A few ml of water is heated in a 12-ounce soda can (photograph at left below) until the
water boils, producing steam as evidenced by the condensation coming out of the can
(photograph at the center below). The steam-filled can is then grabbed by a pair of tongs
and quickly placed upside down on a dish of room-temperature water, as seen in the
photograph at the right below, taken just before the top of the can reaches the water. You
can see how the demonstration is set up in an mpeg video by clicking your mouse on the
photograph at the right below.
The question this week
involves exactly what happens after the can hits the water.
When the can reaches the surface of the water:
(a) The can will RAPIDLY implode.
(b) The can will SLOWLY implode.
(c) Water will be pulled up into the can, filling the can with water.
(d) Nothing will happen.
Explain your answer!
The answer is (a) The can will RAPIDLY implode, as seen in an mpeg video by clicking
your mouse on the photograph below.
When the can is removed from the hot plate and rapidly cooled by the water, the steam in
the can IMMEDIATELY condenses, creating a partial vacuum in the can, thus causing it
to collapse. This process occurs much too rapidly for water to be drawn up into the can
by the low pressure relative to the outside atmosphere. The only way for the pressure in
the can to be equalized so rapidly is for the can to collapse.
After the can rapidly collapses with a loud "thunk," the remaining volume of the
collapsed can does in fact fill with water, due to the reduced pressure in the can.
V vs. T for a Gas (Charles’ Law I)
How does temperature affect the volume occupied by a trapped gas maintained at a
constant pressure?
You should set up the apparatus shown in the diagram that follows and fill the coffee
mug with hot water at about 60-70°C. To prevent spills, put the flask in the mug on the
tray. Place the tubing in the rubber stopper hole. Place the thermometer in the mug outside
the flask. Since you will be cooling the water with the ice to reduce the volume of the
trapped air, pull the plunger out to about the 9 cc mark before seating the rubber stopper in
the flask. Mount the syringe as shown in the diagram so the piston is at the 10 cc mark
when the flask is in the hot water.
Diagram of apparatus that can be used to measure volume changes as a function of
temperature in air trapped at a constant pressure.
Note Since the hose has a cylindrical shape, you can calculate the
volume of air contained in it by noting that the equation for the volume
of a cylindrical shape of length L and inner diameter d is given by V =
Activity: Volume vs. Temperature
a. How do you predict that the volume of the air trapped in the flask, hose, and syringe will
change if it is cooled (or heated) at constant pressure? Sketch a graph and explain your
b. Explain why the pressure of the air remains constant even if the volume of the air in the
system changes.
c. You will start measurements of V vs. T with the plunger bottom at the 10 cc mark when
the flask is sitting in 70°C water. Determine the initial volume of the system including
the hose, flask, and syringe. Show your basic measurements and calculations. Hint:
You can use an electric balance to determine how many grams of water it takes to fill the
flask up to the stopper. The number of grams is the same as the number of cubic
centimeters of available flask volume.
Volume of hose:
Volume of flask:
d. Take and record the data needed to determine the total volume of the trapped air as a
function of temperature of the water surrounding the gas inside the flask. Note: Strictly
speaking, you should measure the temperature of the air in the flask, but it’s much faster to measure
the surrounding water temperature which is roughly proportional to that of the air.
V flask (cm3)
V syringe(cm3)
V total (cm3)
e. Use the Kelvin temperature scale and display a graph of your results (or your summary of
the results of others) for V vs. T using graphing software. How are V and T related
The Ideal Gas Law
We have seen how pressure depends on temperature at constant volume, and how volume
depends on temperature at constant pressure, and how pressure and volume are related at
a constant temperature. Let’s summarize these relationships using simple mathematics.
Activity: Summarizing Boyle’s and Charles’ Laws
a. Write down an equation that describes the relationship between volume and temperature
when the pressure is held constant. Express this in terms of V, T, and C1, where C1 is a
b. Write down an equation that describes the Charles’ law II relationship you discovered in
in terms of P, T, and another constant, C2.
c. Multiply the two relationships from parts a. and b. together and summarize the results in
terms of P, V, T, and a third constant denoted C3.
d. Is the new product you created in part c. consistent with the Boyle’s law
relationship you discovered? Explain. Hint: Recall that the temperature was a constant.
