Download ME12001 Thermodynamics T6

Thermodynamics T6
Simple Ways of Expanding
A plot of pressure as a function of volume is known as a pV diagram. pV diagrams are often used in analyzing
thermodynamic processes. Consider an ideal gas that starts in state O, as indicated in the diagram. Your task is to
describe how the gas proceeds to one of four different
states, along the different curves indicated.
Although there are an infinite number of such curves,
several are particularly simple because one quantity or
another does not change:
Adiabatic: ΔQ = 0. No heat is added or
subtracted.
Isothermal: ΔT = 0 . The temperature does not
change. (The prefix "iso" means equal or alike.)
Isobaric: Δp = 0. The pressure does not change.
(A barometer measures the pressure.)
Isochoric: ΔV = 0 . The volume does not change.
(This process is infrequently used.)
The key idea in determining which of these processes is occurring from a pV plot is to recall that an ideal gas must
obey the ideal gas equation of state: pV = N kB T (where the constant kB is the Boltzmann constant, which has
the value 1.381 × 10 −23 J/K in SI units). Generally N , the number of gas particles, is held constant, so you can
determine what happens to T at various points along the curve on the pV diagram.
Note that in this problem, as is usually assumed, the processes happen slowly enough that the gas remains in
equilibrium without hot spots, without propagating pressure waves from a rapid change in volume, or without
involving similar nonequilibrium phenomena. Indeed, the word "adiabatic" is often used by scientists to describe a
process that happens slowly and smoothly without irreversible changes in the system.
Part A
What type of process does curve OA represent?
Hint 1. What quanitity remains constant?
Curve OA is horizontal; the pressure of the system remains constant throughout this process. What is the
name of a process in which pressure remains constant?
ANSWER:
adiabatic
isobaric
isochoric
isothermal
Correct
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Part B
What type of process does curve OC represent?
Hint 1. Relationship between pressure and volume
When V /V0 = 2, p/p0 = 1/2; when V /V0 = 4 , p/p0 = 1/4 . Evidently, pressure is proportional to
the inverse of the volume. What is the name of a process in which p is proportional to 1/V ?
Hint 2. Use the ideal gas law
Recall the ideal gas equation of state for a fixed amount of gas: pV = cT , where c is some constant.
Solving this equation for p yields p = cT /V . If p is to be proportional to 1/V , the temperature T must
remain constant throughout the process. What is the name of a process in which temperature remains
constant?
ANSWER:
adiabatic
isobaric
isochoric
isothermal
Correct
Detailed analysis of curve OB
The following questions refer to the process represented by curve OB, in which an ideal gas proceeds from state O
to state B.
Part C
The pressure of the system in state B is __________ the pressure of the system in state O.
ANSWER:
greater than
less than
equal to
Correct
Part D
The work done by the system is __________.
Hint 1. How to find the work done using the pV diagram
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On a pV diagram, the work done during a particular process is represented by the area under the curve
describing that process. Is the area under curve OB greater than, less than, or equal to zero?
ANSWER:
greater than zero
less than zero
equal to zero
Correct
Part E
The temperature of the system in state B is __________ the temperature of the system in state O.
Hint 1. Compare curve OB to an isothermal process
Does curve OB lie above or below a curve representing an isothermal (constant-temperature) process?
Use this, along with the ideal gas equation of state, to figure out how T changes as the system proceeds
from state O to state B.
Hint 2. Computing the change in temperature mathematically
The ideal gas equation of state for a fixed amount of gas is pV = cT , where c is some constant. To find
the sign of the change in temperature, you could find (p/p0 )(V /V0 ) at point B, then subtract the value
of (p/p0 )(V /V0 ) at point O. Although this method will not yield the actual change in temperature, it will
give a number proportional to the change in temperature (with the proper sign).
ANSWER:
greater than
less than
equal to
Correct
Part F
The internal energy of the system in state B is __________ the internal energy of the system in state O.
Hint 1. Relationship between internal energy and temperature
The change in internal energy of an ideal gas is proportional to its change in temperature; the constant of
proportionality is the heat capacity at constant volume.
ANSWER:
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greater than
less than
equal to
Correct
Various Gas Expansions: pV Plots and Work
An ideal monatomic gas is contained in a cylinder with a movable piston so that the gas can do work on the outside
world, and heat can be added or removed as necessary. The figure shows various paths that the gas might take in
expanding from an initial state whose pressure, volume,
and temperature are p0 , V0 , and T0 respectively. The gas
expands to a state with final volume 4V0 . For some
answers it will be convenient to generalize your results by
using the variable Rv = Vfinal /Vinitial , which is the ratio
of final to initial volumes (equal to 4 for the expansions
shown in the figure.)
