Phys260 (Gupta) SOLUTIONS
HW6: Ideal Gases and First Law of Thermodynamics
Required Reading
This week’s HW is based on more of Ideal Gases. Now adding in the first law of
thermodynamics. Chapter 17 in Knight and Chapter 18 in Tipler. The next exam
covering ideal fluids, gases and first law of thermodynamics is on October 29th: so
mark your calenders.
Some things you might need to know: 1 liter (L) = 1000 cm3 = 10-3 m3 . 1 Pascal (Pa) =
105 N/m2 . 1 mole = 6.02x1023 . In pV=NkT, N is the number of particles, k =1.38x10-23 J/
K, and T is measured in Kelvin. In the alternate formulation, pV = nRT, n is the number
of moles, R = 8.31 J/mole/K, and T is measured in Kelvin. Helium and Argon are
monoatomic gases. Benzene is polyatomic.
1. In everyday life, we use the words, heat, temperature, and sometimes energy
interchangeably. This question highlights the need to distinguish them. Describe a
physical situation in which:
My examples are all about pots of water.
(a) a system’s internal energy increases but there is no heat.
Imagine that you have a pot of water at room temperature. You take a stirring rod (also
at room temperature) and quickly stir up the water. The internal energy of the water is
going to increase, because you did work on the system: the stirring rod hit the water
molecules and sped them up.
(b) two systems have the same temperature but different internal energies.
Imagine you had two pots at the same temperature, one with 1 L of water and one with
1000 L of water (big pot!). All of the water molecules have the same energy, but there is
more internal energy in the 1000 L pot, because there are more molecules. (Recall that
internal energy sums up the energy of all of the molecules)
(c) two systems get the same amount of heat but undergo different changes in
temperature.
Now, take your two pots of water and heat them up. If you put the same amount of heat
into both (like if they were both on the stove for the same amount of time at the same
setting), then the 1 L pot would be at a higher temperature. This is because the heat
that goes into the 1000 L pot is split up among a lot more water molecules, so each one
gets less energy and therefore there’s a smaller change in temperature.
2. 60 moles of atomic argon is in a tube sealed by a piston. The gas is initially at
pressure 4.0x105 Pa and volume 2.0x10-4 m3. With the piston locked in place, the gas
is heated by adding 10.0 J. Then, the gas is maintained at its new temperature as the
piston is slowly allowed to move until the volume is doubled.
Phys260 (Gupta) SOLUTIONS
HW6: Ideal Gases and First Law of Thermodynamics
Required Reading
(a) Draw a numerically accurate p-V graph
for this two-part process.
In the first part, the volume is constant and
the temperature increases from heat. By
the ideal gas law: PV = nRT, this means that
the pressure increases. To find out how
much, first I found the change in
temperature in the first part (see part b) is
0.013 K. Then using the ideal gas law, the
change in pressure is
ΔP = nR ΔT / Vi = 3.2 x 104 Pa. Then the
new pressure P2 = Pi + ΔP = 4.3 x 105 Pa.
Then in the second part, the volume is
doubled, but the temperature stays the
same. This means (by the ideal gas law) that the pressure is halved. The shape of this
second part is curvy, because it’s at constant temperature (also PV are inversely
related).
(b)By how much did the neon’s temperature increase?
Since the temp. is constant for the second part, I only have to look at the temp. increase
in the first part. I know that the only way to increase the energy of the gas (which is
proportional to temperature) is to do work on it or to add/take away heat. (Also known
as the first law: ΔEgas = Wdone on the gas + Qadded to gas) Since the piston is locked, there is
no work being done. So the heat causes the energy of the gas to increase by 10 J. I
also know that the change in energy of the gas can be related to temperature by
ΔE = (3/2) nR ΔT. The 3/2 is because argon is monatomic, so has only 3 translational
degrees of freedom. The increase in temperature is then ΔT = (2 ΔE) / (3 nR) = 0.013 K
(c)During the second part of the process (the expansion), is there a net adding of
energy (as heat) to the gas? A net taking away of energy? Neither? Explain?
Here’s my gut reaction: If the temperature doesn’t change, then the internal energy
of the gas doesn’t change since internal energy is proportional to temperature, so there
should be no adding of heat to the gas.
Ok, so let’s look at what is changing the energy of the system. The gas does
positive work by lifting the piston as it doubles its volume. The work of lifting the piston
had to come from the energy of the gas, so the gas loses some energy from work. If
the temperature says the same, then the total energy of the gas stays the same, so
there needs to be heat energy coming in to compensate for the loss of energy to work.
So I had to modify my gut reaction a little bit.
(d)How much heat-energy in Joules was added to the gas in the second process?
If I can figure out how much energy the gas loses by doing work on the piston, then I
can find out how much heat energy would have to come in to compensate for this.