The Ideal Gas Law describes all three relationships mathematically in an idealized
fashion. It is given by:
PV = nRT
n = the number of moles of gas
R = the Universal Gas Constant given by 8.31 J/mol•K
An alternative statement of the Ideal Gas Law is:
N = the number of gas molecules
kB = Boltzmann’s Constant given by 1.38  10 J/K
Activity: The Ideal Gas Law
a. Write the 3 relationships that we found in the past three experiments. Show that all three
relationships observed are compatible with the Ideal Gas Law.
b. Describe the Ideal Gas Law in words
Ideal Gas Law:
P = Pressure of the Gas (SI: Pa = M/m2; 1 atm =101.3 kPa)
V = Volume of the Gas (SI: m3; 1 liter = 10-3 m3 = 1000 cm3)
T = Temperature of the Gas (SI: K; oC = K - 273.15)
N = Number of molecules in the Gas
k = Boltzmann's Constant (SI: k = 1.38x10 -23 J/K)
n = number of gram moles in the Gas
R = Universal Gas Constant ( R = 8.314 J/mole/K)
NA = 6.022x1023 molecules/mole (Avogadro's Number)
M = Molecular Weight of the Gas in grams per mole
m = mass of the Gas (SI: kg; 1 g = 10-3 kg)
Useful Relations
Forms of the Ideal Gas Law
P = Pressure of gas in N/m2 or Pa.
Usually Measure with some device.
V = Volume occupied by gas in m3.
Depends upon the shape of the container.
N = Number of molecules in the gas.
N = nNA = (m/M)NA
NA = 6.022x1023 molecules/mole
(Avogadro's number)
Number of molecules per gram mole.
n = Number of gram moles of the gas. n = N/NA = m/M
m = Mass of the gas.
m = nM = (N/NA)M
M = Molecular weight of gas
(converted to grams/mole).
Look up values in Atomic Tables.
k = 1.380x10 -23 J/K
(Boltzmann's Constant)
R = 8.314 J/mole-K
= NAk (Universal Gas Constant).
Rs = Gas Constant for a Specific gas. Rs = R/M = NAk/M
N = number density in molecules/m3. N = N/V
n = molar density in moles/m3.
n = n/V
 = mass density in kg/m3.
 = m/V
Triple Point of a substance:
For ordinary substances like water there is a specific pressure and temperature were a
substance can exist in all three states at the same time. This means that a mixture of ice
and water lowered to a pressure of 611 Pa (about .6% atm) will boil at a temperature very
close to 0 0 C.
Critical Point of a substance :
At high enough pressure and temperature a substance like water may not have a separate
liquid and gaseous phase. This means that the value of physical properties of the
substance such as its heat capacity vary continuously in this region. There is no
discontinuous jump associated with a phase change. Moreover, the heat needed to boil
the substance (i.e. its heat of vaporization) goes to zero at the critical point.
Diving Bell Problem
A cylindrical diving bell 1.50 m in radius and 4.25 m tall with
an open bottom is submerged in the ocean (density = 1025
kg/m3). As the diving bell descends, the level of seawater in
the diving bell rises. The diving bell stops when the surface
of the seawater in the diving bell is 225 m below the surface
of the ocean. At this depth the temperature of the seawater is
15.0oC. The temperature and pressure of the air in the diving
bell at the surface is 25.0oC and 1 atm. Note that the
molecular weight of air is 28.8 g/mole.
(A) What is the pressure of the air in the diving bell
when it is submerged ?
(B) How high will the seawater rise in the diving bell
when the diving bell is submerged to a depth of 225
m, if the temperature of the air drops to 15.0oC ?
(C) At what depth will the diving bell be half full of
(D) How many grams of air are in the diving bell ?
(E) When the bell is submerged, all the water in the
diving bell is forced out by adding more air from a
pressurized air-tank. How many grams of air have
been added ? (Assume the air temperature remains at
Sketch and Process:
The air in an open-ended diving bell is compressed
due to pressure of seawater when the bell is submerged.
