Vapours Stefan Ivanov

This note will cover a lot of ground, but there aren’t too many prerequisites. You should be familiar
with the notions of entropy, reversible processes, and heat engines. If you’re not completely clear on
what those are or need a refresher, I strongly recommend Chapter 24 from Halliday, Resnick, Krane.

1. Phase Equilibrium
Any substance can take on a variety of phases, and under certain conditions two phases of a substance
can coexist when they’re put in contact. Consider, for example, a closed vessel with liquid water and
water vapour inside, kept at some fixed temperature T . There are mechanisms through which the
water molecules can pass from one state to the other. We have evaporation, where the molecules at
the surface of the liquid may break away if their thermal energy happens to be large enough (hence,
there is more evaporation at higher temperatures). Likewise, we also have condensation, where the
molecules in the vapour rejoin the liquid – many of these molecules will actually reflect off the interface,
but some will always get swallowed up.
The interaction between the vapour and the liquid can be quantified by the pressure p at the
interface due to both of these processes. Naturally, the push from the vapour on the liquid is the same
thing as the push from the liquid on the vapour. To belabor the point, if you think about moist air
above the ocean, the pressure of the water will be equal to the total atmospheric pressure p0 . This is
emphatically not the interaction pressure p of the two water phases. Instead, p0 is the sum of p and
the pressure due to the interaction of the liquid water and the dry air. You can define an interaction
pressure p for other setups as well, like liquid water pushing on a block of ice. If this happens out
in the open, p should be equal to p0 . But if you’re pressing water and ice together in the lab, p is
whatever you want it to be. The upshot is, you always need to mind the external constraints on the
system.
Let’s now go back to the example of liquid water and vapour in a closed vessel. This situation is,
in general, unstable. The value of p at our temperature T may correspond to the rate of evaporation
being larger than that of condensation. For other values of p, the condensation might dominate. At
any given T , there’s one specific value of p for which the two rates are equal (and I’m not claiming
that this is obvious!). We then say that the two phases are in equilibrium. That is, as long as you
hold the system at T , it will evolve so that the interaction pressure is p, and then it will stay like that
forever.
It follows from the above that there will be some curve p(T ) along which both phases can coexist.
We will try to derive p(T ). Consider a system of two phases (1 and 2) with total mass m. The
surroundings of this system won’t matter in what follows. Assume that we’ve found some point of
phase equilibrium (p, T ), where the densities of the phases are ρ1 and ρ2 , and the transition from 1 to
2 requires you to supply a latent heat1 Q = λm. Suppose the system is initially in phase 1 – call this
state A. We will now carry out a cyclic process on this system. Here is how:
• Process A-B: First, we transfer everything to phase 2 at (p, T ). We do this very slowly, i.e. re-
versibly. Note that we are able to do that because the system is in equilibrium at those p and
T to begin with, such that it won’t spontaneously convert to phase  1 or phase
 2 by itself. We
m m
have put in heat Q = λm, and the system has done work AAB = p ρ2 − ρ1 .
• Process B-C: We move adiabatically and reversibly from (p, T ) to another equilibrium state
(p − dp, T − dT ). The system remains in phase 2 throughout. This is feasible as there is no other
phase to oppose us.
• Process C-D: We transfer everything back to phase 1 at this new equilibrium
 state. Again, we
m m
do this reversibly. The system has done work ACD = (p − dp) ρ′ − ρ′ . Note that the densities
1 2

1
The latent heat goes towards: 1. Changing the binding energy between the molecules. 2. Expanding the substance
in accordance with its new density. Mathematically, Q = ∆U + p∆V . Looking at this, there doesn’t seem to be
any reason why λ should be constant. And indeed, it isn’t! But we’ll deal with this separately. You’ll get to work
out how λ depends on T by yourself in Problem 7.

1
 ρ′1 and ρ′2 are the same as ρ1 and ρ2 up to a first-order differential. This doesn’t really matter,
especially considering that ρ1 and ρ2 are themselves markedly different.
• Process D-A: While the system is in phase 1, we move adiabatically and reversibly back to the
initial state.
The processes are shown below on a pT diagram and a pV diagram.

