is rho=81 kg/
m3
6.- Melting temperature Tm=13.99+P/3.3 (Tm in
K, P in MPa=106 N/m2)
7.- Vapor pressure equation: P=P0exp(-Lvap/RT),
with P0=90 MPa.
(a) What is the volume difference Vliq-Vsol per mole when solid H2 is
melted by increasing T at constant P?
(b) Calculate the latent heat of melting
(c) Calculate the latent heat of sublimation
(d)Compute the slope and sign of the vapor pressure curve (dP/dT) for
H2 near the triple point, in units of MPa/K. Assume that H2 can be
treated as an ideal gas.
Wednesday, April 10, 13
(a) What is the volume difference Vliq-Vsol per mole when solid H2 is
melted by increasing T at constant P?
Vliq
Vsol
mliq
(2g/mole)(1kg/1000g)
=
=
=
2.817
·
10
⇢liq
71kg/m3
msol
(2g/mole)(1kg/1000g)
=
=
=
2.469
·
10
⇢sol
81kg/m3
V = VS
VL = 0.348 · 10
5
5
5
3
m /mole
m3 /mole
3
m /mole
During the melting the volume increases, which is in
agreement with the negative slope f the P-T (L/S)
coexistence line
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(b) Calculate the latent heat of melting
We use Clausius-Clapeyron:
Lmelt
Ttr (Vl Vs )
=
dTm /dP
1K
6
2
dTm /dP =
=
0.303K/(10
N/m
)
3.3 · 106 N/m2
Lmelt
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(14K)(0.348 · 10 5 m3 /mole)
=
=
160.8J/mole
0.303K/(106 N/m2 )
(c) Calculate the latent heat of sublimation:
Sublimation: Solid-->gas
We do not have enough data to compute it from the information provided,
however, the latent heat of sublimation, can be computed form the sum of
latent heats of vaporization and evaporation, because we are near the triple
point:
Lvap=Ttr(SGas-Sliq); Lmelt=Ttr(Sliq-SSol) ; Lsub=Ttr(SGas-SSol)
Lsub=Ttr(SGas-SSol)= Lmelt + Lvap=160.8 J/mole + 1010 J/mole = 1170.8 J/mole.
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(d)Compute the slope and sign of the vapor pressure curve (dP/dT) for H2 near
the triple point, in units of MPa/K. Assume that H2 can be treated as an ideal gas.
Use Clausius-Clapeyron:
Ideal gas:
Lvap
Lvap
dTm /dP =
⇠
Ttr (VG Vs )
Ttr (VG )
RTtr
P Ttr
VG =
=
P
P0 e Lvap /RT
Replacing numbers we obtain VG=7.7 10-3m3/mole and dP/dT=0.0094 MPa/K
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6.- The vdW gas undergoes an isothermal expansion from volume V1 to volume V2.
Calculate the change in the Helmholtz free energy.
In the isothermal process, the change of the Helmholtz free energy is
We can compare with
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