The document provides examples and problems related to thermodynamics. Example 1 involves calculating the final temperature of a system consisting of a brass block and ice water. Example 2 calculates the root mean square speeds of nitrogen and oxygen molecules in air. Example 3 determines the new temperature of an ideal gas that is compressed.

1. Review , Examples
and
Problems
Thermodynamics
221PHYS

2. HeatHeat

3. Example-1: An m1 = 485-gram brass block sits in boiling water (T1 = 100° C). It is
taken out of the boiling water and placed in a cup containing m2 = 485 grams of
ice water (T2 = 0° C). What is the final temperature, TF, of the system (i.e.,
when the two objects have the same T)? (cbrass = 380 J/kg.
K; cwater = 4184
J/kg.
K)
a. TF < 50° C b. TF = 50° C c. TF > 50° C
Solution:
Heat flows from the brass to the water. No work is done,
and we assume that no energy is lost to the environment.
Remember: Q = C∆T = mc∆T
Brass (heat flows out): Q1 = ∆U1 = m1c1(TF-T1)
Water (heat flows in): Q2 = ∆U2 = m2c2(TF-T2)
Energy is conserved: Q1 + Q2 = 0
Solve for TF: TF = (m1c1T1+m2c2T2) / (m1c1+m2c2)
= (c1T1+ c2T2) / (c1+ c2) = 8.3° C
We measured TF = _____° C.
QT1
m1
T2
m2

4. Kinetic theoryKinetic theory
of the idealof the ideal
gasgas

5. Example2 : The Speed of Molecules in Air
Air is primarily a mixture of nitrogen N2 molecules (molecular mass 28.0 u) and oxygen O2
molecules (molecular mass 32.0 u). Assume that each behaves as an ideal gas and determine
the rms speeds of the nitrogen and oxygen molecules when the temperature of the air is 293K.
2 31
2 2
3
rms rms
kT
mv kT v
m
= ⇒ =Q
Solution
For Nitrogen…
kg1065.4g1065.4
mol106.022
molg0.28 2623
123
−−
−
×=×=
×
=m
( )( ) sm511
kg1065.4
K293KJ1038.133
26
23
=
×
×
== −
−
m
kT
vrms

6. Example 3:
A container of an ideal gas has a
moveable top. The top has an area of
0.01 m2
and is 50 cm above the bottom of
the cylinder. A mass of 200 kg is placed
on the container, which compresses the
gas by 20 cm. The gas in the container is
initially at atmospheric pressure (1.01 x
105
Pa) and 20o
C. What is the new
temperature of the gas?
nRTPV = nR
T
PV
=
2
22
1
11
T
VP
T
VP
=
Solution

7. 1 2
1 1 2 2
V Ah and A A
V Ah and V Ah
= =
⇒ = =
Q
15.27311 += cTT K15.293=
1 0 101P P kPa= =
A
F
PP += 02
A
mg
PP += 02 kPa2.297=
The following relations have been used
11
122
2
VP
TVP
T = K6.517=
10
122
AhP
TAhP
=
10
122
hP
ThP
= CT 
2442 =
The new temperature

8. ThermodynamicThermodynamic
lawslaws

9. Ex.4 : The process shown on the Pressure-Volume diagram is an
(A) adiabatic expansion.
(B) isothermal expansion.
(C) isometric expansion.
(D) isobaric expansion.
P
V
o

10. Ex.5: In an isochoric process, there is no change in
(A) pressure.
(B) temperature.
(C) volume.
(D) internal energy.

11. Ex-6: The process shown on the Temperature-Volume graph is an
(A) adiabatic compression.
(B) isothermal compression.
(C) isochoric compression.
(D) isobaric compression.
T
V
o

12. Ex-7: When the first law of thermodynamics, Q = ΔU +
W, is applied to an ideal gas that is taken through an
isothermal process,
(A) ΔU = 0
(B) W = 0
(C) Q = 0
(D) none of the above

13. Ex-8: An ideal gas is compressed to one-half its original
volume during an isothermal process. The final pressure
of the gas
(A) increases to twice its original value.
(B) increases to less than twice its original value.
(C) increases to more than twice its original value.
(D) does not change.