Figure 1 Figure 2

Let’s summarise.
  The system has gone back to its initial state, having done work dA = AAB +
1 1
ACD = dp ρ2 − ρ1 m. Since the entropy of the system is a function of state, its overall entropy
change is ∆S = 0. Then we have
Q Q − dA
∆S = ∆SAB + ∆SBC + ∆SCD + ∆SDA = +0− + 0 = 0,
T T − dT
where we R made use of the fact that all of the processes are reversible, and hence the entropy change is
exactly dQ T . We also required that all the heat we put in is either converted to work or drawn out,
for otherwise the system cannot return to state A, which has a fixed internal energy. Now we can do
the following:
dA dT dp λ
= , Q = λm =⇒ =  .
Q T dT T 1
− 1
ρ2 ρ1

This result is called the Clausius-Clapeyron equation, even though it was derived by Carnot decades
earlier. Generally, λ, ρ1 and ρ2 are not constants, but we can nevertheless integrate this and obtain
the coexistence curve p(T ). We have obtained something tangible through an abstract derivation. It
seems as we’ve pulled the equation out of thin air! But this should instead convince you that entropy
is as real a quantity as pressure, temperature, or internal energy.
To illustrate, the graph of p(T ) for liquid water and water vapour is shown on Figure 3 (note the
logarithmic scale). Of course, you can study the phase boundaries of other phases as well, and this
leads to phase diagrams (Figure 4).

2. Vapours
From now on we will limit ourselves to working with liquid-vapour equilibrium. Firstly, we need to
introduce some terms. If the vapour is in equilibrium with the adjacent liquid, we call it saturated
vapour. The pressure of a saturated vapour corresponds to the coexistence curve p(T ), meaning that
it depends only on temperature.
We state without delay that in most everyday situations the vapours are not saturated. In the
case of evaporation from water on a hob, there’s not enough time for thermal equilibrium to settle
in. The vapours are lighter than air, so they’d rather rise to the ceiling. And even if this wasn’t the
case, they’d get carried away by the fresh air in the room. This is the reason why vapours at a given
temperature T generally have only a fraction of the pressure expected for a saturated vapour at T .

2
 Temperature
0K 50 K 100 K 150 K 200 K 250 K 300 K 350 K 400 K 450 K 500 K 550 K 600 K 650 K 700 K 750 K
1 TPa 10 Mbar

100 GPa 1 Mbar

10 GPa 100 kbar

1 GPa 10 kbar

100 MPa Critical point 1 kbar

10 MPa
Solid Liquid 647 K, 22.064 MPa

100 bar

Pressure
1 MPa 10 bar

100 kPa 1 bar

Freezing point at 1 atm Boiling point at 1 atm
273.15 K, 101.325 kPa 373.15 K, 101.325 kPa
10 kPa 100 mbar

1 kPa 10 mbar

Solid/Liquid/Gas triple point

273.16 K, 611.657 Pa
100 Pa 1 mbar

Gas
10 Pa 100 µbar

1 Pa 10 µbar
-250 °C -200 °C -150 °C -100 °C -50 °C 0 °C 50 °C 100 °C 150 °C 200 °C 250 °C 300 °C 350 °C 400 °C 450 °C

Figure 3 Figure 4

In the case of water, this fraction φ = p/psat is known as the relative humidity. Normally you can’t
get φ > 100%, because at that point the two phases are not in equilibrium, but unlike the case of
φ < 100%, this time around external factors like air flows cannot facilitate a quasi-equilibrium state.
In contrast to the conditions in big open spaces, vapours in closed containers want to remain
saturated. As you increase the temperature of the vessel, psat rises sharply (again, refer to Figure 3).
If the vapours are to stay at psat , there would have to be more and more molecules in the gaseous
state. These can only be supplied by the liquid, and eventually that bunch of liquid runs out of moles
to give out. From that point on, there is only vapour in the vessel, and it won’t be saturated.
Next, let’s solve some problems to get our head around these concepts.

Example 1. Juice (Estonia 2003). A juice bottle is pasteurised at a temperature of t1 =

80 ◦ C while a non-deformable light cap lies freely on the rim of the bottle (excess gas can exit
freely from the bottle, but outside air cannot come in). The cap is then tightly fixed so that
air can no longer enter or leave the bottle. The bottle is cooled down to t0 = 20 ◦ C. What is
the pressure under the bottle cap at that instant? The atmospheric pressure is p0 = 101 kPa.
The dependence of the saturated vapour pressure on temperature is shown on Figure 3.

Solution. In the beginning, the inside of the bottle and the outside atmosphere are in mech-
anical equilibrium. The total pressure inside is p0 . The vapour above the juice is saturated, so
this p0 is the sum of the saturated vapour pressure (SVP) at t1 and the pressure of the dry air
inside the bottle:
p0 = psat (t1 ) + pair .
After the bottle is sealed and cooled down, there will be some condensation, because the vapours
remain saturated, but the SVP has changed. Nonetheless, since the liquid is much denser than
the gas, the condensation will not change the volume available to the gas significantly. Hence,
from the dry air’s point of view, the cooling is done at constant volume. The final pressure of
the dry air is then
T0 T0
p′air = pair = (p0 − psat (t1 )) ,
T1 T1
and the total pressure would be
T0
p = psat (t0 ) + (p0 − psat (t1 )) = 47 100 Pa.
T1

3
Example 2. Ice in a cube (Physics Brawl 2023). We enclose ice with mass m = 15 g and
temperature t0 = 0 ◦ C under normal conditions (pn = 101 kPa, tn = 20 ◦ C) in a hermetically
sealed cubic container with volume V = 0.10 l. We start heating the container until it reaches
a temperature t = 120 ◦ C. What is the pressure p in the container at that point? Use this
website for any reference data that you need.

