PHY 171
Homework 8 solutions
(Due by beginning of class on Wednesday, February 29, 2012)
Submit neat work, with answers or solutions clearly marked by the question number.
Unstapled, untidy work will be charged a handling fee of 20% penalty. Writing only
an answer without showing the steps used to get to that answer will fetch very few
points, even if the answer is correct. Late homework will not be accepted.
1. When 20.0 J of thermal energy was added to an ideal gas, the volume of the gas changed
from 50.0 cm3 to 100.0 cm3 , while the pressure remained constant at 1.00 atm.
(a) By how much did the internal energy of the gas change?
Solution:
We are given that Q = 20.0 J, and P remains constant at 1 atm.
Since P remains constant, we can take it outside the integral for W , to obtain
Z VB
Z VB
W =
P dV = P
dV = P VB − VA = P ∆V
VA
VA
The change in internal energy is then given by
h
i
∆Eint = Q − W = Q − P ∆V = 20.0 J − (1 × 105 Pa) (100 − 50) × 10−6 m3 = 15.0 J
(b) If the quantity of gas present is 2.00 × 10−3 mol, find the molar specific heat of the gas at
constant pressure CP , and the molar specific heat at constant volume CV .
Solution:
To find the molar specific heat, we will need the change in temperature ∆T ; we can find
that from the ideal gas law P V = nRT , from which we obtain ∆(P V ) = ∆(nRT ), which
gives P (∆V ) = nR(∆T ), because the pressure P is constant (and so are n and R). So
CP =
Q
Q
=
n(∆T ) Z
n(P ∆V /Z
nR)
Therefore
CP =
(20.0 J)[8.31 J/(mol.K)]
QR
=
= 33.2 J/(mol.K)
P ∆V
(1 × 105 Pa)(50 × 10−6 m3 )
and
CV = CP − R = 33.2 − 8.31 = 24.9 J/(mol.K)
PHY 171 (Winter 2012)
Homework 8 solutions
2. We give 70.0 J of heat energy to a diatomic gas, which then expands at constant pressure.
The gas molecules rotate but do not vibrate. By how much does the internal energy of the
gas increase?
Solution: Given that the diatomic gas molecules rotate but do not vibrate, CV =
5
2
R.
The molar specific heat at constant pressure is QP = nCP (∆T ), so
7
5
R + R (∆T ) = n
R (∆T )
70 J = n CV + R (∆T ) = n
2
2
From the above, we get
70
∆T =
n
Therefore
∆Eint
2
7R
2A
70
5
5
n@
R
= nR(∆T ) = Z
= 50 J
@
2A
n 7@
2
R
Z
@
3. 14 g of nitrogen gas (N2 ) at STP (i.e., T = 0◦ C, P = 1 atm) are adiabatically compressed
to a pressure of 20 atm.
(a) What is the final temperature of the gas?
Solution: The first step is to write down which of P, V, T we know for each of the two
states.
P1 = 1 × 105 Pa,
T1 = 273 K,
V1 =
and
P2 = 20 × 105 Pa,
T2 =
,
V2 =
The long method of solving this part of the problem would then be to:
• Use the ideal gas law P V = nRT to find V1 .
• Use the relation for adiabatic processes, P V γ = constant, to find V2 .
• Use the ideal gas law to find T2 .
For those of you comfortable with symbolic manipulation, though, a much shorter method
would be to use the two relations for adiabatic processes, P V γ = constant, and T V γ−1 =
constant, to write a relation involving only P1 , P2 , T1 , and T2 .
I will solve using the long method to enable all students to follow along. The solution
starts on the next page.
Page 2 of 6
PHY 171 (Winter 2012)
Homework 8 solutions
Problem 3 (a) — continued from previous page
For our calculations, we will need the number of moles, n =
14 g
= 0.5 mol.
28 g/mol
7
5
R+R
R
CP
7
CV + R
2
2
Also, since N2 is diatomic, CV = R, and γ =
= 5 = = 1.4
=
= 5
CV
CV
5
R
R
2
2
5
2
In all calculations below, I will retain several more digits than significant to avoid rounding
off errors. The significant digits will be implemented only in the final answer.
First, use the ideal gas law to solve for V1 :
V1 =
nRT1
(0.5 mol)(8.31 J/mol.K)(273 K)
=
= 1.1343 × 10−2 m3
P1
1 × 105 Pa
Then use the relation for adiabatic processes, P V γ = constant, to solve for V2 .
So, P1 V1γ = P2 V2γ gives
V2γ
P1 V1γ
(1 × 105 Pa)(1.1343 × 10−2 m3 )1.4
=
=
= 9.4534 × 10−5
P2
20 × 105 Pa
and therefore
1/1.4
V2 = 9.4534 × 10−5
= 1.3348 × 10−3 m3
Finally, use the ideal gas law to solve for T2 :
T2 =
(20 × 105 Pa)(1.3348 × 10−3 m3 )
P2 V2
=
= 642.5 K
nR
(0.5 mol)(8.31 J/mol.K)
Therefore, the final temperature of the gas is 643 K
(b) What is the work done on the gas?