Relevant Physics:
We assume that the air behaves like an ideal gas.
The pressure of the air in the bell is the same as the hydrostatic pressure of the seawater
at the given depth.
The volume of the air in the cylindrical bell can be calculated from the volume of a
A) Find P2 when d = 225 m
The pressure of the air inside the diving bell must be holding the water from completely
flooding the bell, thus the pressure of the air must equal the hydrostatic pressure of the
water at depth of 225 m plus the pressure of the atmosphere pushing down on the ocean's
This is 23.3 atm !
Observe that this problem would have been much harder if only the depth of the bottom
of the diving bell were given. Then we would have to determine the height of the air in
the diving bell which depends upon the pressure we are looking for.
(B) Find the height of the seawater in the diving bell.
If h is the height of the air in the bell, then H - h is the height of the water in the bell. To
apply the Ideal Gas Law to the air inside the bell we need to know the pressure,
temperature, and volume of the air.
Since V2 contains the unknown h, let us find V2 using the ideal gas law.
Alternately, we first find the number of moles in the diving bell from the values given for
the initial conditions of the air.
Then use the value of n in the same equation applied to the second state.
Since the volume of the air is equal to the cross-sectional area times the height of the air
in the cylinder,
(C) Find d when V2 = V1/2.
We know (from part A) that the pressure of the seawater depends upon the depth d.
If we can find the value of the pressure P2, we can find d. We can find the pressure P2
using the Ideal Gas law when V2 = V1/2.
Not very deep, is it ?
(D) Find the mass of the air in the diving bell initially m1.
Knowing that the molecular weight of air is 28.8 g/mole we can find the mass of the air
using the Ideal Gas Law, PV = nRT, to find the number of mole of gas in the bell.
As a consistency check we can also find n at d = 225 m. Since no air is lost or added at
this point in the problem they should be the same. From part B we knows the values
pressure, temperature, and volume at d = 225 m.
(E) Find the amount of air that has to be added to expel all of the water in the bell.
If m2 is the mass of the air after the bell has been pressurized so that the bell is empty of
water, then m2 - m1 is the amount of air that was added. We found m1 = 35.4 kg in part
To find m2 we will again use the ideal gas law, PV = nRT, as we did in part D. We know
that the final volume of air is the same as the initial volume, V2 = V1 = 30.041 m3. We
assume that this part of the question implies that the final state is in equilibrium so that
the final temperature of the gas in the bell is same as the sea water at that depth, given as
15.0oC. (If the gas came from a compressed air cylinder at 15.0oC, then the added gas
would be at a lower temperature because of its expansion. As this gas heats up to 15.0oC
it would expand some more so that some gas might escape out the bottom. The amount of
gas left after equilibrium is reached is what we will find.)
To find P2 we must calculate the final pressure of the air whose surface is now at the
bottom of the diving bell. This is not the 225 m given for part A, but 225m plus the
height of the water in the bell found in part B.
This is consistent with the pressure we found in part A of 2.36 MPa in that it is slightly
Now that we knows the values for the final state we can find m2.
Thus the mass added is,
Note the density of the air would now be (832 kg)/(30.0 m3) = 27.7 kg/m3 compared to its
initial density of 1.18 kg/m3 = (35.4 kg)/(30.0 m3). That is about 23 times as dense as
normal air.
Vacuum Chamber Demos.
A balloon is placed into a vacuum chamber, what will happen
when the pressure is reduced and why? What will happen
when the pressure is allowed back into the chamber and
A marshmallow is placed into
a vacuum chamber, what will
happen when the pressure is
reduced and why? What will
happen when the pressure is
allowed back into the
chamber and why?
High Altitude Balloon Problem
A high altitude balloon contains helium whose molecular mass is 4 grams/mole. At its
maximum altitude the balloon's volume is 830 m3. The outside temperature and pressure
are -51.0 oC and 5.40 kPa at the balloon's maximum altitude. Assuming that the helium in
the balloon is in equilibrium with the outside air temperature and pressure, i.e., they are
the same:
(A) What is the mass of the helium in the balloon at its maximum altitude?