Solution. The density of the ice at 0 ◦ C is ρice = 916 kg/m3 , so the available volume is Vi =
V − ρmice
. The SVP at the initial temperature is negligible, so the amount of dry air inside the
container can be found as
pn Vi
nair = = 3.48 × 10−3 mol.
RTn
Now consider the amount of water in the vessel,
m
nw = = 0.833 mol.
µw
If all of this has evaporated at t = 120 ◦ C, the pressure of that vapour would be
nw RT
p′ = = 27 MPa.
V
This greatly exceeds the SVP at that temperature, which is psat = 198.9 kPa. Hence, rather
than all of the water evaporating, only a small part of it has turned to gas. So, in the final state
we have liquid water of mass ≈ m and density ρ120◦ C = 943 kg/m3 . The remaining volume
Vf = V − ρ m◦ is occupied by dry air and saturated vapour at pressure psat . And the answer is
120 C

 
Vi T
p = psat + pn = 334 kPa.
Vf Tn

Example 3. Cloud formation (Russia 2017). Assuming an adiabatic atmosphere, estim-

ate:
(a) The total height H of Earth’s atmosphere.
(b) The lowest height h0 at which clouds can form.
The surface temperature of the Earth is t0 = 27 ◦ C, and the relative humidity there is φ = 80 %.
Assume h0 ≪ H. Also assume that air is an ideal diatomic gas of molar mass µ = 29 g/mol.
The dependence of the saturated vapour pressure of water pH on the temperature t is given in
the table below.

t, ◦ C 6 8 10 12 14 16 18 20 22 24 26 28
pH , mmHg 7.01 8.05 9.21 10.5 12.0 13.6 15.5 17.5 19.8 22.4 25.2 28.4

Hint: In an adiabatic atmosphere, a parcel of gas moving vertically with no heat transferred to
it will remain in mechanical equilibrium throughout.

Solution. (a) This is a problem on gas flows. We will examine how a column of air (‘the
system’) can slowly rise through the atmosphere (‘the surroundings’), occupying a new position
where the local parameters of the atmosphere are different. Let this column of air start from
the ground, z = 0, and end at some arbitrary z. As it rises from the ground, molecules that
covered a volume V0 at the ground will clear the space up for the surrounding atmosphere to
come in. The pressure there is p0 . Meanwhile, at the top of the column, where the pressure is
p, molecules from the system will push out the surroundings and occupy a volume V :

4
 Figure 5

When this happens, the atmosphere has done work Aext = p0 V0 − pV on the air column, so the
air column has done work Agas = −Aext = pV − p0 V0 . Now let us think about how the internal
energy of the air column has changed. Since the adiabatic atmosphere has a fixed temperature
distribution, the final state of the column is almost the same as the initial state, the exceptions
being the packets V and V0 at the top and the bottom. The change in the internal energy ∆U
comes only from the internal energies of those two packets. Note that the packets have the
same mass m and the same number of moles n. This follows from mass conservation. Then,
treating air as an ideal diatomic gas, we can write
5
∆U = nR(T − T0 ) + mg(z − 0).
2
It’s more accurate to account for the gravitational potential energy change in the work term,
but I don’t want to dwell on formalities right now. Anyway, since the process is adiabatic, we
can write the first law as
Agas + ∆U = 0.
Finally, using pV = nRT and p0 V0 = nRT0 , we reach
7
nR(T − T0 ) + (µn)gz = 0,
2
2µg
T = T0 − z.
7R
We’ll consider the height of the atmosphere to be the height z at which T turns to zero. Thus
7RT0
H≈ = 30 km.
2µg
Of course, there’s no reason why this way of estimating H is better than any other, but we got
a decent numerical value, and we should leave it at that.