Since the pressure actually decreases as the volume increases, we have to do the
integral Wdone by the gas = ∫Pgas dV from the initial volume to the final volume.
Phys260 (Gupta) SOLUTIONS
HW6: Ideal Gases and First Law of Thermodynamics
Required Reading
Pgas is NOT a constant, it depends on V through the ideal gas law: P = nRT / V. So if we
know nRT at the beginning of the second process, then we can do the integral
nRT ∫1/V dV. We know n is always the same, but we don’t know what the temperature
is at the beginning of the second process: we know how much it changes by in the first
process, but we don’t know what the initial temperature was, so we have to find that
now. Ti = Pi Vi / nR = (4.0 x 105 Pa)(2 x 10-4 m3) / (60 mol)(8.31 J/ mol K) = 0.16 K. Ok,
so now I think there was a mistake in the initial numbers of the problem, because this
temperature is really cold and I’m having trouble imagining it. Outer space is only 2.7 K.
But anyways, we can solve this problem and then complain about the numbers
afterwards. This means that the temperature for the second process is 0.173 K. Now,
let’s do an integral: Wgas = nRT ∫1/V dV = nRT [ln(Vf) - ln(Vi)]
= (60 mol)(8.31 J/ mol K)(0.173K) [ln(4.0 x 10-4) - ln (2.0 x 10-4) = 60 J.
The work done by the gas is 60 J. That means the gas would lose 60 J of internal
energy, unless 60 J of heat comes into the system. Of course, this whole problem is
crazy, because the initial conditions say that the temperature was 0.16 K, which is hard
for me to believe.
3. 2 moles of benzene vapor at 200° C occupies a 2.0-liter tube sealed by a piston. A
constant pressure is maintained as the gas slowly cools down to 100° C. Then a
constant volume is maintained as the gas is slowly cooled to 0° C.
(a) In the lab, how could cool at constant pressure? How could you cool at constant
volume?
Just from the ideal gas law: PV = nRT, I can see that I can cool at constant pressure if
the volume decreases. There’s a very easy way to do this, just let the piston slide. The
pressure needed to hold the piston at the same position is always the same, because
the opposing weight of the piston and the atmospheric pressure pushing down on the lid
is always the same. So if the inside pressure decreases by any amount (because
you’re cooling) then the net force on the lid will be down and the piston will slide down
until the pressure inside is back to its original value.
If you want to cool at constant volume, just lock the lid in place, ensuring constant
volume.
(b) Sketch a rough p-V graph of this process. Your graph need not be numerically
accurate, but should qualitatively capture all the
aspects of the two processes.
Two processes: 1) constant pressure and decreasing
volume and 2) constant volume and decreasing pressure.
Both parts have something decreasing (by the ideal gas
law) because the temperature is decreasing and
everything else except for either volume or pressure is
constant.
Phys260 (Gupta) SOLUTIONS
HW6: Ideal Gases and First Law of Thermodynamics
Required Reading
(c) During which part of the cooling, if either, does the gas lose more internal energy?
Explain.
What I’m thinking:
The change in the gas’ energy is proportional to the change in temperature, so both
parts must lose the same amount of internal energy since the temperatures decrease by
the same amount.
Alternatively, I could argue that, in the first process, the piston is moving down so the
piston is doing work on the gas, so the gas is getting some energy also. So in the first
process, you should lose less internal energy since you also get some energy from the
piston doing work on the gas.
Resolving the debate of ideas: so my first idea must be correct: the change in the gas’
energy is proportional to the change in temperature, assuming that the number of moles
is constant, since ΔE = d/2 nR ΔT. The second idea is incomplete, since I don’t know
what the heat is doing. It must be that in the first process, the gas gives off more heat
to compensate for the extra energy it gets from work.
4. Consider the two-part cooling process described in question#3 above.
(a) Find the volume when the gas is at 100°C.
In the first process, the pressure is constant. So PV = nRT tells me that if the
temperature decreases by 100°C, then the volume decreases by ΔV = nR ΔT / P. The
only thing not given to me in the problem is P, but since P is constant I can find it initially
by Pi = [nRT /V ]i = (2 mol)(8.31 J/ mol K)(473 K) / (.002 m3) = 3.9 x 106 Pa. Now, I can
find the change in volume:
ΔV = nR ΔT / P = (2 mol)(8.31 J/ mol K)(-100 K) / (3.9 x 106 Pa) = -0.0004 m3 = - 0.4 L
The minus sign is a good sign, because we know volume is decreasing.
So the volume at the end of the first process is V2 = Vi + ΔV = 1.6 L
(b)Find the pressure when the gas is at 0° C.
With volume constant now, use the ideal gas law to find the change in pressure:
ΔP = nR ΔT / V2 = (2 mol)(8.31 J/ mol K)(-100 K) / (0.0016 m3) = -1.0 x 105 Pa
So the pressure at the end of the second process is P2 = Pi + ΔP = 3.8 x 106 Pa.
