Tutorial 2
By
Miss Anis Atikah Ahmad
Question 1
A heat engine uses reservoirs at 800°C and 0°C.
a) Calculate maximum possible efficiency
b) If qH is 1000 J, calculate the maximum -w and the
minimum value of -qC
Question 1
a) Calculate maximum possible efficiency
TH= 800°C = 1073.15K, TC =0°C =273.15K
 rev 
Work output per cycle
Energy input per cycle
273.15K
 1
 0.745
1073.15K
qC
TC
 1
 1
qH
TH
Question 1
b) If qH is 1000 J, calculate the maximum -w and the
minimum value of -qC
 rev 
Work output per cycle
Energy input per cycle
 1
w
0.745 
1000 J
 wmax  0.745 1000 J  745 J
qC
qH
qC
0.745  1 
qH
qC
0.745  1 
1000 J
 qC  1000  0.7451000 J 
 qC  255 J
Question 2
Calculate ΔS for each of the following changes
in state of 2.50 mol of a perfect monoatomic
gas with CV,m =1.5R for all temperatures:
a) (1.50 atm, 400K)  (3.00 atm, 600K)
b) (2.50 atm, 20.0L)  (2.00 atm, 30.0L)
c) (28.5L, 400K)  (42.0L atm, 400K)
Question 2
• For perfect gas;
dU  dqrev  dwrev
dS  d qrev T
dqrev  dU  dwrev
dqrev  CV dT  PdV  CV dT  nRT T dV
dS  CV dT T  nRdV V
T2
V2
T1
V1
S   CV dT T   nRdV V
Question 2
• Since CV,m is constant (same for all temperatures) ;
S  CV ln T2 T1  nR ln V2 V1
 n  CV ,m ln T2 T1  nR ln V2 V1
• Given, CV , m  1.5 R, thus:
S  1.5nRln T2 T1  nR ln V2 V1
 nR1.5 ln T2 T1  ln V2 V1 
Question 2
a) P1 = 1.5 atm, P2= 3 atm, T1 = 400 K, T2= 600K
S  nR1.5 ln T2 T1  ln V2 V1 
 2.5mol8.314 J mol  K 1.5 ln 600 400  ln T2 P1 P2T1 
 2.5mol8.314 J mol  K 1.5 ln 600 400  ln 600K 1.5atm 3atm 400K 
 6.66 J K
Question 2
b) P1 = 2.5 atm, P2= 2 atm, V1 = 20L, V2= 30L
S  nR1.5 ln T2 T1  ln V2 V1 
 2.5mol8.314 J mol  K 1.5 ln P2V2 P1V1   ln 30L 20L
 2.5mol8.314 J mol  K 1.5 ln 2atm 30L 2.5atm 20L  ln 30L 20L
 14.1 J K
Question 2
c) V1 = 28.5L, V2= 42L, T1= T2 = 400K
S  nR1.5 ln T2 T1  ln V2 V1 
 2.5mol8.314 J mol  K 1.5 ln 400K 400K   ln 42L 28.5L
 8.06 J K
Question 3
After 200 g of gold [cP = 0.0313 cal/(g °C)] at 120°C is dropped into
25.0 g of water at 10°C, the system is allowed to reach
equilibrium in an adiabatic container. Calculate:
a) The final temperature
b) ΔSAu
c) ΔSH2O
d) ΔSAu + ΔSH2O
Question 3
(a)
T2
dqrev
S  
T
T1
• At constant pressure;
dqrev  dqP  CP dT  mcP dT
• At equilibrium;
m1cP,1 T2  T1  m2cP, 2 T2  T1
200g  0.0313 cal
g C T2 120  m2cP, 2 T2 10
200 g  0.0313 cal
g C 120  T2   m2 cP , 2 T2  10
Question 3
• Solving for T2;
200g  0.0313 cal
gC 120  T2   25g 1.00 cal gC T2 10
6.26 cal C 120  T2   25 cal C T2 10
751.2  6.26T2  25T2  250
31.26T2  1001.2
T2  32C
Question 3
(b)
T2
dqrev
S  
T
T1
dqrev  dqP  CP dT  mcP dT
T2
mcP dT
S  
T
T1
S  mcP ln T2 T1
S  200g 0.0313cal / g  K ln 305.15 393.15
 1.59cal / K
Question 3
(c)
T2
dqrev
S  
T
T1
dqrev  dqP  CP dT  mcP dT
T2
mcP dT
S  
T
T1
S  mcP ln T2 T1
S  25g 1.00cal / g  K ln 305.15 283.15
 1.87cal / K
Question 3
(d)
Stotal  S Au  S water
 1.59  1.87cal / K  0.28cal / K
Question 4
A sample consisting of 2.00 mol of diatomic perfect
gas molecules at 250 K is compressed reversibly
and adiabatically until its temperature reaches 300
K. Given that CV,m = 27.5 JK-1mol-1, calculate:
a)
b)
c)
d)
e)
q
w
ΔU
ΔH
ΔS.
Question 4
n = 2.00 mol
CV,m = 27.5 JK-1mol-1
T1= 250 K
T2 = 300 K
Process: reversible adiabatic of a perfect gas
a) qrev =0
b) w = ?
Recall first law:
ΔU = q + w
For perfect gas,
CV dT  0  w
U  CV dT


