1. By Miss Anis Atikah Ahmad
Tutorial 2 By
Miss Anis Atikah Ahmad

2. Question 1 A heat engine uses reservoirs at 800°C and 0°C.
Calculate maximum possible efficiency
If qH is 1000 J, calculate the maximum -w and the minimum value of -qC

3. Question 1 TH= 800°C = 1073.15K, TC =0°C =273.15K
Calculate maximum possible efficiency
TH= 800°C = K, TC =0°C =273.15K
Work output per cycle
Energy input per cycle

4. Question 1
b) If qH is 1000 J, calculate the maximum -w and the minimum value of -qC
Work output per cycle
Energy input per cycle

5. 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:
(1.50 atm, 400K)  (3.00 atm, 600K)
(2.50 atm, 20.0L)  (2.00 atm, 30.0L)
(28.5L, 400K)  (42.0L atm, 400K)

6. Question 2
For perfect gas;

7. Question 2 Since CV,m is constant (same for all temperatures) ;
Given, , thus:

8. Question 2
a) P1 = 1.5 atm, P2= 3 atm, T1 = 400 K, T2= 600K

9. Question 2
b) P1 = 2.5 atm, P2= 2 atm, V1 = 20L, V2= 30L

10. Question 2
c) V1 = 28.5L, V2= 42L, T1= T2 = 400K

11. Question 3
After 200 g of gold [cP = 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:
The final temperature
ΔSAu
ΔSH2O
ΔSAu + ΔSH2O

12. Question 3
(a)
At constant pressure;
At equilibrium;

13. Question 3
Solving for T2;

14. Question 3
(b)

15. Question 3
(c)

16. Question 3
(d)

17. 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: q w ΔU ΔH
ΔS.

18. Question 4 qrev =0 n = 2.00 mol CV,m = 27.5 JK-1mol-1
T1= 250 K T2 = 300 K
Process: reversible adiabatic of a perfect gas
qrev =0
b) w = ?
Recall first law:
ΔU = q + w
For perfect gas,
Thus,

19. Question 4 ΔU d) ΔH ΔU = q + w = 0 + w = w = 2750 J For a perfect gas;
is not given. However, we know that,
Thus;

20. Question 4
e) ΔS
Since qrev =0 (reversible adiabatic),

21. 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 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: q w ΔU ΔT ΔS

22. 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 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: q w ΔU ΔT ΔS
Pext = 1.5 atm
T1 = 15°C
P1 = 9 atm
A = 100 cm2 V1 V2
15 cm

23. Question 5 n = 1.5 mol CO2 CV,m = 28.8 JK-1mol-1 A perfect gas
Adiabatic
q = 0 (adiabatic)
w = ?
Pext = 1.5 atm
T1 = 15°C
P1 = 9 atm
A = 100 cm2 V1 V2
15 cm

24. Question 5 ΔU = ? ΔT = ? n = 1.5 mol CO2 CV,m = 28.8 JK-1mol-1
A perfect gas
Adiabatic
ΔU = ?
ΔT = ?
For perfect gas,
Thus,
Pext = 1.5 atm
T1 = 15°C
P1 = 9 atm
A = 100 cm2 V1 V2
15 cm

25. Question 5 e) ΔS=? n = 1.5 mol CO2 CV,m = 28.8 JK-1mol-1 A perfect gas
Adiabatic
e) ΔS=?
From part d)
Pext = 1.5 atm
T1 = 15°C
P1 = 9 atm
A = 100 cm2 V1 V2
15 cm
The gas undergoes constant-volume cooling followed by isothermal expansion

26. Question 5 e) ΔS=? n = 1.5 mol CO2 CV,m = 28.8 JK-1mol-1 A perfect gas
Adiabatic
e) ΔS=?
Pext = 1.5 atm
T1 = 15°C
P1 = 9 atm
A = 100 cm2 V1 V2
15 cm
Constant volume cooling
Find this first!

27. Question 5 e) ΔS=? n = 1.5 mol CO2 CV,m = 28.8 JK-1mol-1 A perfect gas
Adiabatic
e) ΔS=?
Pext = 1.5 atm
T1 = 15°C
P1 = 9 atm
A = 100 cm2 V1 V2
15 cm