الديناميكا الحرارية – المشاكل والحلول
القانون الأول للديناميكا الحرارية
1. Based on graph P-V below, what is the ratio of the العمل done by the gas in the process I, to the work done by the gas in the process II?
معروف :
Process 1 :
الضغط (P) = 20 N/m2
الحجم الأولي (V)1) = 10 لتر = 10 دسم3 = 10 × 10-3 m3
المجلد الأخير (V)2) = 40 لتر = 40 دسم3 = 40 × 10-3 m3
Process 2 :
Process (P) = 15 N/m2
الحجم الأولي (V)1) = 20 لتر = 20 دسم3 = 20 × 10-3 m3
المجلد الأخير (V)2) = 60 لتر = 60 دسم3 = 60 × 10-3 m3
مطلوب: The ratio of the work done by gas
الحل:
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 م3
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 م3
The ratio of the work done by gas in the process I and the process II :
0.6 م3 : 0.6 م3
1: 1
2.
Based on the graph below, what is the work done by helium gas in the process AB?
معروف :
Pressure (P) = 2 x 105 N / م2 = 2 × 105 محدة ضغط
الحجم الأولي (V)1) = 5 سم3 = 5 × 10-6 m3
المجلد الأخير (V)2) = 15 سم3 = 15 × 10-6 m3
مطلوب: Work done by gas in process AB
الحل:
W = ∆P ∆V
W = P (V2 - ف1)
W = (2 × 105)(15 × 10-6 - 5 × 10-6)
W = (2 × 105)(10 × 10-6) = (2 × 105)(1 × 10-5)
W = 2 جول
3.
Based on the graph below, what is the work done in process a-b?
معروف :
الضغط الأولي (P)1) = 4 باسكال = 4 نيوتن/متر2
الضغط النهائي (P)2) = 6 باسكال = 6 نيوتن/متر2
الحجم الأولي (V)1) = 2 م3
المجلد الأخير (V)2) = 4 م3
مطلوب: work done I process a-b
الحل:
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 جول
4. Based on graph below, what is the work done in process A-B-C-A.
الحل:
Work (W) = Area of the triangle A-B-C
W = ½ (20-10)(6 × 105 - 2 × 105)
W = ½ (10)(4 × 105)
W = (5)(4 × 105)
W = 20 × 105
W = 2 × 106 جول وحدة طاقة
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?
معروف :
مدخلات الحرارة (Q)H) = 2000 جول
ناتج الحرارة (Q)L) = 1200 جول
Work done by engine (W) = 2000 – 1200 = 800 Joule
مطلوب: efficiency (e)
الحل:
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.
معروف :
درجة حرارة عالية (TH) = 960 ك
درجة حرارة منخفضة (TL) = 576 ك
مطلوب: efficiency (e)
الحل:
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?
معروف :
Work (W) = 6000 Joule
درجة حرارة عالية (TH) = 800 كلفن
درجة حرارة منخفضة (TL) = 300 كلفن
مطلوب: heat discharged by the engine
الحلول :
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 = س1 - ث
Q2 = 9600 - 6000
Q2 = 3600 جول
8. The efficiency of a Carnot engine is 40%. If heat absorbed at 727°C then what is the low temperature.
معروف :
Efficiency (e) = 40% = 40/100 = 0.4
درجة حرارة عالية (TH) = 727oج + ٢٧٣ = ٢٩٣ كلفن
مطلوب: درجة حرارة منخفضة
الحل:
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.
معروف :
درجة حرارة عالية (TH) = 600 كلفن
درجة حرارة منخفضة (TL) = 250 كلفن
مدخلات الحرارة (Q)1) = 800 جول
مطلوب: اعمل مع)
الحل:
The efficiency of Carnot engine :
Work was done by the engine :
W = e Q1
W = (7/12)(800 جول)
W = 466.7 جول
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.
معروف :
درجة حرارة منخفضة (TL) = 400 ك
درجة حرارة عالية (TH) = 600 ك
مدخلات الحرارة (Q)1) = 600 جول
مطلوب: تم إنجاز العمل بواسطة محرك كارنو (W)
الحل:
كفاءة محرك كارنو:
Work was done by Carnot engine :
W = e Q1
الشغل = (1/3)(600) = 200 جول
- What is the primary focus of thermodynamics? إجابة: 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? إجابة: 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? إجابة: 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? إجابة: 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? إجابة: 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? إجابة: 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? إجابة: 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? إجابة: 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? إجابة: 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.
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What is a Carnot cycle, and why is it significant in thermodynamics? إجابة: 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.