(b) Clouds will form at the instant when condensation becomes favourable, i.e. at phase
equilibrium. This happens at a height z for which the pressure of the water vapour in
the air pvap equals the saturated vapour pressure psat . The latter will depend on tem-
perature, so it certainly changes with z. We want to derive pvap (z). Firstly, note that
pvap (0) = φpsat (0) = φpH (T0 ) ≈ 21.4 mmHg, where we interpolated from the table. Now,
the main point here is that the vapour column must have reached mechanical equilibrium by
itself (in an ideal gas mixture, all species interact separately). Balancing forces on a cylinder of
height dz and cross-section S, we find

S dpvap = −ρvap g 
 S dz.
µH2O pvap 2µg
We substitute ρvap = RT with T = T0 − 7R z to find

5
     7µH2O
dpvap 7µH2O dT T 2µ
=− ⇐⇒ pvap = pvap (0) .
pvap 2µ T T0
2µg
Since we’re looking for a value of h0 much smaller than H, we’ll have T ≈ T0 , or 7R z ≪ T0 .
We can then apply the binomial approximation, which yields
 
µH2O g
pvap (z) = φpsat (t0 ) 1 − z .
RT0

We’ll now plot this dependence on the same graph as psat (z), the data points for which are
taken from the table of pH (T ):

Figure 6

The intersection is at h0 = 0.43 km, and we are done. Note that, indeed, h0 ≪ H, so our
approximations above were reasonable.

3. Boiling
In a previous example, we looked at a vessel in which water was heated up to 120 ◦ C and yet remained
in a liquid state. If you’re still young and innocent, you might have asked yourself, why doesn’t all
the water boil away when it reaches 100 ◦ C? In reality, boiling is something entirely separate from
your usual evaporation, and it plays by different rules. Conventional (nucleate) boiling is associated
with bubbles which form within the liquid phase. These are filled with molecules of the substance
that have transitioned to a gaseous state. This sort of phase transition requires putting in heat, and
it will block the system from increasing its temperature.
Imagine that a bubble has randomly appeared in the bulk of the liquid. This is exceedingly
unlikely to happen purely due to thermal fluctuations. Instead it occurs because of the uneven surface
of the vessel or because of the small particles suspended in the liquid. These can act as nucleation sites
which make the presence of bubbles more energetically favourable. Anyway, given that the bubble is
in contact with the liquid, the vapour inside is saturated. So, at a given temperature T , the vapour
pushes on the surface of the bubble with a pressure psat (T ). On the other hand, the surroundings
want the bubble to collapse. Let the pressure above the liquid be p′ . Then, if the bubble is at depth
h and has a radius r, the pressure on the bubble is
2σ
pout = p′ + ρgh + ,
r
where ρ is the density of the liquid and σ is the surface tension at the liquid-vapour interface2 . The
bubble can exist only if psat (T ) exceeds pout . It follows that bubbles will start to appear only when
2
If you’re feeling underprepared on surface tension, Ashmit Dutta has a made a nice handout which goes into a lot
of detail.

6
the temperature is high enough. When it comes to pout , we can usually neglect the hydrostatic term.
We can do the same for the surface tension term, because the initial r of the bubbles is usually large
enough. Also, in the case of a vessel open to the atmosphere (such as a pot filled with water), the
pressure p′ is simply the atmospheric pressure p0 . Note that even if there’s lots of vapour above the
liquid, mechanical equilibrium tends to take precedence, so we still have p′ = p0 . In conclusion, the
nice and simple way to write the condition for boiling is then

psat (T ) = p0 .

And as long as boiling remains possible, it will take away any heat that you put in. Only after
everything has evaporated can you increase the temperature further. Still, you can get around this
by purifying the liquid and smoothing out the vessel, which removes the nucleation sites. You can
then go above the usual boiling point, albeit in an unstable state. This is called superheating, and
you should probably try to avoid it.
Going back to the condition for boiling, notice that if the pressure of the surroundings were lower,
the boiling would commence at a lower temperature. If you want to make some soup fast, you need
a high temperature, though of course you also don’t want the food to explode in your face, so you
must stay below the boiling point. Consequently, if the boiling threshold is lower, cooking gets harder!
Paying a premium for food when vacationing in the mountains is probably justified.
Now, let’s see what would happen if the vessel were closed. The pressure above the liquid is
simply psat (T ). The outside pressure near any potential bubble is therefore larger than psat (T ). That
is why bubbles will always collapse under these circumstances. In a perfectly sealed pressure cooker,
water would never boil. This allows you to cook at temperatures higher than 100 ◦ C, so you’d be done
with it faster. Please understand – in general, substances want to remain in phase equilibrium. But in
certain conditions boiling will get in the way, and you have to complete the phase transition all at once.

Example 4. Unknown liquid under a piston (Russia 2022).

A vertical cylindrical vessel of cross-section S = 0.01 m2 and height
2H (H = 1 m) is separated from the atmosphere (P0 = 105 Pa)
by a piston, as shown on Figure 7. The piston initially rests on
narrow supports at a height H above the bottom of the vessel.
The piston can slide along the walls of the vessel smoothly while
maintaining a hermetic seal. The piston and the vessel do not
conduct heat. Figure 7

We drain all the air from the chamber below the piston and we put some small amount of liquid
there, of negligible volume compared to SH. The temperature of the system after it has reached
thermal equilibrium is T0 = 350 K. We then slowly heat up the vessel through the bottom, and
we track the pressure P and the temperature T below the piston until it reaches the top of the
vessel at height 2H. The graph of P (T ) is given on Figure 8.