CV dT  2.0mol  27.5 J  K 1mol  300  250K  2750 J
Thus,
w  2750J
Question 4
c) ΔU
ΔU = q + w
= 0+w
= w = 2750 J
d) ΔH
For a perfect gas;
H  CP dT
C P is not given. However, we know that, CP  CV  nR
Thus;
H  nR  CV dT
 nR  nCV ,m dT


 2.0mol 8.314 J  K 1mol 1  27.5J  K 1 300  250K
 3581.4J
Question 4
e) ΔS
2
dqrev
S  
T
1
Since qrev =0 (reversible adiabatic),
S  0
Question 5
A system consisting of 1.5 mol CO2 (g), initially at 15°C and
9 atm and confined to a cylinder of cross-section 100.0
cm2. It is allowed to expand adiabatically against an
external pressure of 1.5 atm until the piston has moved
outwards through 15 cm. Assume that carbon dioxide may
be considered a perfect gas with CV,m = 28.8 JK-1mol-1,
calculate:
a) q
b)
c)
d)
e)
w
ΔU
ΔT
ΔS
Question 5
A system consisting of 1.5 mol CO2 (g), initially at 15°C and
9 atm and confined to a cylinder of cross-section 100.0
cm2. It is allowed to expand adiabatically against an
external pressure of 1.5 atm until the piston has moved
outwards through 15 cm. Assume that carbon dioxide may
be considered a perfect gas with CV,m = 28.8 JK-1mol-1,
calculate:
a) q
b)
c)
d)
e)
w
ΔU
ΔT
ΔS
Pext = 1.5 atm
T1 = 15°C
P1 = 9 atm
A = 100 cm2
V1
V2
15 cm
Question 5
•
•
•
•
Pext = 1.5 atm
n = 1.5 mol CO2
CV,m = 28.8 JK-1mol-1
A perfect gas
Adiabatic
T1 = 15°C
P1 = 9 atm
A = 100 cm2
V1
V2
15 cm
a) q = 0 (adiabatic)
b) w = ?
W   Pext V

 1.5atm 100cm2 15cm

1.01105 Pa
1m3
 2250atm  cm 
 6 3  227.25 J
1atm
10 cm
3
Question 5
•
•
•
•
Pext = 1.5 atm
n = 1.5 mol CO2
CV,m = 28.8 JK-1mol-1
A perfect gas
Adiabatic
T1 = 15°C
P1 = 9 atm
A = 100 cm2
V1
V2
15 cm
c) ΔU = ?
U  q  w
U  0  w  w  227.25J
d) ΔT = ?


For perfect gas, U  CV dT  n  CV ,m dT
Thus, T  U n  CV ,m




  227.25 J 1.5mol  28.8 JK 1mol 1  5.26 K
Question 5
•
•
•
•
n = 1.5 mol CO2
CV,m = 28.8 JK-1mol-1
A perfect gas
Adiabatic
Pext = 1.5 atm
T1 = 15°C
P1 = 9 atm
A = 100 cm2
V1
V2
15 cm
e) ΔS=?
From part d) T  5.26K
S  S1  S2
2
3
CV
CV
V3
V2

dT  nR ln  
dT  nR ln
T
V1 2 T
V2
1
2
CV
V
dT nR ln 3
T
V2
1

The gas undergoes
constant-volume
cooling followed by
isothermal expansion
Question 5
•
•
•
•
n = 1.5 mol CO2
CV,m = 28.8 JK-1mol-1
A perfect gas
Adiabatic
Pext = 1.5 atm
T1 = 15°C
P1 = 9 atm
A = 100 cm2
V1
V2
15 cm
e) ΔS=?
2
CV
V3
S  
dT nR ln
T
V2
1
V3
T2
 CV ln  nR ln
T1
V2
T2  T1  T
 (15  273.15)  5.26 K  282.89 K
V1  V2 Constant volume cooling
nRT1
V1 
P1

Find this first!

1.5mol 8.206 10 2 L  atm  K 1mol 1 288.15K 

9atm
 3.941L
Question 5
•
•
•
•
n = 1.5 mol CO2
CV,m = 28.8 JK-1mol-1
A perfect gas
Adiabatic
Pext = 1.5 atm
T1 = 15°C
P1 = 9 atm
A = 100 cm2
V1
V2
15 cm
e) ΔS=?
V3  V2  V
2
CV
V3
S  
dT nR ln
T
V2
1
V3
T2
 CV ln  nR ln
T1
V2

 103 L 
  5.44 L
 3.941L  100cm 15cm 
3 
 1cm 


 1.5mol  28.8 JK 1mol 1 ln
2



282.89 K
5.44 L
 1.5mol  8.314 JK 1mol 1 ln
288.15K
3.941L
 0.79588  4.01996  3.224 JK 1