(B) Assuming no loss of helium, what would be the volume of the balloon when it
was launched from the ground where the air temperature and pressure are 22.0 oC
and 101 kPa?
(C) If the volume of the balloon when it was launched was 61.0 m 3, how many atoms
of helium were lost during the balloon's assent?
(D) Show that the buoyant force of the helium gas is approximately constant as the
balloon rises. Calculate the mass that the helium would lift initially if the
molecular weight of air is 28.8 grams/mole.
Sketch and Process:
The gas in a balloon expands as the balloon rises.
Relevant Physics:
The helium that does not escape as well as the air surrounding the balloon can treated as
an ideal gas.
The buoyant force due to Helium is equal to the weight the air displaced by the balloon
less the weight of the Helium in the balloon
(A) Find m at maximum altitude.
Since we know the temperature, pressure, and volume at the max. altitude we can find the
number of moles.
Since Helium’s molecular weight is 4 g/mole,
(B) Find initial volume if no Helium were lost during the assent.
Apply the ideal gas law to the initial and final states, and solve for the initial volume.
Alternately, since we know the number of moles from part A and since none of the
Helium is assumed lost,
(C) Find the number of helium atoms lost during the assent.
Initially we know the volume of the balloon was 61.0 m 3. From part B we know that it
would have been 58.958 m 3 if no gas escaped. Thus 61.0 – 58.958 = 2.0414 m 3 of
Helium must have escaped. Using the ideal gas law we can find the number of moles in
2.0414 m 3 of Helium at the ground conditions.
(D) Find the mass that the Helium could lift.
The buoyant force due to the Helium is equal to the weight of the air displaced less the
weight of the Helium in the balloon.
Thus the lifting force is approximately equal to weight of the mass of the air displaced
less the weight of the mass of the helium. At the start, the lifting force would be
Since the air and the Helium are assumed to be in equilibrium with each other at any
altitude, the number of moles of air displaced will always equal to the number of moles
of Helium in the balloon.
If no helium escaped, the buyout force would remain constant. The density of both the
Helium and the air will become less as the balloon rises, but the number of moles of air
displace has to equal to the number moles in the balloon.
The upper limit for balloon flights is dictated by the ultimate size to which the balloon
can expand. Once the balloon reaches it maximum volume, the density of the Helium
remains fixed while that of the out size air drops as the balloon rises.
The height of this balloon is about 20 km
For a fixed number of molecules/moles The Ideal Gas Law forms the surface of a
three-dimensional plot where the axis are Pressure, Volume, and Temperature.
Lines of constant pressure, constant volume, and constant temperature form a
coordinate system labeling the location of an ideal gas.
Robert Boyle showed that the pressure of a low-density gas is inversely
proportional to the volume of a gas when the temperature is held constant, P 
1/V for constant temperature.
Jacques Charles and Gay-Lussac showed that the pressure of a low-density gas is
proportional to the temperature of the gas when the volume is held constant, P 
T for constant volume.
The ideal gas law is only valid for low-density gas. Fortunately, most ordinary
gases behave like an ideal gas because the sizes of the molecules are small
compared to their separation. None the less, there is still a PVT surface for highdensity gases only it is not the ideal equation of state PV = nRT. One such
equation is that of the van der Waals equation.
The Ideal Gas Law, PV = nRT, can be derived by assuming that the molecules
that make up the gas have negligible sizes, that their collision with themselves and
the wall are perfectly elastic, and that the molecules have no interactions with
each other.
The van der Waal's equation is a second order approximation of the equation of
state of a gas that will work even when the density of the gas is not low.
Here a and b are constants particular to a given gas.
Some van der Waals Constants
(J m /mole2)
(m /mole)
133 K
Carbon Dioxide (CO2)
304.2 K
Nitrogen (N2)
126.2 K
Hydrogen (H2)
33.2 K
Water (H2O)
647.3 K
Ammonia (NH3)
Helium (He)
Freon (CCl2F2)
406 K
5.2 K
385 K
The parameter b is related to the size of each molecule. The volume that the
molecules have to move around in is not just the volume of the container V, but is
reduced to ( V - nb ).
