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A rigid insulated tank is divided into two compartment contains 1 kmol of O, and the other compartment contains 4 kmol of N. Both gases are initially at 300 K and 200 kPa pressure. Now the partition is removed, the two gases are allowed to mix. Assuming both gases are ideal, the total entropy change in kJ/K of the gases is (A) 0 (B) 20.8 (C) 0.718 (D) 2.502?
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A rigid insulated tank is divided into two compartment contains 1 kmol...
Initial conditions:
- Initial temperature (T): 300 K
- Initial pressure (P): 200 kPa
- Number of moles of O: 1 kmol
- Number of moles of N: 4 kmol
Entropy change during mixing:
- When the partition is removed, O and N mix until they reach equilibrium.
- Since both gases are ideal gases, the entropy change during mixing is given by ΔS = nR ln(Vf/Vi), where n is the total number of moles, R is the gas constant, Vi is the initial total volume, and Vf is the final total volume.
- The total number of moles (n) = 1 kmol + 4 kmol = 5 kmol
- Initial total volume (Vi) can be calculated using ideal gas law: Vi = nRT/P = 5*8.314*300/200 = 62.71 m^3
- Final total volume (Vf) is the sum of the individual volumes of O and N after mixing. Since volume is additive for ideal gases, Vf = Vo + Vn, where Vo and Vn are the individual volumes of O and N.
- Final pressure (Pf) is the same for both gases: Pf = P = 200 kPa
- Using ideal gas law for each gas, Vo = 1*8.314*300/200 = 12.471 m^3 and Vn = 4*8.314*300/200 = 49.886 m^3
- Therefore, Vf = 12.471 + 49.886 = 62.357 m^3
- Substituting these values into the entropy change equation, ΔS = 5*8.314*ln(62.357/62.71) = -20.8 kJ/K
Conclusion:
- The total entropy change of the gases during mixing is -20.8 kJ/K.
- The correct answer is (B) 20.8.
- Initial temperature (T): 300 K
- Initial pressure (P): 200 kPa
- Number of moles of O: 1 kmol
- Number of moles of N: 4 kmol
Entropy change during mixing:
- When the partition is removed, O and N mix until they reach equilibrium.
- Since both gases are ideal gases, the entropy change during mixing is given by ΔS = nR ln(Vf/Vi), where n is the total number of moles, R is the gas constant, Vi is the initial total volume, and Vf is the final total volume.
- The total number of moles (n) = 1 kmol + 4 kmol = 5 kmol
- Initial total volume (Vi) can be calculated using ideal gas law: Vi = nRT/P = 5*8.314*300/200 = 62.71 m^3
- Final total volume (Vf) is the sum of the individual volumes of O and N after mixing. Since volume is additive for ideal gases, Vf = Vo + Vn, where Vo and Vn are the individual volumes of O and N.
- Final pressure (Pf) is the same for both gases: Pf = P = 200 kPa
- Using ideal gas law for each gas, Vo = 1*8.314*300/200 = 12.471 m^3 and Vn = 4*8.314*300/200 = 49.886 m^3
- Therefore, Vf = 12.471 + 49.886 = 62.357 m^3
- Substituting these values into the entropy change equation, ΔS = 5*8.314*ln(62.357/62.71) = -20.8 kJ/K
Conclusion:
- The total entropy change of the gases during mixing is -20.8 kJ/K.
- The correct answer is (B) 20.8.
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A rigid insulated tank is divided into two compartment contains 1 kmol of O, and the other compartment contains 4 kmol of N. Both gases are initially at 300 K and 200 kPa pressure. Now the partition is removed, the two gases are allowed to mix. Assuming both gases are ideal, the total entropy change in kJ/K of the gases is (A) 0 (B) 20.8 (C) 0.718 (D) 2.502?
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A rigid insulated tank is divided into two compartment contains 1 kmol of O, and the other compartment contains 4 kmol of N. Both gases are initially at 300 K and 200 kPa pressure. Now the partition is removed, the two gases are allowed to mix. Assuming both gases are ideal, the total entropy change in kJ/K of the gases is (A) 0 (B) 20.8 (C) 0.718 (D) 2.502? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A rigid insulated tank is divided into two compartment contains 1 kmol of O, and the other compartment contains 4 kmol of N. Both gases are initially at 300 K and 200 kPa pressure. Now the partition is removed, the two gases are allowed to mix. Assuming both gases are ideal, the total entropy change in kJ/K of the gases is (A) 0 (B) 20.8 (C) 0.718 (D) 2.502? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rigid insulated tank is divided into two compartment contains 1 kmol of O, and the other compartment contains 4 kmol of N. Both gases are initially at 300 K and 200 kPa pressure. Now the partition is removed, the two gases are allowed to mix. Assuming both gases are ideal, the total entropy change in kJ/K of the gases is (A) 0 (B) 20.8 (C) 0.718 (D) 2.502?.
A rigid insulated tank is divided into two compartment contains 1 kmol of O, and the other compartment contains 4 kmol of N. Both gases are initially at 300 K and 200 kPa pressure. Now the partition is removed, the two gases are allowed to mix. Assuming both gases are ideal, the total entropy change in kJ/K of the gases is (A) 0 (B) 20.8 (C) 0.718 (D) 2.502? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A rigid insulated tank is divided into two compartment contains 1 kmol of O, and the other compartment contains 4 kmol of N. Both gases are initially at 300 K and 200 kPa pressure. Now the partition is removed, the two gases are allowed to mix. Assuming both gases are ideal, the total entropy change in kJ/K of the gases is (A) 0 (B) 20.8 (C) 0.718 (D) 2.502? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rigid insulated tank is divided into two compartment contains 1 kmol of O, and the other compartment contains 4 kmol of N. Both gases are initially at 300 K and 200 kPa pressure. Now the partition is removed, the two gases are allowed to mix. Assuming both gases are ideal, the total entropy change in kJ/K of the gases is (A) 0 (B) 20.8 (C) 0.718 (D) 2.502?.
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