(a) Find the mass of the piston M .

(b) Find the mass m0 of the substance in the chamber under the piston (accounting for all
states of matter).
(c) Find the heat Q supplied to the vessel from the moment when the piston begins rising
until the moment it has reached height 2H.

The heat of vaporisation of the liquid at temperature 1.1T0 is equal to L = 2.2 MJ/kg. The
molar mass of the liquid is µ = 18 g/mol. The gas constant is R = 8.31 J/mol K, and the
acceleration due to gravity is g = 9.8 m/s2 . The vapour of the substance can be treated as an
ideal polyatomic gas.

7
 Figure 8

Solution. (a) Initially, below the piston there is the liquid and its saturated vapour, which
maintains some pressure Psat that depends only on the temperature. As long as this pressure
is less than P0 + (M g/S), the piston will not rise. Even when the piston does move upwards,
this happens very slowly because the heat is supplied gradually too. So, as this happens, the
pressure below the piston should be constant and equal to P0 + (M g/S). On the graph, this
corresponds to 1.5P0 , so
P0 S
M= = 51 kg.
2g
(b) Because of the atmosphere and the weight of the piston, the pressure in the chamber is not
allowed to exceed P0 + (M g/S). As mentioned above, this pressure initially comes from the
saturated vapour of the substance. As we raise the temperature T from T0 to 1.1T0 , we might
reach a situation where there isn’t enough substance to produce a vapour of pressure Psat (T ).
After that we would only have an ideal gas beneath the piston, and it will obey
 
m0 RT
P = .
µ SH
On the graph, this would correspond to a straight line passing through (0, 0), which we don’t
observe. Hence, when we first reach 1.1T0 , we still have some liquid and its saturated vapour in
the chamber. If this is to remain so at even higher temperatures, the pressure in the chamber
must always follow the ever-increasing Psat (T ). However, there is a constraint on the pressure,
so the only option left is for the supplied heat to go towards boiling away the remaining liquid.
Thus, everything is converted to vapour at 1.1T0 , after which the temperature can actually
start increasing. From that point until the end, we simply have expansion of an ideal gas
under constant pressure. Note that the piston will start rising as soon as the boiling begins,
because the boiling increases the amount of vapour in the chamber while the temperature and
the pressure remain constant (meaning that the volume available to the vapour has to change).
One important conclusion is that there’s only vapour beneath the piston at P/P0 = 1.5, T /T0 =
1.5, and it occupies a volume of 2SH. Therefore,
 
m0
(1.5P0 )(2SH) = R(1.5T0 ),
µ

8
 2P0 SHµ
m0 = = 12.4 g.
RT0
(c) The mass of the vapour before right before boiling sets in can be obtained from the ideal
gas equation, given that the saturated vapour is still an ideal gas:

(1.5P0 )(SH)µ
mi = = 8.4 g.
R(1.1T0 )

Then, the total heat necessary to boil the remaining liquid is

7P0 SHµL
Q1 = L(m0 − mi ) = .
11RT0
We also need to supply some heat in order to expand the vapour of mass m0 isobarically from
a temperature of 1.1T0 to a temperature of 1.5T0 . This requires heat
 
m0
Q2 = Cp (1.5T0 − 1.1T0 ),
µ

where Cp = i+22 R = 4R is the molar heat capacity of the vapour at constant pressure. The
vapour is polyatomic, so we used i = 6. In total, we have
 
7µL 4
Q = Q1 + Q2 = + P0 SH = 11.9 kJ.
11RT0 5

There’s one subtle problem left to take care of, though. The latent heat, as mentioned in
Footnote 1, is spent on unbinding the molecules and expanding the resulting vapour out in the
open. The L from the problem statement is probably taken from some reference book, so it
corresponds to expanding the vapour in the atmosphere, not in a closed vessel of some fixed
volume. In the former case the necessary work is P0 (Vgas, atim − Vliquid ), while in the latter it is
1.5P0 (Vgas, vessel − Vliquid ). Luckily, if the volume of the liquid is neglibigle compared to that of
the gas (no matter if that gas is in the atmosphere or in the vessel), these turn out to be the
same, because the atmosphere and the vessel have the same temperature:
 
m0
P0 (Vgas, atm − Vliquid ) ≈ P0 Vgas, atm = R(1.1T0 ),
µ
 
m0
1.5P0 (Vgas, vessel − Vliquid ) ≈ 1.5P0 Vgas, vessel = R(1.1T0 ).
µ
This means that our answer for Q is good enough. If you redo the calculations without approx-
imating, you’d get Q = 12.6 kJ, but this isn’t necessary.