The figure shows several possible paths of the system in
the pV plane. Although there are an infinite number of
paths possible, several of those shown are special
because one of their state variables remains constant
during the expansion. These have the following names:
Adiabiatic: No heat is added or removed during the
expansion.
Isobaric: The pressure remains constant during the
expansion.
Isothermal: The temperature remains constant during the expansion.
Part A
Which of the curves in the figure represents an isobaric process?
ANSWER:
A
B
C
D
Correct
Part B
What happens to the temperature of the gas during an isobaric expansion?
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Hint 1. Implications of pV
= nRT
The ideal gas law, pV = nRT , must hold at all points on the curve; furthermore, n is constant since no
gas is added or escapes. For an isobaric process p does not change, so the volume and the temperature
are simply related. Find the final temperature TA for the expansion along path A.
Express TA in terms of T0 , the initial temperature.
ANSWER:
TA = 4T0
ANSWER:
Temperature increases.
Temperature remains constant.
Temperature decreases.
Correct
Here is an explanation of why the temperature increases based on kinetic molecular theory. When the
volume is increased, the mean time between collisions with the wall would tend to increase, since
molecules need to travel longer distances before hitting the wall. Keep in mind that the pressure on the
walls of the vessel is given by force/area, where force = Δ(momentum)/Δ(collision time) . If this
increase in collision time were the only thing to change, the pressure exerted on the walls would decrease.
To keep the pressure the same, the molecules must travel faster, which increases the momentum they
impart to the wall at each collision. The molecules are made to travel faster by heating the gas and
increasing its internal energy.
Part C
Which of the curves in the figure represents an isothermal process?
Hint 1. Ideal gas law again
The ideal gas law is pV = nRT . Temperature is constant here so the equation says pV = constant.
The other way to state this is to say that p0 V0 = pA VA . Which of the curves shown has that property?
ANSWER:
A
B
C
D
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Correct
In cyclic thermodynamic processes such as the one that occurs in heat engines and refrigerators, the
system traverses a closed path in the pV plane. In this case there will be curves like those depicted here for
which the system traverses from right to left, that is, on which the gas undergoes compression. The sign of
the work for compression is the opposite of that for expansion. For compression, the work done by the gas
is negative; that is, the world does work on the gas to compress it.
Thermodynamic processes like the three discussed here can also be shown in a plot of p vs. T or T vs. V .
These are useful for showing changes of phase of the gas--for example the line where gas changes to
liquid. However, there is no simple interpretation of the work on such plots, so the pV plot is the most
commonly used when discussing the work done by a system.
Part D
Graphically, the work along any path in the pV plot ____________.
ANSWER:
is the area to the left of the curve from p0 to pfinal
is the area under the curve from V0 to Vfinal
requires knowledge of the temperature T (V )
Correct
Part E
Calculate WA , the work done by the gas as it expands along path A from V0 to VA = Rv V0 .
Express WA in terms of p0 , V0 , and Rv .
Hint 1. Expression for W
Remember that the work along the pV curve is W
V
= ∫ V0A p(V ) dV .
Hint 2. Find an expression for p(V )
The total work is the integral of dW , which is an integral with respect to volume of the pressure p(V ).
What is the pressure along curve A?
Express p(V ) in terms of p0 .
ANSWER:
p(V ) = p0
Hint 3. Do the integral
Calculate the integral from V0 to Vfinal of p = p0 .
Express the integral in terms of p0 , V0 , and Vfinal , the final volume.
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ANSWER:
V
V
∫ V0final p dV =∫ V0final p0 dV
= p0 (Vfinal − V0 )
ANSWER:
WA = p0 (Rv V0 − V0 )
Correct
Remember that p = p0 is not always true. For this particular process, the pressure was constant. Now
consider a process in which the temperature and not the pressure is constant.
Part F
Calculate the work WC done by the gas during the isothermal expansion.
Express WC in terms of p0 , V0 , and Rv .
Hint 1. Find an expression for p(V )
Find the expression for p(V ) using the fact that for this type of expansion pV = constant. To find the
constant, evaluate p ⋅ V at the beginning of the process.
