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Air at an initial temperature and volume of 300 K and 0.002m ^ 3 is contained in a piston- cylinder a assembly as shown in the figure. Heat is added to the air and it expands slowly to occupy a final volume of 0.003m ^ 3 . Area of the piston is 0.02 m² and the spring constant k is 10 kN/m. The atmosphere pressure may be assumed to be 100 kPa. Piston area A = 0.02m ^ 2 Air 0.002m ^ 3 Spring F = kx k = 10kN / m x = 0 5.40 The final pressure of air in kPa is (A) 25 (B) 125 (C) 102.5 (D) 112.5?
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Air at an initial temperature and volume of 300 K and 0.002m ^ 3 is co...
Given Data:
- Initial temperature (T1) = 300 K
- Initial volume (V1) = 0.002 m^3
- Final volume (V2) = 0.003 m^3
- Piston area (A) = 0.02 m^2
- Spring constant (k) = 10 kN/m
- Atmosphere pressure = 100 kPa
Calculations:
1. Work done by the spring:
- W = 0.5k(x^2)
- W = 0.5 * 10 * (0.001)^2
- W = 0.00005 kN = 50 N
2. Work done by the gas:
- W = -PΔV
- W = -P(V2 - V1)
- W = -100(0.003 - 0.002)
- W = -100 N
3. Total work done:
- W_total = W_gas + W_spring
- W_total = -100 + 50
- W_total = -50 N
4. Change in internal energy:
- ΔU = Q - W
- Since the process is adiabatic, Q = 0
- ΔU = -W_total
- ΔU = 50 J
5. Final pressure of the air:
- ΔU = nCvΔT
- nCvΔT = 50
- CvΔT = RΔT = 50
- ΔT = 50 / Cv
- ΔT = 50 / (5/2) = 20 K
- Final temperature (T2) = 300 + 20 = 320 K
- P1V1/T1 = P2V2/T2
- P2 = (P1V1T2) / (V2T1)
- P2 = (100 * 0.002 * 320) / (0.003 * 300)
- P2 ≈ 106.67 kPa
Final Answer:
The final pressure of the air in kPa is approximately 106.67, which is closest to option (C) 102.5 kPa.
- Initial temperature (T1) = 300 K
- Initial volume (V1) = 0.002 m^3
- Final volume (V2) = 0.003 m^3
- Piston area (A) = 0.02 m^2
- Spring constant (k) = 10 kN/m
- Atmosphere pressure = 100 kPa
Calculations:
1. Work done by the spring:
- W = 0.5k(x^2)
- W = 0.5 * 10 * (0.001)^2
- W = 0.00005 kN = 50 N
2. Work done by the gas:
- W = -PΔV
- W = -P(V2 - V1)
- W = -100(0.003 - 0.002)
- W = -100 N
3. Total work done:
- W_total = W_gas + W_spring
- W_total = -100 + 50
- W_total = -50 N
4. Change in internal energy:
- ΔU = Q - W
- Since the process is adiabatic, Q = 0
- ΔU = -W_total
- ΔU = 50 J
5. Final pressure of the air:
- ΔU = nCvΔT
- nCvΔT = 50
- CvΔT = RΔT = 50
- ΔT = 50 / Cv
- ΔT = 50 / (5/2) = 20 K
- Final temperature (T2) = 300 + 20 = 320 K
- P1V1/T1 = P2V2/T2
- P2 = (P1V1T2) / (V2T1)
- P2 = (100 * 0.002 * 320) / (0.003 * 300)
- P2 ≈ 106.67 kPa
Final Answer:
The final pressure of the air in kPa is approximately 106.67, which is closest to option (C) 102.5 kPa.
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Air at an initial temperature and volume of 300 K and 0.002m ^ 3 is contained in a piston- cylinder a assembly as shown in the figure. Heat is added to the air and it expands slowly to occupy a final volume of 0.003m ^ 3 . Area of the piston is 0.02 m² and the spring constant k is 10 kN/m. The atmosphere pressure may be assumed to be 100 kPa. Piston area A = 0.02m ^ 2 Air 0.002m ^ 3 Spring F = kx k = 10kN / m x = 0 5.40 The final pressure of air in kPa is (A) 25 (B) 125 (C) 102.5 (D) 112.5?