The parameter a is related to intermolecular attractive force between the
molecules, and n/V is the density of molecules. The net effect of the
intermolecular attractive force is to reduce the pressure for a given volume and
When the density of the gas is low (i.e., when n/V is small and nb is small
compared to V) the van der Waals equation reduces to that of the ideal gas law.
One region where the van der Waals equation works well is for temperatures that
are slightly above the critical temperature Tc of a substance
Observe that inert gases like Helium have a low value of a as one would expect
since such gases do not interact very strongly, and that large molecules like Freon
have large values of b.
There are many more equations of state that are even better approximation of real
gases than the van der Wall equation.
One system we will meet often in our study of thermodynamics is a mass of gas confined
in a cylinder with a movable plunger or piston. The use of a gas-filled cylinder to study
thermodynamics is not surprising since the development of thermodynamics in the
eighteenth and nineteenth centuries was closely tied to the development of the steam
engine, which employed hot steam confined in just such a cylinder.
You are to observe the relationship between expansion and compression of a gas and
work done on or by the gas. To do this, you will need the following:
• 1 glass syringe, 10 cc
• 1 length Tygon tubing, 5 cm (1/8” ID)
• 1 tubing clamp
Raise the syringe plunger about halfway up and insert the short tube and clamp at
the end of the syringe to seal it.
Try compressing the air in the syringe gently. Do you have to do work on the gas
to compress it? What happens when the plunger springs back?
In thermodynamics, pressure (defined as the component of force that is
perpendicular to a given surface per unit area of that surface) is often a more useful
quantity than force alone. It can be represented by the equation:
P 
Let’s extend our definition of work developed earlier in
the course and this new definition of pressure to see if we can
calculate the work done by a gas on its surroundings as it expands
out against the piston with a (possibly changing) pressure P.
Activity: Relating Work and Pressure Mathematically
Are you doing work when you compress the gas in a
Gas at pressure P exerts
a force on the piston F =
PA as it moves a
distance dx.
b. You know that work can be written as W = Fdx when F
represents the force you exert on the syringe plunger. Show the
mathematical steps to verify that work can be written as W = PdV
for the situation shown at right.
If you lift a mass in the presence of a gravitational force or compress a spring, it is obvious
that the system gains a potential energy, Epot. Is the concept of “potential energy” useful in
discussing what happens if work is done on a system such as gas confined in a syringe
where there is no apparent change in potential energy? The answer is yes, but we have to
give a new meaning to our potential energy Epot. In thermodynamics, Epot is called the
internal energy or Eint and represents any way of storing energy inside a system. The
internal energy of a system is the sum of the helter-skelter kinetic energies of atoms in a
gas, the vibration of atoms in a crystal of quartz, or the spinning of gas molecules. One way
to increase the internal energy of a system is to do work on it.
Heat Energy and Work Together
Transferring heat energy to a system could serve to increase its internal energy, but it might
result instead in the system doing work on its surroundings. In thermodynamics we are
interested in the relationship between heat energy transfer to a system’s internal energy and
work done by the system on its surroundings. What do you think would happen if you
attach a cylinder with a low-friction movable piston (or plunger) to a small flask and place
the flask in hot water? Would its piston experience a force? Can the air in the system do
work on the plunger?
Activity: The Heated Syringe
a. Predict what happens to a plunger (piston) if a flask attached to it
is put into hot tap water that is about 40-50° C as shown at right
b. Watch what happens when you do it while holding the plunger fixed,
submerge the flask in hot tap water. Explain what might be
happening to the gas while you hold the plunger to keep the gas
c. After a minute or so, release the plunger and let it move freely.
What happens
Setup to transfer heat
energy to air in a
syringe with a low
friction plunger (or
piston) in it.
You should have concluded in doing the last activity that the transfer of heat energy to a
system can either cause it to do work on its surroundings or serve to increase its internal
energy. What is the relationship between heat energy transfer, changes in a system’s
internal energy, and the work done by the system? We picture Eint as the “true” energy of
the system. In the theory of thermodynamics, Eint is called a “state” variable, a quantity
that tells us how to calculate useful things about a system such as its temperature. To help
us understand how Eint is related to work and heat energy, we will take a closer look at
what happened to the gas confined in a syringe.