Give your answer in terms of p0 , V0 , and V .
ANSWER:
p(V ) = p0
V0
V
Correct
= ∫ p(V ) dV = p0 V 0 ∫
dV
. Integrate this expression from the initial
V
to the final volume to find the total work done in this process.
So you can rewrite W
Hint 2. Doing the integral
Vfinal dV
Remember that ∫ V
V
0
ANSWER:
WC = p0 V0 ln(Rv )
Correct
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= ln(
Vfinal
).
V0
ME12001 Thermodynamics T6
Part G
Which of the curves shown represents an adiabatic expansion?
Hint 1. Pressure and volume in an adiabatic expansion
Recall that for an adiabatic expansion pV γ = constant, where γ = Cp /Cv = 5/3 for an ideal
monatomic gas.
ANSWER:
A
B
C
D
Correct
± Isothermal Expansion: Work and Heat
Learning Goal:
To understand how various thermodynamic quantities describing a gas change during the process of isothermal
expansion.
An ideal monatomic gas is contained in a cylinder with a movable piston. Initially, the gas is at pressure p0 and
temperature T0 , and the cylinder has volume V0 .
Part A
The diagram shows several possible processes of expansion. Note that the "coordinates" on this diagram are
somewhat unusual: They represent the relative
change in pressure and in volume.
Which of the curves in the figure represents an
isothermal process?
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Hint 1. Use the ideal gas law
The ideal gas law states that pV = nRT . An isothermal process is a process that takes place at
constant temperature. From the ideal gas law, if T = constant, then pV = constant. Which of the
curves shown corresponds to such a case?
ANSWER:
A
B
C
D
Correct
Part B
Calculate the change in internal energy dU of the gas as it expands isothermally from V0 to the final volume
RV V0 . Here, RV is the ratio of final volume to initial volume; RV = 4.0 in this problem.
Express your answer in terms of given quantities.
Hint 1. Find the formula for internal energy
The total internal energy of the ideal gas is equal to the total kinetic energy of all the individual gas
atoms. According to the Equipartition Theorem, for each atom
1
m⟨v2x ⟩
2
where kB is Boltzmann's constant.
=
1
2
m⟨v2y ⟩ =
1
2
m⟨v2z ⟩ =
1
2
kB T ,
What is the total internal energy U for a three-dimensional gas of N particles at temperature T , pressure
p0 , and volume V0 ?
Express your answer in terms of kB , N , and T , or in terms of p0 and V0 .
ANSWER:
U = 1.5N kB T
ANSWER:
dU = 0
Correct
Part C
The pressure, temperature, and volume of a system change differently during the process of isothermal
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expansion. Indicate whether each quantity increases (I), remains constant (C), or decreases (D).
Your answers should describe the changes of pressure, temperature, and volume (in that order).
Separate your answers with commas (e.g., I,C,D means pressure increases, temperature remains
constant, and volume decreases).
ANSWER:
pressure, temperature, volume = D,C,I
Correct
Part D
Calculate the work W done by the gas as it expands isothermally from V0 to RV V0 .
Express the work done in terms of p0 , V0 , and RV . Use ln for the natural logarithm.
Hint 1. Find an expression for incremental work
If the gas, at a constant pressure p, changes volume by a small amount dV , how much work dW is
done by the gas on the outside world? (This is the standard expression for work in thermodynamics.)
Give your answer in terms fo p and dV .
ANSWER:
dW = pdV
Hint 2. Express pressure as a function of volume
The total work is the integral of dW , which is an integral with respect to volume. Using the ideal gas law,
find the pressure as a function of volume V , that is, p(V ).
Express your result in terms of p0 , V0 , and V .
ANSWER:
p(V ) = p0
V0
V
Hint 3. Evaluate the integral
What is the integral from V0 to V1 of dV /V ? (You will need an integral of this form, with V1 = RV V0 , to
find W .)
Express your answer in terms of V0 and V1 . Use ln for the natural logarithm.
ANSWER:
V dV
=
V
∫ V01
ANSWER:
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ln( VV1 )
0
W = p0 V0 ln(RV )
Correct
Part E
Calculate the quantity of heat Q that must be delivered to the gas as it expands from V0 to RV V0 .
Express your answer in terms of p0 , V0 , and RV . Use ln for the natural logarithm.
Hint 1. Use the first law of thermodynamics
The first law of thermodynamics relates work done ( ΔW ), heat added ( ΔQ), and change in internal
energy ( ΔU ). What is the general expression for ΔQ?