R
Solution: While it is possible in principle to integrate W = P dV to find the work done
by the gas, this can turn out to be uncomfortably long for the case of adiabatic processes.
There is an easier way, since Q = 0 for an adiabatic process.
So, applying the 1st law of thermodynamics, ∆Eint = Q − W , with Q = 0 for an adiabatic
process, we get
5
5
W = −∆Eint = − nR ∆T = − nR T2 − T1
2
2
or
5
W = − 0.5 mol 8.31 J/mol.K 642.5 − 273 = −3838.15 ≈ −3850 J
2
The minus sign indicates that the work is done on the gas, which we would expect since the
gas is being compressed.
Page 3 of 6
PHY 171 (Winter 2012)
Homework 8 solutions
Problem 3 — continued from previous page
(c) What is the heat input to the gas?
Solution:
Since this is an adiabatic process, there is no heat flow into or out of the system, so the heat
input to the gas is zero.
(d) Show the process on a P − V diagram. You may sketch by hand, but you must show the
proper scales on both axes for full credit.
Solution: The graph is shown below.
Page 4 of 6
PHY 171 (Winter 2012)
Homework 8 solutions
4. A copper rod and an iron rod with exactly the same dimensions (i.e., same area of cross
section and same length) are welded together end to end. The outside end of the copper rod
is held at 100◦ C, and the outside end of the iron rod is held at 0◦ C. What is the temperature
at the midpoint where the rods are joined together?
Solution: The geometry of the problem is shown in the figure below.
The key to solving this problem is to understand
that the rate of heat (in J/s) flowing into the CuFe junction on the copper side must be equal to
the rate of heat flowing out of the Cu-Fe junction
on the iron side. In other words, we need to assume that there is no heat sink at the junction,
in order to be able to solve this problem.
TH − TC
∆Q
= kA
The rate of heat conducted through a rod is given by
∆t
L
where k is the conductivity of the rod, A is the area of cross section, TH is the temperature
at the hot end of the rod, TC is the temperature at the cold end of the rod, and L is the
length of the rod.
∆Q
∆Q
=
As discussed above, to solve this problem we need to set
∆t Cu
∆t Fe
and so
kCu A
373 K − Tj
L
= kFe A
Tj − 273 K
L
where we have kept A and L the same for copper and iron because it is stated in the problem
that they have exactly the same dimensions. Also, the left end of the copper rod is at 100◦ C
(see figure), so we have set TH = 373 K for the copper rod, and the right end of the iron
rod is at 0◦ C (see figure), so we have set TC = 373 K for the iron rod. Also, we will set the
temperature of the junction (for which we are solving) as Tj (see figure), and so we have set
both TC for the copper rod and TH for the iron rod as Tj .
After canceling terms, we get
kCu 373 − Tj = kFe Tj − 273
From Table 17.5 (page 530) in your text, kCu = 400 W/m.K, and kFe = 80 W/m.K, so
400 373 − Tj = 80 Tj − 273
Rearranging terms
400 373 + 80 273 = 400 + 80 Tj
Solving for Tj , we find the temperature at the junction to be 356 K or 83◦C
Page 5 of 6
PHY 171 (Winter 2012)
Homework 8 solutions
5. A sphere of radius 0.500 m, temperature 27.0◦ C, and emissivity 0.850 is located in an environment of temperature 77.0◦ C.
Solution: Given R = 0.500 m, T = 27◦ C = 300 K, ǫ = 0.850, Tenv = 77◦ C = 350 K
(a) At what rate does the sphere emit thermal radiation?
Solution: Rate at which the sphere emits thermal radiation is given by
h
4
2 i
∆Q
= ǫσAT 4 = 0.850 5.6703 × 10−8 W/m2 4π 0.500 m
300 K
∆t emit
After carrying out the calculation, we obtain
∆Q
= 1.23 × 103 watts
∆t emit
(b) At what rate does the sphere absorb thermal radiation?
Solution: Rate at which the sphere absorbs thermal radiation is obtained from
4
∆Q
4
= ǫσATenv = ǫσA 350 K = 2.27 × 103 watts
∆t absorb
(c) What is the sphere’s net rate of energy exchange?
Solution: Net rate of energy exchange is
∆Q
∆Q
∆Q
=
−
= 1.05 × 103 watts
∆t net
∆t absorb
∆t emit
where I have put the absorbed rate first in order to get the answer with a positive sign; this
also means that the sphere is absorbing energy, and this makes sense since the temperature
of the external environment is greater than that of the sphere.
Page 6 of 6