Suppose the plunger on the syringe is clamped while the flask attached to it is
immersed in hot water. The clamped plunger can’t move, so no work is done. However,
since the water is at a higher temperature than the syringe, we expect that heat energy is
transferred from the water to the syringe. This causes the temperature of the water to
decrease and the temperature of the syringe to increase. The heat energy transfer can be
calculated using the equation Q = cwm T where m is the mass of the water, cw is its
specific heat, and T is the temperature change of the water.
Assuming that no heat energy can be transferred to the surroundings, the transferred
heat energy Q, must equal the increase in the internal energy of the gas. This assumption
is based on a belief that energy is conserved in the interaction between the hot water and
the gas.
Suppose we release the plunger and allow the gas to expand and do work on the
plunger when it’s placed in the hot water. How can we calculate the work done by the gas
and its change in internal energy?
As the trapped gas expands we can calculate the amount of work it did against its
surroundings by evaluating the integral W = PdV. Where did the energy to do this work
come from? The only possible source is the internal energy of the gas, which must have
decreased by an amount W. The total change in the internal energy of our trapped air
must be
Eint = Q  W
This relationship between absorbed heat energy, work done on surroundings, and
internal energy change is believed to hold for any system, not just a syringe and a flask
filled with air. It is known as the first law of thermodynamics.
The concepts of work, heat energy transfer, and internal energy are subtle and
complex. For example, work is not simply the motion of the center of mass of a rigid
object or the movement of a person in the context of the first law. Instead, we have to
learn to draw system boundaries and total the mechanical work done by the system inside
a boundary to its surroundings.
The first law of thermodynamics is a very general statement of conservation of
energy for thermal systems. It is not easy to verify it in an introductory physics
laboratory, and it is not derivable from Newton’s laws. Instead, it is an independent
assertion about the nature of the physical world.
More Comments About the First Law
There are many ways to achieve the same internal energy change, Eint. To achieve a
small change in the internal energy in a syringe, you could transfer a large amount of heat
energy to it and then allow the gas to do work on its surrounding. Alternatively, you
could transfer a small amount of heat energy to the gas and not let it do any work at all.
The change in internal energy, Eint, could be the same. Since Eint depends only on Q 
W and not Q or W alone, it is said to be “path independent.”
Activity: The First Law
a. Write down in words only your understanding of the First Law of Thermodynamics.
b. Can you think of any situations where W is negligible and Eint = Q? (Hint: Is it necessary
to do work on a cup of hot coffee to cool it? Can you think of similar situations?)
c. How could you arrange a situation where Q is negligible and in which Eint =  W?
First Law of Thermodynamics
The Conservation of Energy
The net amount of energy added to a system (in the form of heat or work) =
net increase in the energy stored internally in the system plus any change in
the mechanical energy of system's center of mass .
U = Internal Energy of the System (SI: J)
W = Work done by or on the System (SI: J)
Q = Heat flow in or out of the System (SI: J)
MEcm = Mechanical Energy of the center-of-mass (SI: J)
= KEcm + PEcm + REcm
* For systems where the change in mechanical is unimportant, the first law of
thermodynamics is written as:
It is important to observe that the  in U is absolutely necessary because both
work W and heat Q represent a transfer of energy where as the internal energy U
is a quantity of energy that a system contains.
The quantities Qin, Qout, Win, and Wout are all taken to be positive quantities
where as Q and W can be either positive or negative.
A 1.0-kg bar of copper is heated at atmospheric pressure so that its temperature increases from
20°C to 50°C.
(A) What is the work done on the copper bar by the surrounding atmosphere?
(B) How much energy is transferred to the copper bar by heat?
(C) What is the increase in internal energy of the copper bar?
A) Assume that the pressure is constant.
=-1.7*10-2 J
B) .
C) .
Most of the energy transferred into the system by heat goes into increasing the internal
energy of the copper bar. The fraction of energy used to do work on the surrounding
atmosphere is only about 10-6 . Hence, when the thermal expansion of a solid or a liquid
is analyzed, the small amount of work done on or by the system is usually ignored.