Express your answer in terms of ΔU and ΔW .
ANSWER:
ΔQ = ΔU + ΔW
ANSWER:
Q = p0 V0 ln(RV )
Correct
Part F
Which of the following statements are true?
A. Heat is converted completely into work during isothermal expansion.
B. Isothermal expansion is reversible under ideal conditions.
C. During the process of isothermal expansion, the gas does more work than during the isobaric
expansion (at constant pressure) between the same initial and final volumes.
Carefully consider all options and choose the best one.
Hint 1. Use the pV diagram to compare the work done
Since dW = p dV , work can be found by integrating the p(V ) function with respect to volume (as you
did in Part D). However, that integral also represents the area under the p(V ) curve on the pV diagram.
Compare the areas under the isobaric and isothermal curves to determine in which case more work is
done.
Hint 2. Identify the isobaric curve
The diagram shows several possible processes of expansion. Note that the "coordinates" on this diagram
are somewhat unusual: They represent the
relative change in pressure and in volume.
Which of the curves in the figure represents an
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isobaric process?
Hint 1. Isobaric processes
An isobaric process is a process that takes place at constant pressure.
ANSWER:
A
B
C
D
ANSWER:
A only
B only
C only
A and B only
A and C only
B and C only
A and B and C
Correct
Cylinder of Ideal Gas with a Piston
A cylinder contains 0.250 mol of carbon dioxide (CO2 ) gas at a temperature of 27.0∘ C. The cylinder is provided
with a frictionless piston, which maintains a constant pressure of 1.00 atm on the gas. The gas is heated until its
temperature increases to 127.0∘ C. Assume that the CO2 may be treated as an ideal gas.
Part A
How much work W is done by the gas in this process?
Express your answer in joules.
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Hint 1. Definition of the work done by an ideal gas
The work done by an ideal gas when its volume is changed is given by the expression
V
W = ∫ Vi f p dV
.
where Vi is the initial volume, Vf is the final volume, and p is the pressure (which is constant in this
particular case).
Hint 2. Calculate the initial volume
Using the ideal gas law and the constants given, determine the initial volume Vi of the gas. Note that
1 atm = 101, 325 N/m2 .
Express your answer in cubic meters
Hint 1. The ideal gas law
Recall that the ideal gas law is given by
pV = nRT ,
where p is pressure, V is volume, n is the number of moles of gas, R = 8.3145 J/(mol ⋅ K) is
the gas constant, and T is the gas temperature in kelvins.
ANSWER:
Vi = 6.15×10−3 m3
Hint 3. Calculate the final volume
Using the ideal gas law and the constants given, determine the final volume Vf of the gas. Note that
1 atm = 101, 325 N/m2 .
Express your answer in cubic meters.
Hint 1. The ideal gas law
Recall that the ideal gas law is given by
pV = nRT ,
where p is pressure, V is volume, n is the number of moles of gas, R = 8.3145 J/(mol ⋅ K) is
the gas constant, and T is the gas temperature in kelvins.
ANSWER:
Vf = 8.21×10−3 m3
ANSWER:
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W = 208 J
Correct
Part B
What is the change in internal energy ΔU of the gas?
Express your answer in joules.
Hint 1. Dependence of internal energy on temperature
One of the unique features of an ideal gas is that its internal energy is independent of volume or
pressure. It depends only on the temperature of the gas. This can be seen if we look at the internal
energy Eint of one atom, which, from equipartition, has internal energy given by Eint = (3/2)kB T ,
where kB = 1.38 × 10 −23 J/K is the Boltzmann constant and T is the temperature in kelvins.
The 3/2 comes from the three translational degrees of freedom (x, y, and z). If the gas consists of
diatomic molecules there can also be two independent degrees of rotational freedom
(Eint = (5/2)kB T ) and, if the gas is polyatomic, there can also be two independent degrees of
vibrational freedom from bending (Eint = (7/2)kB T ). At higher temperatures both the diatomic and
polyatomic gases can have additional modes due to longitudinal vibrations. (At low temperatures there is
not enough energy to excite these modes.)
Hint 2. The molar heat capacities of CO2
For CO2 (which is a polyatomic gas) at low pressure the molar heat capacity at constant volume is
CV = 28.46 J/(mol ⋅ K).
The molar heat capacity at constant pressure is Cp = 36.94 J/(mol ⋅ K).