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Air at an initial temperature and volume of 300 K and 0.002m ^ 3 is contained in a piston- cylinder a assembly as shown in the figure. Heat is added to the air and it expands slowly to occupy a final volume of 0.003m ^ 3 . Area of the piston is 0.02 m² and the spring constant k is 10 kN/m. The atmosphere pressure may be assumed to be 100 kPa. Piston area A = 0.02m ^ 2 Air 0.002m ^ 3 Spring F = kx k = 10kN / m x = 0 5.40 The final pressure of air in kPa is (A) 25 (B) 125 (C) 102.5 (D) 112.5? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Air at an initial temperature and volume of 300 K and 0.002m ^ 3 is contained in a piston- cylinder a assembly as shown in the figure. Heat is added to the air and it expands slowly to occupy a final volume of 0.003m ^ 3 . Area of the piston is 0.02 m² and the spring constant k is 10 kN/m. The atmosphere pressure may be assumed to be 100 kPa. Piston area A = 0.02m ^ 2 Air 0.002m ^ 3 Spring F = kx k = 10kN / m x = 0 5.40 The final pressure of air in kPa is (A) 25 (B) 125 (C) 102.5 (D) 112.5? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Air at an initial temperature and volume of 300 K and 0.002m ^ 3 is contained in a piston- cylinder a assembly as shown in the figure. Heat is added to the air and it expands slowly to occupy a final volume of 0.003m ^ 3 . Area of the piston is 0.02 m² and the spring constant k is 10 kN/m. The atmosphere pressure may be assumed to be 100 kPa. Piston area A = 0.02m ^ 2 Air 0.002m ^ 3 Spring F = kx k = 10kN / m x = 0 5.40 The final pressure of air in kPa is (A) 25 (B) 125 (C) 102.5 (D) 112.5?.
Air at an initial temperature and volume of 300 K and 0.002m ^ 3 is contained in a piston- cylinder a assembly as shown in the figure. Heat is added to the air and it expands slowly to occupy a final volume of 0.003m ^ 3 . Area of the piston is 0.02 m² and the spring constant k is 10 kN/m. The atmosphere pressure may be assumed to be 100 kPa. Piston area A = 0.02m ^ 2 Air 0.002m ^ 3 Spring F = kx k = 10kN / m x = 0 5.40 The final pressure of air in kPa is (A) 25 (B) 125 (C) 102.5 (D) 112.5? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Air at an initial temperature and volume of 300 K and 0.002m ^ 3 is contained in a piston- cylinder a assembly as shown in the figure. Heat is added to the air and it expands slowly to occupy a final volume of 0.003m ^ 3 . Area of the piston is 0.02 m² and the spring constant k is 10 kN/m. The atmosphere pressure may be assumed to be 100 kPa. Piston area A = 0.02m ^ 2 Air 0.002m ^ 3 Spring F = kx k = 10kN / m x = 0 5.40 The final pressure of air in kPa is (A) 25 (B) 125 (C) 102.5 (D) 112.5? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Air at an initial temperature and volume of 300 K and 0.002m ^ 3 is contained in a piston- cylinder a assembly as shown in the figure. Heat is added to the air and it expands slowly to occupy a final volume of 0.003m ^ 3 . Area of the piston is 0.02 m² and the spring constant k is 10 kN/m. The atmosphere pressure may be assumed to be 100 kPa. Piston area A = 0.02m ^ 2 Air 0.002m ^ 3 Spring F = kx k = 10kN / m x = 0 5.40 The final pressure of air in kPa is (A) 25 (B) 125 (C) 102.5 (D) 112.5?.
Solutions for Air at an initial temperature and volume of 300 K and 0.002m ^ 3 is contained in a piston- cylinder a assembly as shown in the figure. Heat is added to the air and it expands slowly to occupy a final volume of 0.003m ^ 3 . Area of the piston is 0.02 m² and the spring constant k is 10 kN/m. The atmosphere pressure may be assumed to be 100 kPa. Piston area A = 0.02m ^ 2 Air 0.002m ^ 3 Spring F = kx k = 10kN / m x = 0 5.40 The final pressure of air in kPa is (A) 25 (B) 125 (C) 102.5 (D) 112.5? in English & in Hindi are available as part of our courses for UPSC.
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