Example 5. Glub glub (Russia 2021). A chemical reaction

in the water solution of some substance in a vat supplies heat at
a constant rate of N = 5 kW. In order to maintain a constant
temperature, we send in air through a perforated pipe at the
bottom of the vat (Figure 9). The pressure of this air is equal
to the atmospheric pressure P0 = 105 Pa, while its temperature
is equal to that of the room, T0 = 22 ◦ C. Find the volumetric
flow rate of air q required to maintain a temperature T < TB
in the vat, where TB = 100 ◦ C is the boiling point of the water
solution at a pressure of P0 . Assume TB − T ≪ TB . Evaluate q
for T = 95 ◦ C. Figure 9

9
 The molar heat of vaporisation of water at T is λ = 40 kJ/mol. The saturated vapour pressure
near TB varies linearly with a rate of α = 3.5 kPa/◦ C. The pressure of the saturated vapour
above the water solution is precisely equal to the saturated vapour pressure of water. Neglect
the pressure difference on both sides of the pipe connecting the vat to the atmosphere. Also
neglect the heat that goes towards raising the temperature of the incoming air. The system
does not exchange heat with its surroundings when air isn’t flowing in.

Solution. This device carries off heat from the water by forcing the water to boil, which
is associated with a latent heat. Without the air supply, we can’t have boiling because the
pressure above the liquid is P0 (most of it from the saturated vapour, the rest from dry air),
while the pressure of the bubbles in the bulk in the liquid would be Psat = P0 + α(T − TB ).
This is less than P0 , so the bubbles are immediately suppressed. However, when we send in air
through the pipe, the air will take part in the bubbles, and it can provide the necessary extra
pressure α(TB − T ) to keep them stable.
Working with partial pressures, we see that the ratio of air to vapour in the bubbles must be

α(TB − T )
k= .
P0 − α(TB − T )
To pump out heat at a rate N = 5 kJ/s from the water, we need the water to leave through
the bubbles at a rate of N/λ = 0.125 mol/s. The air would then need to come in at a rate of
kN/λ (mol/s). All that is left is to find the volume that corresponds to a mole of gas. This is
V /n = P0 /RT0 = 40.8 mol/m3 . We then obtain

N RT0 α(TB − T )
q= = 6.5 × 10−4 m3 /s.
λ P0 (P0 − α(TB − T ))

4. Evaporation and kinetic theory

In this section, we will proceed to estimate the rate of evaporation from a liquid in idealised conditions.
Unfortunately, evaporation is a rather complicated process. Usually, right above the surface of the
liquid there is a thin layer of stagnant saturated vapour. For a molecule to leave the liquid, it first
needs to have enough energy to overcome the attraction from its neighbours at the surface. After that,
it has to diffuse through the stagnant layer. The thinner the layer, the faster the evaporation. If we
set up a flow of dry air above the liquid with a fan, the layer will thin out. The mean free path of the
molecules from the liquid could then end up much larger than the width of the layer, meaning these
molecules essentially don’t perceive it. This is the reason why we blow on our cup of hot tea. It’s not
because our breath will cool the tea down directly; rather, it’s mostly to speed up the evaporation
(and hence the cooling, as vapours carry away heat)3 . This is also why hot weather feels worse when
it’s humid. Your sweat evaporates slower, so its phase transition takes away heat less efficiently.
Now onto working with numbers. We’ll assume that there’s a dry air flow above the liquid, such
that we can ignore the stagnant layer. Assume that the temperature of the system is T , at which
point the saturated vapour pressure would be psat (T ). We’ll now use a trick. Imagine that the vessel,
instead of being out in the open, is placed in an enclosure, so that the liquid is in equlibrium with its
saturated vapour. The evaporation occurs in the same way as before, but now there’s condensation
in the other direction as well. More importantly, the rate of evaporation dN dt is the same as the rate
at which vapour molecules enter the liquid. And this latter quantity is easier to estimate.
If the surface of the liquid has area S, the molecular flux Φ of the vapour through that surface

3
Another trick is to pour the tea in a bowl, which increases the surface area available for evaporation. Likewise,
you should cut your potatoes into wedges if you want them to cool down quicker. In the same vein, there’s a good
reason why many desert plants have spines instead of large leaves.