Hint 3. Internal energy and heat capacity
The relation between the internal energy and the heat capacity of an ideal gas is given by the equation
dU = nCV dT , where dU is the differential energy, n is the number of moles, CV is the heat capacity
at constant volume, and dT is the differential temperature.
The the heat capacity at constant volume is used instead of that at constant pressure because of the first
law of thermodynamics, which states that dU = dQ − dW .
The heat capacity basically means that the heat goes as dQ = nCV dT (constant volume) or
dQ = nCp dT (constant pressure). If the system is at constant volume then dW = 0 and thus
dU = dQ = nCV dT . Now, since the internal energy for any process depends only on temperature,
then its change in one type of process (constant volume) must be the same as that in another type of
process (constant pressure).
ANSWER:
ΔU = 712 J
Correct
Part C
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How much heat Q was supplied to the gas?
Express your answer in joules.
Hint 1. The first law of thermodynamics
The first law of thermodynamics states that dU = dQ − dW . In the case we are dealing with, it can
also be written as ΔU = Q − W , where ΔU is the total change in internal energy, Q is the total
amount of heat put in, and W is the total amount of work that the gas does on its surroundings. Since we
know W from Part A and ΔU from Part B, it should be relatively straightforward to find Q.
ANSWER:
Q = 924 J
Correct
Part D
How much work W would have been done if the pressure had been 0.50 atm?
Express your answer in joules.
Hint 1. How work depends on pressure
Recall that this gas obeys the ideal gas law,
pV = nRT ,
and that the equation for work for a system under constant pressure is given by
V
W = ∫ Vi f p dV = p(V f − V i ).
As a result, one can eliminate the dependence on pressure of the amount of work a gas does, as long as
the pressure is kept constant.
ANSWER:
W = 208 J
Correct
Exercise 19.18
During an isothermal compression of an ideal gas, 335 J of heat must be removed from the gas to maintain constant
temperature.
Part A
How much work is done by the gas during the process?
ANSWER:
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W = -335 J
Correct
Exercise 19.20
A cylinder contains 0.0100 mol of helium at 27.0 ∘ C.
Part A
How much heat is needed to raise the temperature to 67.0 ∘ C while keeping the volume constant?
ANSWER:
Q = 4.99 J
Correct
Part B
If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from
27.0 ∘ C to 67.0 ∘ C?
ANSWER:
Qp = 8.31 J
Correct
Part C
What accounts for the difference between your answers to parts A and B? In which case is more heat required?
What becomes of the additional heat?
ANSWER:
3515 Character(s) remaining
In Part A there is no work done by the gas as there is
constant volume so Q is just delta U. While in Part B the
volume does change so there is work done. Q is a sum of
work and delta U. So more heat is required in Part B and
Part D
If the gas is ideal, what is the change in its internal energy in part A?
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ANSWER:
ΔUA = 4.99 J
Correct
Part E
If the gas is ideal, what is the change in its internal energy in part B?
ANSWER:
ΔUB = 4.99 J
Correct
Exercise 19.23
Heat Q flows into a monatomic ideal gas, and the volume increases while the pressure is kept constant.
Part A
What fraction of the heat energy is used to do the expansion work of the gas?
Express your answer using two significant figures.
ANSWER:
W
= 0.400
Q
Correct
Relationships between Molar Heat Capacities
The amount of heat needed to raise the temperature of 1 mole of a substance by one Celsius degree (or,
equivalently, one kelvin) is called the molar heat capacity of the system, denoted by the letter C . If a small amount of
heat dQ is put into n moles of a substance, and the resulting change in temperature for the system is dT , then
1 dQ
n dT .
This is the definition of molar heat capacity--the amount of heat Q added per infinitesimal change in T , per mole.
C=
A heated gas tends to expand, and the heat capacity depends on whether the gas is held at constant volume or
allowed to expand so that it remains at constant pressure. The molar heat capacities are denoted Cv and Cp for
constant volume and constant pressure, respectively.
When a gas is held at constant volume and heated, dV = 0, so the work, p dV done by the gas is zero. The first
law of thermodynamics, dQ = dU + dW , therefore implies that all of the heat must go into increasing the gas's
internal energy U .
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When a gas is held at constant pressure as it is heated, the same amount of heat is required to increase the gas's
internal energy. In addition, the gas expands and does work. The first law therefore demands addition of extra heat to
do the work. Consequently, for a gas held at constant pressure, it takes more heat for a given increase in
temperature than it does for a gas held at constant volume.
