Thermodynamics – problems and solutions
The first law of thermodynamics
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 :
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
- What is the primary focus of thermodynamics? Answer: Thermodynamics focuses on the study of energy, its transformations, and its relationship with matter, especially in systems at equilibrium.
- How is the zeroth law of thermodynamics related to temperature? Answer: The zeroth law states that if two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other. This implies the existence of a property called temperature, which is the same for all systems in thermal equilibrium.
- What does the first law of thermodynamics describe? Answer: The first law, also known as the law of energy conservation, states that energy cannot be created or destroyed, only converted from one form to another. In a closed system, the change in internal energy is equal to the heat added to the system minus the work done by the system on its surroundings.
- Why is the second law of thermodynamics crucial for understanding the direction of natural processes? Answer: The second law states that the entropy (or disorder) of an isolated system always increases or remains constant. This dictates that energy spontaneously disperses if not hindered from doing so, providing a direction to natural processes and essentially explaining why certain processes occur spontaneously while others do not.
- What is entropy, and how does it relate to disorder in a system? Answer: Entropy is a measure of the amount of energy in a system that is unavailable to do work. It is also often described as a measure of the system’s disorder or randomness. In general, higher entropy corresponds to greater disorder or randomness.
- How does the third law of thermodynamics describe the entropy of a perfect crystal at absolute zero? Answer: The third law states that the entropy of a perfect crystal is exactly zero at absolute zero temperature (0 Kelvin). This means that at this temperature, the system is perfectly ordered.
- Why can’t heat flow from a colder body to a hotter body on its own? Answer: This behavior is a consequence of the second law of thermodynamics. If heat were to flow from a colder body to a hotter one spontaneously, it would lead to a decrease in the overall entropy of the system, which is not favored by natural processes.
- What is the difference between an isolated, closed, and open system in thermodynamics? Answer: An isolated system does not exchange energy or matter with its surroundings. A closed system can exchange energy but not matter with its surroundings. An open system can exchange both energy and matter with its surroundings.
- How is the concept of “work” in thermodynamics different from the everyday use of the term? Answer: In thermodynamics, “work” refers to the process of energy transfer where forces applied to an object move it in a direction parallel to the force. For example, when a gas expands against a piston, it does work on the piston. This is a more specific definition compared to the everyday use of “work,” which might simply mean any task or activity.
What is a Carnot cycle, and why is it significant in thermodynamics? Answer: The Carnot cycle is an idealized thermodynamic cycle that provides an upper limit on the efficiency that any classical thermodynamic engine can achieve during the conversion of heat into work (or vice versa). It’s significant because it sets a fundamental efficiency limit based on the temperatures of the heat reservoirs between which an engine operates.
You must be logged in to post a comment.