10
can be estimated from the standard formula
s
1 1 8RT
Φ = nS⟨v⟩ = nS ,
4 4 πµ

where n is the number density of the molecules in the vapour, and µ is their molar mass. We also
need to account for the fact that only a small fraction q of the molecules actually enter the liquid,
whereas the rest are reflected by the surface (for water, q ≈ 3 %). Then, finally,
s
dN qpsat S 8RT
= .
dt 4kB T πµ

We can then deduce the mass flow rate from the liquid from multiplying this by the mass of a single
molecule:   r
dM µ dN µ
= = qpsat S .
dt NA dt 2πRT
And if the liquid has density ρ, the level of liquid in the vessel will change as
r
dh qpsat µ
=− .
dt ρ 2πRT

Evaluating this for water at 20 ◦ C (where psat = 2.3 kPa), we find dh

dt ≈ −4 mm/min. This seems like a
lot, but bear in mind that if you remove the air flow above the liquid, the evaporation will slow down
significantly.
Let’s address one more question – what is the pressure on the vessel due to the evaporating
molecules if we ignore the stagnant layer? To answer this, we recycle the trick from above. If we
had evaporation and condensation at the same time, the pressure on the liquid would be psat . Both
processes transfer an equal number of molecules, so they contribute equally to the pressure. Hence,
in the case where we have evaporation only, the answer is psat /2. This brings us to a famous problem.

Example 6. Ice propulsion (USSR 1980, IOAA 2014). In a science fiction novel, an as-
tronaut of mass M0 = 100 kg is left stranded at a distance of l = 100 m from his spaceship. He
carries has a cup full of ice, and he hopes to return to the spaceship using the sublimation of
the ice. Check if this is realistic by estimating the time τ necessary for the astronaut to return.
Assume that the ice sublimates at a constant temperature of T = 272 K. The saturated vapour
pressure of water at that temperature is psat = 550 Pa. Estimate the dimensions of the cup and
the mass of the ice by yourself.

Solution. We’ve essentially solved this already. When the ice sublimates in a vacuum, there
isn’t any condensation and there’s no stagnant layer to complicate things. We estimate the
cross-section of the cup as S = 50 cm2 and the mass of the ice as M = 200 g. The ice will then
evaporate in time s

dM M 2πRT
T0 = M = = 2150 s.
dt qpsat S µ
While the ice evaporates, there is a push on the cup equal to psat S/2, and the astronaut
maintains a constant acceleration
psat S
a= = 0.014 m/s2 .
2M0
By the time all the ice has sublimated, the astronaut will have covered a distance

aT02
L0 = = 31 km ≫ l.
2

11
This means that the astronaut will move with a constant acceleration right until he reaches the
spaceship. Thus, s
aτ 2 M0 l
= l =⇒ τ = 2 = 120 s.
2 psat S

Last updated: 28th December 2024

Send comments and corrections to sivanov.mail@proton.me

12
Problems
The problems are chosen to go along with the handout. Starting with Problem 4, you will need to
have read the section on boiling.

Problem 1. Dry ice (Estonia 2010). A disk of radius r = 1 cm is made out of dry ice (solid carbon
dioxide). The disk is pressed against a thermally conductive plate with a force F = 100 N applied
uniformly perpendicular to the plate. When the temperature of the plate is low, the disk cannot move
laterally due to the friction between the disk and the plate. However, above some critical temperature
T0 , the friction force almost vanishes, and the disk can start to slide along the plate. Find T0 . The
atmospheric pressure is 101 kPa, and the saturated vapour pressure of CO2 is given on Figure 10. The
triple point of CO2 corresponds to pressure pt = 5.1 × 105 Pa.

Figure 10

Problem 2. From Lord Kelvin’s archives (Russia 2017). As Mr. Bug was digging through Lord
Kelvin’s archives, he found some graph (Figure 11) with a note scribbled on it, mentioning something
about studying isochoric processes. With time, the ink has faded away, and the numbers on the
vertical pressure axis and the horizontal temperature axis have disappeared. We only know that the
point at the top right corner of the grid corresponds to pressure p = 2000 mmHg and temperature
t = 127 ◦ C.
Mr. Bug realised that the graph shows the dependence of pressure on temperature for the contents
of a closed vessel. He figured that apart from air, there was some other substance in the vessel which
underwent a phase transition. In order to find out what this substance was, he decided to calculate
the saturated vapour pressure at the point marked with ‘?’.
(a) What are the total pressure and temperature at that point?
(b) Can you deduce what the unknown substance is?
(c) Find the temperature in the vessel at the instant when 30 % of the unknown substance has
evaporated.
Note: To solve this problem, you’ll probably have to take a screenshot of the graph, import it in
e.g. GeoGebra, and do your work there. There isn’t enough space for that on this handout.

Problem 3. Physics in the mountains (IZhO 2018-2). A chunky problem on humidity in the
atmosphere. See here.

13
 Figure 11

Problem 4. Runaway liquid (Russia 2007 Regionals). One leg

of a tall U -shaped pipe of cross-section S is in contact with the at-
mosphere, whereas the other leg is kept closed. The pipe is filled with
some liquid of density ρ such that the liquid completely fills the open
leg, while its level in the other leg is lower by h, leaving an air pocket
there. The pipe is heated up from the initial room temperature T1 to
the boiling temperature of the liquid T2 at the atmospheric pressure P0 .
Find the volume of liquid ∆V that spills from the pipe by the time the
liquid starts to boil.
The level of the liquid at the closed end of the pipe remains above
the horizontal segment. You can neglect the evaporation from the open
leg while the pipe is heated up. The saturated vapour pressure of the Figure 12
liquid at the initial temperature is also negligible.