This problem concerns the important relationship between molar heat capacity at constant volume Cv and the molar
heat capacity at constant pressure Cp .
This discussion shows that the molar heat capacity of a gas depends on the circumstances under which the gas is
heated. The molar heat capacity for a gas held at constant pressure is greater than that for a gas held at constant
volume. In this problem, you will derive a single equation that relates Cp and Cv for an ideal gas.
Molar heat capacity at constant volume
Part A
Consider an ideal gas being heated at constant volume, and let Cv be the gas's molar heat capacity at constant
volume. If the gas's infinitesimal change in temperature is dT , find the infinitesimal change in internal energy dU
of n moles of gas.
Express the infinitesimal change in internal energy in terms of given quantities.
Hint 1. Find the heat added
Find an expression for dQv , the infinitesimal heat put into n moles of this gas.
Express your answer in terms of some or all of the quantities n, Cv , and dT .
ANSWER:
dQv = nCv dT
Hint 2. Effect of constant volume
The first law of thermodynamics states that dQ = dU + dW , where dQ is the heat put into the gas,
dU is the gas's change in internal energy, and W = p dV is the work done by the gas. For a gas held at
constant volume, p dV = 0 (since dV = 0 ).
ANSWER:
dU = nCv dT
Correct
Part B
Rewrite this equation, solving for Cv .
Give your answer in terms of dU and dT and n.
ANSWER:
Cv = n1 dU
dT
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Correct
Molar heat capacity at constant pressure
Part C
Now suppose the ideal gas is being heated while held at constant pressure p. The infinitesimal change in the
gas's volume is dV , while its change in temperature is dT . Find the gas's molar heat capacity at constant
pressure, Cp .
Express Cp in terms of some or all of the quantities Cv , p, dV , n, and dT .
Hint 1. Find the heat added
Let dQp be the infinitesimal heat added to the system. Use the first law of thermodynamics to express
dQp .
Express yout answer in terms of some or all of the quantities dU , p, and dV .
ANSWER:
dQp = dU + pdV
ANSWER:
Cp = pdV +Cv ndT
ndT
Correct
This result is quite general and holds for any type of gas.
Part D
Suppose there are n moles of the ideal gas. Simplify your equation for Cp using the ideal gas equation of state:
pV = nRT .
Express Cp in terms of some or all of the quantities Cv , n, and the gas constant R.
Hint 1. Find dV /dT from the ideal gas law
Using the ideal gas equation of state, pV = nRT , and the fact that p is constant, find an expression for
dV /dT .
Give your answer in terms of some or all of the quantities n, R, and p.
ANSWER:
dV
=
dT
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nR
p
ANSWER:
Cp = R + Cv
Correct
It is clear from this result that Cp is indeed greater than Cv .
Part E
The ratio of the specific heats Cp /Cv is usually denoted by the Greek letter γ. For an ideal gas, find γ.
Give your answer in terms of some or all of the quantities n, R, and Cv .
ANSWER:
γ=
Cv +R
Cv
Correct
Note that, for an ideal gas, γ > 1 . It is important to remember that the equation found here for γ is only
true for ideal gases. In general, for all materials that expand when heated, γ > 1 , but the specific form for
γ will always depend on the equation of state of the material.
Exercise 19.26
Propane gas (C3 H8 ) behaves like an ideal gas with γ = 1.127.
Part A
Determine the molar heat capacity at constant volume.
ANSWER:
cV = 65.5 J/mol ⋅ K
Correct
Part B
Determine the molar heat capacity at constant pressure.
ANSWER:
cp = 73.8 J/mol ⋅ K
Correct
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A Flexible Balloon
A flexible balloon contains 0.355 mol of an unknown polyatomic gas. Initially the balloon containing the gas has a
volume of 7000 cm3 and a temperature of 21.0 ∘ C. The gas first expands isobarically until the volume doubles.
Then it expands adiabatically until the temperature returns to its initial value. Assume that the gas may be treated as
an ideal gas with Cp = 33.26 J/mol ⋅ K and γ = 4/3.
Part A
What is the total heat Q supplied to the gas in the process?
Hint 1. Which parts of the process have heat flow?
The expansion of the balloon is split into two different parts. In the first part the gas is expanding while
the pressure is staying the same (an isobaric process). As a result, there must be heat flowing into the
balloon or (from the ideal gas law) the pressure would decrease as the balloon expanded. The second
part of the expansion is adiabatic, which means no heat flows into or out of the balloon. As a result, we
only have to worry about the isobaric expansion to determine the total heat supplied to the gas.