14. Ex-10: When the first law of thermodynamics, Q = ΔU
+ W, is applied to an ideal gas that is taken through an
adiabatic process,
(A) ΔU = 0.
(B) W = 0.
(C) Q = 0.
(D) none of the above

15. Ex-11:
A gas is taken through the cycle illustrated here. During
one cycle, how much work is done by an engine operating
on this cycle?
(A) PV
(B) 2PV
(C) 3PV
(D) 4PV
P
2P
V 2V 3V 4V

16. Ex-12: An ideal gas initially
has pressure Po, at volume Vo
and absolute temperature To.
It then undergoes the following
series of processes:
Po
2Po
Vo 2Vo 3Vo
I. Heated, at constant volume to pressure 2Po
II. Heated, at constant pressure to pressure 3Vo
III. Cooled, at constant volume to pressure Po
IV. Cooled, at constant pressure to volume Vo
I
II
III
IV

17. Po
2Po
Vo 2Vo 3Vo
I
II
III
IV
•
••
•
2To
Find the temperature at each end point in terms of To
nRTPV =
nR
VP
T oo
o =
To
6To
3To

18. Po
2Po
Vo 2Vo 3Vo
I
II
III
IV
•
••
•Find the net work done by the
gas in terms of Po and Vo
ooVP2W =
Net work equals net
area under curve

19. Po
2Po
Vo 2Vo 3Vo
I
II
III
IV
•
••
•
Find the net change in internal
energy in terms of Po and Vo
0U =∆

20. Ex-13: A sample of gas expands from 1.0 m3
to 4.0 m3
while its
pressure decreases from 40 Pa to 10 Pa. How much work is done
by the gas if its pressure changes with volume via each of the three
paths shown in the Figure below?
Path A: W = +120 J
Path B: W =+75 J
Path C: W =+30 J

21. Example-14Example-14
A cylinder of radius 5 cm is kept at pressure with a piston
of mass 75 kg.
a) What is the pressure inside the
cylinder?
b) If the gas expands such that the
cylinder rises 12.0 cm, what work was
done by the gas?
c) What amount of the work went into
changing the gravitational PE of the
piston?
d) Where did the rest of the work go?

22. Solution
Given: M =75, A = π× 0.052
, ∆x=0.12, Patm = 1.013x105
Pa
a) Find Pgas
atmgas P
A
Mg
P += = 1.950x105
Pa
b) Find Wgas
VPW ∆=
xAPW ∆= gas = 183.8 J
c) Find Wgravity
mghW = = 88.3 J
d) Where did the other work go? Compressing the outside air

23. Example-15 a) What amount of work is performed by the gas in the cycle
IAFI?
b) How much heat was inserted into the gas in the cycle
IAFI?
c) What amount of work is performed by the gas in the cycle
IBFI?
areaenclosed=W
WIAFI = 3Patm
= 3.04x105
J
WQU −=∆
∆U = 0
W = -3.04x105
J
Q = 3.04x105
J
V (m3
)

24. Example-Example-
1616
Consider a monotonic ideal gas which
expands according to the PV diagram.
a) What work was done by the gas from A to B?
b) What heat was added to the gas between
A and B?
c) What work was done by the gas from B to C?
d) What heat was added to the gas beween B and C?
e) What work was done by the gas from C to A?
f) What heat was added to the gas from C to A?
V (m3
)
P (kPa)
25
50
75
0.2 0.4 0.6
A
B
C

25. Solution
a) Find WAB
b) Find QAB
V (m3
)
P (kPa)
25
50
75
0.2 0.4 0.6
A
B
C
WAB = Area = 20,000 J
•First find UA and UB
Monotonic Gas:
3
2
3
2
U nRT
PV nRT
U PV
=
=
=
UA = 22,500 J, UB = 22,500 J, ∆U = 0
•Finally, solve for Q
U Q W∆ = − Q = 20,000 J

26. c) Find WBC
d) Find QBC
V (m3
)
P (kPa)
25
50
75
0.2 0.4 0.6
A
B
C
WBC = -Area = -10,000 J
•First find UB and UC
PVU
2
3
=
UB = 22,500 J, UC = 7,500 J, ∆U = -15,000
•Finally, solve for Q
WQU −=∆
Q = -25,000 J

27. e) Find WCA
f) Find QCA
V (m3
)
P (kPa)
25
50
75
0.2 0.4 0.6
A
B
C
WAB = Area = 0 J
•First find UC and UA
UC = 7,500 J, UA = 22,500 J, ∆U = 15,000
•Finally, solve for Q
Q = 15,000 J

28. g) Net work done by gas in the cycle
h) Amount of heat added to gas
WAB + WBC + WCA = 10,000 J
QAB + QBC + QCA = 10,000 J
This does NOT mean that the engine is 100% efficient!