(c) In which part of the process was there ‘work’ done? Does this work contribute to
increasing or decreasing the gas’s internal energy?
The work is only being done while the piston is moving in the first process. The work
done on the gas was positive: the force of the piston is exerted down and the piston
moved down. So if the piston did positive work on the gas, then the gas should gain
energy. Another way to think about it is that the gas is doing negative work, because
the gas is exerting a force up on the piston, but the piston is going down. So if the gas
is doing negative work, then it is gaining energy. This means that since the overall
energy of the gas decreased, there must of have been more energy lost as heat to the
surroundings than gained as work.
(d)Calculate the total work done in cooling the Benzene from 200° C to 0° C.
The work done by the gas is Wgas = P ΔV. I can say this, because the pressure is
constant while the volume decreases. Otherwise, I’d have to integrate, because
pressure would be changing. In the first part, the pressure is 3.9 x 106 Pa and the
Phys260 (Gupta) SOLUTIONS
HW6: Ideal Gases and First Law of Thermodynamics
Required Reading
volume changes by -0.0004 m3. So the work is Wgas = -1560 J, which means the gas’
internal energy gains 1560 J.
5. In a thermally insulated tube, 0.25 moles of Helium gas is initially at pressure 3.0x105
Pa and volume 1.0 L. The gas is compressed to half its volume.
(a) Is the new pressure greater than, less than, or equal to 6.0x105 Pa (twice its original
pressure)? Explain.
I would think that the new pressure is greater than 6x105 Pa. If the temperature of the
gas remained constant then I should expect the pressure to double, when the volume is
halved. Here, the temperature is not constant. In fact, it goes up (see (b)). This rise in
temperature should contribute to increasing the velocity of the molecules, and thus
increasing the pressure, more than the double.
(b) Is the final temperature of the Helium greater than, less than, or equal to its original
temperature?
Compressing the gas implies, that there is positive work being on the gas, that
contributes to increasing its internal energy. Since the system is insulated, non of this
energy that is added via work can escape the system as heat. This implies, that in
effect, there should be a net increase in the internal energy of the gas equal to the
amount of work done in compressing it. Increase in internal energy implies a increase in
the speed of the molecules (since the number of molecules is constant) and hence a
increase in temperature.
(c)Sketch a rough p-V diagram of this process.
p
Dotted line shows the isothermal curve where
the pressure doubles. Solid line is this problem
V
6. In a thin-walled, non-insulated tube sealed by a piston, 0.75 L of Helium is at 2.00
atm. The gas’s temperature is the same as the room’s temperature, 21° C. The piston is
“unlocked” and allowed to come up very slowly until the pressure inside the tube
matches the outside pressure, 1.00 atm.
(a) Is this process adiabatic? Isothermal? Isobaric? Something else? Explain.
The process is isothermal. Since the walls are thin, I am assuming that energy should
be easily exchanged with the surroundings (so it cannot be adiabatic, by definition).
Also, since the process is taking place slowly, I can assume that the exchange of heat
with the surroundings would cause energy to enter/leave the system in such a way that
the gas will always remain at the same temperature as the surroundings. So the
temperature of the gas does not change - and it is isothermal. If the process is not slow
enough to allow for this thermal equilibration, then the temperature of the gas would
change (it should cool a bit) and the process would not be isothermal.
Phys260 (Gupta) SOLUTIONS
HW6: Ideal Gases and First Law of Thermodynamics
Required Reading
The pressure of the gas changes as it expands (2 atm to 1 atm). So it cannot be
isobaric (constant pressure process).
(b) Sketch a p-V diagram for this
process.
p
2 atm
1 atm
V
(c) During this process, the pressure is halved. Does the volume double to
compensate? More precisely, is the final volume greater than, less than, or equal to
1.50 L (twice the initial volume). Explain.
I am assuming the the process is isothermal - it is being carried out sufficiently slowly. It
is a fairly good assumption. In a real experiment, there might be a small deviation from
this result, but not a huge lot. Well, if the temperature of the gas is constant, and so is
the number of molecules, then the Volume should be inversely proportional to the
pressure. Which would imply that the volume doubles to compensate.
Now, suppose the process was not exactly isothermal, then I should expect that the
gas’s temperature would decrease a bit due to the expansion and then the volume
should be a little bit less than double the original volume.
(d) During this process, does heat flow into the Helium, out of the Helium, or neither?
Heat would flow into the Helium. The expansion implies that work is being done by the
system - the gas loses some energy to expand. If the gas is losing internal energy, it
would cool down. But since the walls are thin, as soon as the gas’s temperature falls a
bit, energy flows into the gas from the surroundings, raising its temperature back to that
of the surroundings. In other words, the internal energy lost in doing the work to raise
the piston (expand) is compensated by the flow of energy into the gas to maintain the
same net internal energy, and same temperature.