Hint 2. Relation of heat capacity and heat
The heat Q supplied to a gas system under constant pressure is Q = nCp ΔT , where n is the number
of moles of gas, Cp is the heat capacity at constant pressure, and ΔT is the change in gas temperature
in kelvins.
Hint 3. Find the change in temperature
Calculate the change in temperature ΔT of the gas as it expands isobarically until its volume doubles.
Express your answer in kelvins.
Hint 1. The ideal gas law
Recall that the ideal gas law is pV = nRT , where p is the gas pressure, V is the volume the
gas occupies, n is the number of moles of gas, R = 8.3145 J/(mol ⋅ K) is the gas constant,
and T is the gas temperature in kelvins. To determine ΔT , use the fact that Vf = 2Vi where Vf
is the final volume and Vi is the initial volume in the isobaric process.
ANSWER:
ΔT = 294 K
Correct
ANSWER:
Q = 3470 J
Correct
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Part B
What is the total change in the internal energy ΔU of the gas?
Hint 1. Dependence of internal energy on temperature
Recall that the change in internal energy of an ideal gas depends only on its change in temperature and
is independent of changes in pressure or volume. It follows the equation ΔU = nCV ΔT .
ANSWER:
ΔU = 0 J
Correct
Part C
What is the total work W done by the gas?
Hint 1. How to approach the problem
Recall that the first law of thermodynamics states that ΔU = Q − W , where ΔU is the change in
internal energy, Q is the net heat flow, and W is the net work done by the system on the outside. In Part
A you found the net heat flow and in Part B you found the change in internal energy. Therefore, you can
solve for W in the first law of thermodynamics and then plug in the values from the first two parts.
ANSWER:
W = 3470 J
Correct
Part D
What is the final volume V ?
Hint 1. Relation between volume and temperature
Recall that, in an adiabatic process with finite changes in volume and temperature, T V γ−1 is a constant.
As a result, T1 V1 γ−1 = T2 V2 γ−1 .
ANSWER:
V = 0.112 m3
Correct
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Fast vs. Slow Tire Pumping
Imagine the following design for a simple tire pump. The pump is filled with a volume Vi of air at atmospheric
pressure pa and ambient temperature Ta . When you push the pump handle, the air is compressed to a new
(smaller) volume Vf , raising its pressure. A valve is then opened, allowing air to flow from the pump into the tire until
the remaining air in the pump reaches the pressure of the air in the tire, pt .
In this problem, you will consider whether you can get more air into the tire per pump cycle by pushing the pump
handle quickly or slowly.
We will make the following simplifying assumptions:
The pressure of the air in the tire, pt , does not change significantly as air flows into the tire from the
pump.
The temperature of the air in the pump does not change significantly while air is flowing into the tire
(i.e., this is an isothermal process).
The air is mainly composed of diatomic molecules with γ = 1.40.
Part A
First imagine that you push the pump handle quickly, so that the compression of air in the pump occurs
adiabatically. Find the absolute temperature Tf of the air inside the pump after a rapid compression from volume
Vi to volume Vf , assuming an ambient temperature of Ta .
Express your answer in terms of Ta , Vi , Vf , and γ
Hint 1. Properties of an adiabatic process
For an adiabatic process the product pV γ is constant. Using the ideal gas law, you can derive the
equivalent condition that the product T V γ−1 is constant throughout the adiabatic process.
ANSWER:
Tf = T a ( Vi )
Vf
γ−1
Correct
Part B
Once the tire and pump pressures have equilibrated at pt , what fraction f of the gas particles initially in the
pump will have ended up in the tire?
Express the fraction in terms of pt , pa , Vf , Vi , and γ. The temperatures Ta and Tf should not appear in
your answer.
Hint 1. How to approach the problem
Find an expression for the number of particles in the pump initially ( Ni ) and another for the number in
the pump after the pump and tire have come into pressure equilibrium ( Nf ). Use these to determine the
fraction of the particles in the pump that are transferred into the tire:
f=
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N i −N f
Ni
=1−
Nf
.
Ni
Hint 2. Find an expression for the number of particles in a gas
Using the ideal gas law, find an expression for the number of particles N in a gas at pressure p, volume
V , and temperature T .