29. Example-17
Imagine that we rapidly compress a sample of air whose initial pressure is 105
Pa
and temperature is 220
C (= 295 K) to a volume that is a quarter of its original volume
(e.g., pumping bike’s tire). What is its final temperature?
γγ
2211
222
111
VPVP
TNkVP
TNkVP
B
B
=
=
=
1
2
1
2
1
2
1
11
21
2
11
2
11
2
T
T
V
V
T
T
VP
TNk
V
VP
V
VP
P B =





⇒===
−
−
γ
γ
γ
γ
γ
constVTVT ==
−− 1
22
1
11
γγ
For adiabatic processes:
Rapid compression – approx. adiabatic, no time for the energy
exchange with the environment due to thermal conductivity
constTP =− γγ
/1
also
KKK
V
V
TT 51474.12954295 4.0
1
2
1
12 ≈×≈×=





=
−γ

30. Example-18: Finding the Work
An ideal gas with γ = 1.4 occupies 4.0 L at 300 K & 100 kPa pressure. It’s compressed
adiabatically to ¼ of original volume, then cooled at constant V back to 300 K, & finally
allowed to expand isothermally to its original V. How much work is done on the gas?
1
A A B B
AB
p V p V
W
γ
−
=
−
741 J= −
AB
(adiabatic):
0BCW =BC isochoric):
ln A
CA
C
V
W n R T
V
=CA
(isothermal):
( ) ( ) ( )1.4 1
100 4.0 1 4
1.4 1
kPa L −
−
=
−
A
B A
B
V
p p
V
γ
 
=  ÷
 
1
1
1
A A A
AB
B
p V V
W
V
γ
γ
−
  
 ÷= −  ÷ ÷−   
ln 4A Ap V= 555 J=
work done by gas: ABCA AB BC CAW W W W= + + 186 J= −

31. Example-19: A thermodynamic system undergoes a process in
which its internal energy decreases by 465 J. Over the same time
interval, 236 J of work is done on the system. Find the energy
transferred from it by heat.
465 236 701
U Q W
Q U W J J J
∆ = +
= ∆ − = − − = −
Note: Sign convention for Q :
Q>0  system gains heat from environment

32. Example-20:. Diesel Power
Fuel ignites in a diesel engine from the heat of compression (no spark plug needed).
Compression is fast enough to be adiabatic.
If the ignit temperature is 500°C, what compression ratio Vmax / Vmin is needed?
Air’s specific heat ratio is γ = 1.4, & before the compression the air is at 20 °C.
1
T V constγ −
=
( )1/ 1.4 1
273 500
273 20
K K
K K
−
 +
=  ÷
+ 
( )1/ 1
max min
min max
V T
V T
γ −
 
=  ÷
 
11=

33. Ex-21: Two cylinders at the same temperature contain the
same gas. If B has twice the volume and half the number of
moles as A, how does the pressure in B compare with the
pressure in A?
(A) PB = 1/2 PA
(B) PB = 2 PA
(C) PB = 1/4 PA
(D) PB = 4 PA
(E) PB = PA

34. Ex-22: A gas cylinder and piston are covered with heavy
insulation. The piston is pushed into the cylinder,
compressing the gas. In this process, the gas temperature
A. doesn’t change.
B. decreases.
C. increases.
D. there’s not sufficient information to tell.

35. Ex-23: During an isothermal process, 5.0 J of heat is removed
from an ideal gas. What is the change in internal energy?
A) zero
B) 2.5 J
C) 5.0 J
D) 10 J
The Laws of Thermodynamics

36. Ex-24: If the gas in a container absorbs 300 J of heat,
has 100 J of work done on it, and then does 200 J
of work on its surroundings, what is the increase in
the internal energy of the gas?
(A) 600 J
(B) 400 J
(C) 0 J
(D) 500 J
(E) 200 J