Q19._SecondLawOfThermodynamics

Q19. Second Law of Thermodynamics
1. An ideal gas, consisting of n moles, undergoes an irreversible
process in which the temperature has the same value at the
beginning and end.
If the volume changes from Vi to Vf ,
the change in entropy is given by:
1.
n R ( Vf – Vi )
2.
n R ln( Vf – Vi )
3.
n R ln( Vi / Vf )
4.
5.
n R ln( Vf / Vi )
none of the above (entropy can't be
calculated for an irreversible process)
Temperature has the same value at the beginning and end

Initial & final states can be related by an isothermal
reversible process.
Ideal gas

Isothermal

f
U  n RT
2
PV n RT
dU  dQ  dW  0
Vf
Vf
Vf
dQ
dW
P
nR
S  

  dV  
d V  nR ln
T
T
T
V
Vi
Vi
Vi
2. The temperature of n moles of a gas is increased from Ti to Tf at
constant pressure. If the molar specific heat at constant
pressure is Cp and is independent of temperature, then change in
the entropy of the gas is:
1.
n Cp ln( Tf / Ti )
2.
n Cp ln( Ti / Tf )
3.
n Cp ln( Tf – Ti )
4.
n Cp ln(1 – Ti / Tf )
5.
n Cp ( Tf – Ti )
dQ

S  
T
Tf

Ti
n CP dT
Tf
 n CP ln
T
Ti
3. Consider the following processes: The temperature of two
identical gases are increased from the same initial temperature
to the same final temperature.
Reversible processes are used.
For gas A the process is carried out at constant volume while
for gas B it is carried out at constant pressure.
The change
in entropy:
1.
is the same for A and B
2.
is greater for A
3.
is greater for B
4.
is greater for A only if the initial temperature is low
5.
is greater for A only if the initial temperature is high
dQ
S  

T
CP  CV

Tf

Ti
C dT
T
 S P   S V
4. A Carnot heat engine runs between a cold reservoir at
temperature TC and a hot reservoir at temperature TH.
You want to increase its efficiency.
Of the following, which
change results in the greatest increase in efficiency?
The value of T is the same for all changes.
1.
Raise the temperature of the hot reservoir by T
2.
Raise the temperature of the cold reservoir by T
3.
Lower the temperature of the hot reservoir by T
4.
Lower the temperature of the cold reservoir by T
5.
Raise the temperature of the hot reservoir by (1/2)T
and lower the temperature of the cold reservoir by (1/2)T
TC
e  1
TH

TC
1
e1  2 T 
T
TH
TH
TC
e3   2 T
TH
TC
1
e  2 TH 
TC
TH
TH
1
e2   T
TH
1
e4 
T
TH
1
1  TH  TC 
1  TC
1 
e5   2   T  
 T  T T
2
2  TH 
2  TH TH 
H
5. A perfectly reversible heat pump with a coefficient of
performance of 14 supplies energy to a building as heat to
maintain its temperature at 27°C.
If the pump motor does
work at the rate of 1 kW, at what rate does the pump supply
energy to the building?
1.
15 kW
2.
14 kW
3.
1 kW
4.
1.07 kW
5.
1.02 kW
QC
COP 
W

dQC
 14 1 kW   14 kW
dt
Energy pump rate into building :
dQH dQC

 W  14 1 kW   1 kW   15 kW
dt
dt
6. A Carnot heat engine and an irreversible heat engine both
operate between the same high temperature and low
temperature reservoirs.
They absorb the same heat from the
high temperature reservoir as heat.
The irreversible engine:
1.
does more work
2.
rejects more energy to the low temperature reservoir as heat
3.
has the greater efficiency
4.
5.
rejects less energy to the low temperature reservoir as heat
cannot absorb the same energy from the high temperature
reservoir as heat without violating the second law of
thermodynamics
eCarnot
W TH  TC


QH
TH
QH  QH
QC  QH  W
QC  QH  W 
 eirrev
W  W


QC  QC
W

QH
1 is false
2 is true
4 is false
2nd law has nothing to say about the magnitude of QH

5 is false
3 is false
7. Twenty-five identical molecules are in a box.
Microstates
are designated by identifying the molecules in the left and
right halves of the box.
10–23 J / K.
The Boltzmann constant is 1.38 
The entropy associated with the configuration
for which 22 molecules are in the left half and 3 molecules are
in the right half is on the order of:
1.
10–23 J/K
2.
10–20 J/K
3.
103 J/K
4.
1022 J/K
5.
10–10 J/K
25!
25  24  23


 2300
22!  3!
3 2
S  k ln   1.38  10 –23 J / K   ln 2300   1023
8. Let k be the Boltzmann constant.
If the thermodynamic
state of gas at temperature T changes isothermally and
reversibly to a state with three times the number of
microstates as initially, the energy input to gas as heat is :
1.
Q=0
2.
Q=3kT
3.
Q = –3k T
4.
k T ln 3
5.
– k T ln 3
isothermal
Q  T S  T k  ln  f  ln i   T k ln
f
i
 k T ln 3