2500 Solved Problems In Thermodynamics Pdf
1. Based on graph P-V below, what is the ratio of the work done by the gas in the process I, to the work done by the gas in the process II?
Known :
Process 1 :
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Pressure (P) = 20 N/m2
Initial volume (V1) = 10 liter = 10 dm3 = 10 x 10-3 m3
Final volume (V2) = 40 liter = 40 dm3 = 40 x 10-3 m3
Process 2 :
Process (P) = 15 N/m2
Initial volume (V1) = 20 liter = 20 dm3 = 20 x 10-3 m3
Final volume (V2) = 60 liter = 60 dm3 = 60 x 10-3 m3
Wanted : The ratio of the work done by gas
Solution :
The work done by gas in the process I :
W = P ΔV = P (V2–V1) = (20)(40-10)(10-3 m3) = (20)(30)(10-3 m3) = (600)(10-3 m3) = 0.6 m3
The work done by gas in the process II :
W = P ΔV = P (V2–V1) = (15)(60-20)(10-3 m3) = (15)(40)(10-3 m3) = (600)(10-3 m3) = 0.6 m3
The ratio of the work done by gas in the process I and the process II :
0.6 m3 : 0.6 m3
1 : 1
2.
Based on the graph below, what is the work done by helium gas in the process AB?
Known :
Pressure (P) = 2 x 105 N/m2 = 2 x 105 Pascal
Initial volume (V1) = 5 cm3 = 5 x 10-6 m3
Final volume (V2) = 15 cm3 = 15 x 10-6 m3
Wanted : Work done by gas in process AB
Solution :
W = ∆P ∆V
W = P (V2 – V1)
W = (2 x 105)(15 x 10-6 – 5 x 10-6)
W = (2 x 105)(10 x 10-6) = (2 x 105)(1 x 10-5)
W = 2 Joule
3.
Based on the graph below, what is the work done in process a-b?
Known :
Initial pressure (P1) = 4 Pa = 4 N/m2
Final pressure (P2) = 6 Pa = 6 N/m2
Initial volume (V1) = 2 m3
Final volume (V2) = 4 m3
Wanted : work done I process a-b
Solution :
Work done by gas = area under curve a-b
W = area of triangle + area of rectangle
W = ½ (6-4)(4-2) + 4(4-2)
W = ½ (2)(2) + 4(2)
W = 2 + 8
W = 10 Joule
4. Based on graph below, what is the work done in process A-B-C-A.
Solution :
Work (W) = Area of the triangle A-B-C
W = ½ (20-10)(6 x 105 – 2 x 105)
W = ½ (10)(4 x 105)
W = (5)(4 x 105
)
W = 20 x 105
W = 2 x 106 Joule
5. An engine absorbs 2000 Joule of heat at a high temperature and exhausted 1200 Joule of heat at a low temperature. What is the efficiency of the engine?
Known :
Heat input (QH) = 2000 Joule
Heat output (QL) = 1200 Joule
Work done by engine (W) = 2000 – 1200 = 800 Joule
Wanted : efficiency (e)
Solution :
e = W / QH
e = 800/2000
e = 0.4 x 100%
e = 40%
6. An engine absorbs heat at 960 Kelvin and the engine discharges heat at 576 Kelvin. What is the efficiency of the engine.
Known :
High temperature (TH) = 960 K
Low temperature (TL) = 576 K
Wanted: efficiency (e)
Solution :
Efficiency of Carnot engine = 0.4 x 100% = 40%
7. Based on the graph below, work done by the engine is 6000 Joule. What is the heat discharged by engine each circle?
Known :
Work (W) = 6000 Joule
High temperature (TH) = 800 Kelvin
Low temperature (TL) = 300 Kelvin
Wanted: heat discharged by the engine
Solution :
Carnot (ideal) efficiency :
Heat absorbed by Carnot engine :
W = e Q1
6000 = (0.625) Q1
Q1 = 6000 / 0.625
Q1 = 9600
Heat discharged by Carnot engine :
Q2 = Q1 – W
Q2 = 9600 – 6000
Q2 = 3600 Joule
8. The efficiency of a Carnot engine is 40%. If heat absorbed at 727°C then what is the low temperature.
Known :
Efficiency (e) = 40% = 40/100 = 0.4
High temperature (TH) = 727oC + 273 = 1000 K
Wanted : Low temperature
Solution :
TL = 600 Kelvin – 273 = 327oC
9. Based on graph below, if the engine absorbs 800 J of heat, what is the work done by the engine.
Known :
High temperature (TH) = 600 Kelvin
Low temperature (TL) = 250 Kelvin
Heat input (Q1) = 800 Joule
Wanted: Work (W)
Solution :
The efficiency of Carnot engine :
Work was done by the engine :
W = e Q1
W = (7/12)(800 Joule)
W = 466.7 Joule
10. The high temperature of a Carnot engine is 600 K. If the engine absorbs 600 J of heat and the low temperature is 400 K, what is the work done by the engine.
Known :
Low temperature (TL) = 400 K
High temperature (TH) = 600 K
Heat input (Q1) = 600 Joule
Wanted: Work was done by Carnot engine (W)
Solution :
The efficiency of the Carnot engine :
Work was done by Carnot engine :
W = e Q1
W = (1/3)(600) = 200 Joule
1. 3000 J of heat is added to a system and 2500 J of work is done by the system. What is the change in internal energy of the system?
Known :
Heat (Q) = +3000 Joule
Work (W) = +2500 Joule
Wanted: the change in internal energy of the system
Solution :
The equation of the first law of thermodynamics
ΔU = Q-W
The sign conventions :
Q is positive if the heat added to the system
W is positive if work is done by the system
Q is negative if heat leaves the system
W is negative if work is done on the system
The change in internal energy of the system :
ΔU = 3000-2500
ΔU = 500 Joule
Internal energy increases by 500 Joule.
2. 2000 J of heat is added to a system and 2500 J of work is done on the system. What is the change in internal energy of the system?
Known :
Heat (Q) = +2000 Joule
Work (W) = -2500 Joule
Wanted: The change in internal energy of the system
Solution :
ΔU = Q-W
ΔU = 2000-(-2500)
ΔU = 2000+2500
ΔU = 4500 Joule
Internal energy increases by 4500 Joule.
3. 2000 J of heat leaves the system and 2500 J of work is done on the system. What is the change in internal energy of the system?
Known :
Heat (Q) = -2000 Joule
Work (W) = -3000 Joule
Wanted: The change in internal energy of the system
Solution :
ΔU = Q-W
Thermodynamics Practice Problems Pdf
ΔU = -2000-(-3000)
ΔU = -2000+3000
ΔU = 1000 Joule
Internal energy increases by 4500 Joule.
Conclusion :
– If heat is added to the system, then the internal energy of the system increases
– If heat leaves the system, then the internal energy of the system decreases
– If the work is done by the system, then the internal energy of the system decreases
How To Solve Thermodynamic Problems
– If the work is done on the system, then the internal energy of the system increases