Problem 5. Exo-Earth (MPPP 114). At some time in the distant future, humankind makes con-
tact with the inhabitants of an exoplanet, on which the atmospheric pressure near the surface is the
same as on Earth, i.e. 1 atm ≈ 101 kPa. Further, the atmosphere consists of a mixture of oxygen and
nitrogen gases. Because of these similarities, the planet is called Exo-Earth.
Human researchers and Exo-Earth scientists cross-check the physical and chemical data of their
two atmospheres, and state that, on both planets, the boiling points of liquid nitrogen and liquid
oxygen are 77.4 K and 90.2 K, respectively, at standard atmospheric pressure.
On both planets, local ‘air’ was isothermally compressed at a constant temperature of 77.4 K, and
liquefaction set in when the pressure reached 113 kPa. However, on Earth oxygen, and on Exo-Earth
nitrogen, condensed first.
(a) What is the molar ratio of oxygen to nitrogen in the atmosphere of Exo-Earth?
(b) For what molar ratio would the oxygen and nitrogen begin to liquefy simultaneously under
isothermal compression at 77.4 K, and at what pressure would this happen?
Assume that on Earth, the molar ratio of oxygen to nitrogen is about 1 : 4.

14
Problem 6. Adiabatic anisotropy (Russia 2023). A horizontal
cylindrical vessel is sealed off hermetically with a piston. In the vessel,
there is saturated water vapour at a temperature of T0 = 333 K, and
there isn’t any liquid water.
In what follows, treat water vapour as an ideal polyatomic gas.
The latent heat of evaporation for water is L = 2.36 MJ/kg, and its
temperature dependence can be ignored. The gas constant is R =
8.31 J/mol K, and the molar mass of water is µ = 18 g/mol. Make
use of the fact that at temperatures close to T0 , small variations in
the pressure εp = ∆p/p0 and in the temperature εT = ∆T /T0 are Figure 13
related by εp = αεT , where α = 15.3.
(a) We slowly change the temperature inside the vessel. The volume of the vessel changes such
that the water in the vessel always remains in a gaseous state, whereby the vapour is always
saturated. What is the molar heat capacity of the water vapour for such a process?
Now assume that the vessel is thermally insulated.
(b) Find the change ∆T1 of the temperature in the chamber after slowly decreasing its volume by
a fraction of β = 5 %.
(c) Find the change ∆T2 of the temperature in the chamber after slowly increasing its volume by
a fraction of β = 5 %.

Problem 7. The Clausius-Clapeyron equation (IZhO 2020-2). A nice and difficult problem
on border boiling. The key here is that the pressure inside the bubbles which form at the interface
of the liquids is the sum of the saturated vapour pressures for those two liquids. You can find the
problem statement here. For extra practice on this topic, you can check out IPhO 1989-1.

Solutions
1. The answer is T0 ≈ 212 K. If you got something markedly different, have a look at the solution
here (Problem 78).

2. See the original solution here. You will need to translate the text from Russian, but machine
translation has become eerily good these days anyway. I use a browser extension for this, but there’s
probably a better way.

3. You can find the solution in the same folder as the problem statement.

4. The original solution can be found here.

5. The atmosphere of Exo-Earth consists of 1 part oxygen to 8.4 parts nitrogen. Simultaneous lique-
faction will happen for a molar ratio of 1 part oxygen to 4.5 parts nitrogen, at a pressure of 124 kPa.
See the book for a detailed solution.
If you don’t mind the open-ended style of the questions in the book (I do!), I also recommend reading
through the solutions of 111, 112, and 113.

6. The original solution can be found here.

7. You can find the solution in the same folder as the problem statement.

15
References
A. Varlamov, Evaporation and properties of vapours, Kvant, Volume 19, Issue 6, 1988, https:
//kvant.mccme.ru/1988/06/paroobrazovanie_svojstva_parov.htm

M. Anfimov, A. Chernoutsan, As water evaporates, Kvant, Volume 22, Issue 11, 1991, https:
//kvant.mccme.ru/1991/11/poka_voda_isparyaetsya.htm

A. Korzhuev, Evaporation in nature, Kvant, Volume 24, Issue 6, 1993, https://kvant.mccme.ru/

1993/06/isparenie_v_zhivoj_prirode.htm

J. Kalda, Thermodynamics, 2019, https://www.ioc.ee/~kalda/ipho/Thermodyn.pdf

16