Physics 213 — Problem Set 13 — Solutions
Spring 1998
1. Reading Assignment
Serway Sections 20.4–6, 21.1–3, 22.1–6.
2. Serway 20.28
One mole of an ideal gas is heated slowly so that it goes from the state (P0 , V0 ) to the state (3P0 , 3V0 ).
This change occurs in such a way that the gas pressure is directly proportional to the volume. (a) How
much work is done in the process? (b) How is the temperature of the gas related to its volume during
this process?
SOLUTION:
a)Since the pressure remains proportional to the volume, P =
by the gas is :
Z
W =
b)From the Ideal Gas Law, T =
P0
P dV =
V0
PV
NkB
=
Z
3V0
V0
P0
V dV =
V0
P0
V0 V
throughout the process, the work done
3V0
1 2 V
2
= 4P0 V0 .
V0
P0
2
NkB V0 V .
3. Serway 20.32
An ideal gas undergoes a thermodynamic process that consists of two isobaric and two isothermal steps
as shown in Figure P20.32 of your text. Show that the net work during the four steps is
Wnet = P1 (V2 − V1 ) ln(P2 /P1 ).
(3.1)
SOLUTION: Since the paths from A to B and from C to D are isothermal, we can easily determine the
volume of the gas at points B and C using the ideal gas law. At constant temperatures, T1 and T2 , we have
P1 V1 = P2 VB ,
(3.2)
P1 V2 = P2 VC .
(3.3)
The work along an isotherm from Vi to Vf is given by
Z
Vf
P dV = nRT ln(Vf /Vi ) = Pi Vi ln(Vf /Vi ) = Pf Vf ln(Vf /Vi ).
Vi
Along the isobars we can write P (Vf − Vi ) for the work done by the gas. Altogether
W = WAB + WBC + WCD + WDA
= P1 V1 ln(VB /V1 ) + P2 (VC − VB ) + P1 V2 ln(V2 /VC ) + P1 (V1 − V2 ).
(3.4)
From eqs. (3.2) and (3.3) we see that the second and fourth terms in (3.4) cancel. Also, (3.2) and (3.3)
imply V2 /VC = P2 /P1 and VB /V1 = P1 /P2 . Inserting in (3.4) we find
W = P1 V1 ln(P1 /P2 ) + P1 V2 ln(P2 /P1 ) = P1 (V2 − V1 ) ln(P2 /P1 ).
4. Serway 20.36
An ideal gas undergoes the cyclic process shown in Figure P20.36 from A to B to C and back to A.
(a) Sketch a P V diagram for this cycle, and identify the steps during which heat is absorbed and those
during which heat is evolved. (b) What is the overall result of the cycle in terms of U , Q, and W ?
SOLUTION:
a)On a plot of P vs. V , AB goes straight up, BC goes down and to the right, and CA goes straight left.
• Along AB, V is held constant while P and T increase.
1
– W = 0 since V is constant,
– ∆U > 0 since T increases,
– Q = ∆U + W > 0 : heat is absorbed by the gas.
• Along BC, T is held constant while P decreases and V increases.
– W > 0 since V increases,
– ∆U = 0 since T is constant,
– Q = ∆U + W > 0 : heat is absorbed by the gas.
• Along CA, P is held constant while V and T decrease.
– W < 0 since V decreases,
– ∆U < 0 since T decreases,
– Q = ∆U + W < 0 : heat is released by the gas.
b)Because the gas goes through a complete cycle, ∆U = 0. The cycle encloses a positive area on the P V
diagram (it goes clockwise), so W > 0. Then Q = ∆U + W > 0 as well. The result is that the gas does
work and absorbs heat each cycle.
5. Serway 21.6
In a 30-s interval, 500 hailstones strike a glass window of area 0.60 m2 at an angle of 45◦ to the window.
Each hailstone has a mass of 5.0 g and a speed of 8.0 m/s. If the collisions are elastic, find the average
force and pressure on the window.
SOLUTION: When the hailstone collides with the glass, the component of its momentum perpendicular
to the window goes from mv cos(45◦ ) to −mv cos(45◦ ), so by conservation of momentum it must give up
∆p = 2mv cos(45◦ ) to the window. Since there are 500 such collisions in 30 s, the average force on the
window is :
√
(500)(2)(.005 kg)(8.0 m/s)( 2/2)
Σ∆p
=
≈ 0.94 N.
F =
∆t
30 s
2
0.94 N
The corresponding pressure is : P = F
A = 0.60 m2 ≈ 1.6 N/m .
6. Serway 21.8
Calculate the rms speed of an H2 molecule at 250◦ C.
SOLUTION: Equation 21.7 gives vrms =
r
:
vrms =
p 3RT
M
. Using 250 ◦ C = 523 K and the molar mass of H2 , we find
(3)(8.3 J/mol K)(523 K)
≈ 2550 m/s.
(2.0 × 10−3 kg/mol)
7. Serway 21.26A
An ideal gas (γ = 1.40) confined to a cylinder is put through a closed cycle. Initially the gas is at P0 , V0 ,
and T0 . First its pressure is tripled under constant volume. It then expands adiabatically to its original
pressure and finally is compressed isobarically to its orignal volume. (a) Draw a P V diagram of this
cycle. (b) Determine the volume at the end of the adiabatic expansion. Find (c) the temperature of the
gas at the start of the adiabatic expansion and (d) the temperature at the end of this process. (e) What
was the net work done for this cycle?
SOLUTION:
a)
b)During an adiabatic expansion, P V γ = const. Specifically, for the start and end of this adiabatic expansion
3P0 V0γ = P0 Vxγ ,
Vx = 31/γ V0 .
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c)Compare points 0 and 1. When the pressure of the ideal gas was tripled at constant volume, its temperature
was tripled (ideal gas law, constant volume gas thermometer). The temperature at point 1, which is also at
the start of the adiabatic expansion, is T1 = 3T0 .
d)Compare points 0 and x. We have Px = P0 , and from the ideal gas law
P0 Vx
P0 V0
=
Tx
T0
Tx =
Vx
T0 = 3(1/γ) T0 .
V0
e)To find the net work for the cycle, add up the work done by the gas along each leg of the path. The
constant volume path from Point 0 to Point 1 contributes zero. The other two terms are Wadiab from Point
1 to Point x, and Wisobar from Point x to Point 0.
Wisobar = P0 (V0 − Vx ) = P0 V0 (1 − 31/γ ).
To find Wadiab , remember that P V γ = const. = 3P0 V0γ , so during the adiabatic expansion the formula for
the curve in the P V plane is
3P0 V0γ
P (V ) =
.
Vγ
Then,
Z
Wadiab
Z
Vx
31/γ V0
P dV = 3P0 V0γ
=
V0
V0
dV
Vγ
31/γ V0
1−γ V
P0 V0
= 3P0 V0γ ·
=
(3 − 31/γ ).
1−γ
γ −1
V0
And for the sum of Wisobar and Wadiab
Wnet = P0 V0 (
3 − 31/γ
+ 1 − 31/γ ) = 0.829P0 V0 .
γ−1
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Figure 1: In the diagram and following calculations the subscript 0 is used for the point at P0 , V0 , 1 for the
point at 3P0 , V0 , and x for the point at the end of the adiabatic expansion.
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