Use kB for Boltzmann's constant.
ANSWER:
N=
pV
T kB
Correct
Hint 3. Find Nf /Ni
What is the ratio of the number of particles in the pump after it comes into pressure equilibrium with the
tire to the number of gas particles initially in the pump (before pushing the handle)?
Express the ratio in terms of pt , pa , Vf , Vi , and γ. The temperatures Ta and Tf should not appear
in your answer.
Hint 1. Find the initial number of particles in the pump
How many gas particles Ni were in the pump initially?
Express your answer in terms of quantities given in the problem introduction and
Boltzmann's constant kB .
ANSWER:
Ni =
pa Vi
kB Ta
Hint 2. Find the final number of particles in the pump
How many gas particles Nf are in the pump after the pump and tire have equilibrated?
Express your answer in terms of quantities given in the problem introduction and
Boltzmann's constant kB . Tf should not appear in your answer.
Hint 1. Some helpful quantities
Recall that the final pressure is pt and the temperature continues to be T a (
ANSWER:
ptVf
Nf =
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V
kB Ta ( V i
f
(γ−1)
)
Vi
Vf
)
γ−1
.
ANSWER:
Nf
=
Ni
pt
Vf
V
pa Vi ( V i
f
γ−1
)
Correct
ANSWER:
f=
1−
ptVf
V
pa Vi ( V i
f
γ−1
)
Correct
Part C
Now imagine that you push the pump handle slowly, so that the compression of air in the pump occurs
isothermally. Find the absolute temperature Tf of the air inside the pump assuming an ambient temperature of
Ta .
Hint 1. Properties of an isothermal process
Isothermal means "at constant temperature."
ANSWER:
Tf = Ta
Correct
Part D
Once the tire and pump pressures have equilibrated at pt , what fraction f of the gas particles initially in the
pump will have ended up in the tire?
Express the fraction in terms of pt , pa , Vf , Vi , and γ. The temperatures Ta and Tf should not appear in
your answer.
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Hint 1. How to approach the problem
Find an expression for the number of particles in the pump initially ( Ni ) and another for the number in
the pump after the pump and tire have come into pressure equilibrium ( Nf ). Use these to determine the
fraction of the particles in the pump that are transferred into the tire:
f=
N i −N f
Ni
=1−
Nf
.
Ni
Hint 2. Find an expression for the number of particles in a gas
Using the ideal gas law, find an expression for the number of particles N in a gas at pressure p, volume
V , and temperature T .
Use kB for Boltzmann's constant.
ANSWER:
N=
pV
kB T
Hint 3. Find Nf /Ni
What is the ratio of the number of particles in the pump after it comes into pressure equilibrium with the
tire to the number of gas particles initially in the pump (before pushing the handle)?
Express the ratio in terms of pt , pa , Vf , Vi , and γ. The temperatures Ta and Tf should not appear
in your answer.
Hint 1. Find the initial number of particles in the pump
How many gas particles Ni were in the pump initially?
Express your answer in terms of quantities given in the problem introduction and
Boltzmann's constant kB .
ANSWER:
Ni =
pa Vi
kB Ta
Hint 2. Find the final number of particles in the pump
How many gas particles Nf are in the pump after the pump and tire have equilibrated?
Express your answer in terms of quantities given in the problem introduction and
Boltzmann's constant kB . Tf should not appear in your answer.
ANSWER:
Nf =
ANSWER:
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ptVf
kB Ta
Nf
=
Ni
pt Vf
pa Vi
ANSWER:
f = 1−
ptVf
pa Vi
Correct
Part E
Assume that pt = 3pa and Vf = Vi /6. Which method, fast pumping (adiabatic process) or slow pumping
(isothermal process) will put a larger amount of air into the tire per pump cycle?
Hint 1. Adiabatic process
What is the numerical value of f for the fast (adiabatic) process?
Express your answer numerically, to two significant figures.
ANSWER:
f = 0.75
Hint 2. Isothermal process
What is the numerical value of f for the slow (isothermal) process?
Express your answer numerically, to two significant figures.
ANSWER:
f = 0.50
ANSWER:
fast pumping
slow pumping
Correct
So when you pump quickly, not only do you get more pump cycles per unit time, but you also put more air
in the tire per cycle (at least according to this simplified model). Of course you have to pump not only faster
but also harder, since the average pressure you will pump against will be higher. In fact this higher pressure
is the main reason that more particles are transferred to the tire each time.
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