Mechanical Engineering Exam > Mechanical Engineering Questions > A rigid insulated tank has two equal parts by...
Start Learning for Free
A rigid insulated tank has two equal parts by partition. First part initially contains 5k mol of an ideal gas at
500 kPa, 400C and other part is evacuated. If partition is removed, then the gas fills the entire tank, what will be the total entropy change during this process.
500 kPa, 400C and other part is evacuated. If partition is removed, then the gas fills the entire tank, what will be the total entropy change during this process.
- a)28.81 kJ/K
- b)14.4 kJ/K
- c)7.2 kJ/K
- d)21.6 kJ/K
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
A rigid insulated tank has two equal parts by partition. First part in...
Answer is (a)
Most Upvoted Answer
A rigid insulated tank has two equal parts by partition. First part in...
Solution:
Given data:
Number of moles in the first part of the tank, n1 = 5 kmol
Pressure, P1 = 500 kPa
Temperature, T1 = 400°C = 673.15 K
Volume of each part of the tank, V = V1 + V2
Total number of moles in the tank after removal of the partition = n1 + n2
As the gas is assumed to be ideal, we can use the ideal gas equation to calculate the volume of the gas in the first part of the tank.
PV = nRT
V1 = n1RT1 / P1
V1 = 5 × 8.314 × 673.15 / 500
V1 = 5.629 m³
As both parts of the tank are equal in volume, V2 = V1 = 5.629 m³
The total volume of the tank after removal of the partition is V = 2V1 = 2V2 = 11.258 m³
The final pressure and temperature of the gas in the tank can be calculated using the ideal gas equation and the conservation of mass.
n1RT1 / V1 = (n1 + n2)RT2 / V
T2 = (n1 + n2)T1 / n1
T2 = (5 + n2 / 5) × 673.15 / 5
T2 = (n2 / 5) × 134.63 + 673.15
P2 = nRT / V
P2 = (n1 + n2)RT2 / V
P2 = (5 + n2 / 5) × 8.314 × [(n2 / 5) × 134.63 + 673.15] / 5.629
P2 = (n2 / 5) × 327.7 + 206.3 kPa
The change in entropy of the gas can be calculated using the ideal gas equation and the definition of entropy.
ΔS = ∫(dQ / T)
For an ideal gas, dQ = nCv dT
ΔS = nCv ln(T2 / T1)
Cv for a diatomic gas = 5R / 2
ΔS = (5 + n2 / 5) × (5R / 2) ln[(n2 / 5) × 134.63 + 673.15 / 673.15]
ΔS = (5 + n2 / 5) × (5R / 2) ln(n2 / 5 + 1.013)
As the total number of moles in the tank after removal of the partition is not known, we can assume a small change in the number of moles, dn2, and calculate the corresponding change in entropy, dS.
dS = (5 + (n2 + dn2) / 5) × (5R / 2) ln[(n2 + dn2) / 5 + 1.013] - (5 + n2 / 5) × (5R / 2) ln(n2 / 5 + 1.013)
dS = (5R / 2) ln[(n2 + dn2) / n
Given data:
Number of moles in the first part of the tank, n1 = 5 kmol
Pressure, P1 = 500 kPa
Temperature, T1 = 400°C = 673.15 K
Volume of each part of the tank, V = V1 + V2
Total number of moles in the tank after removal of the partition = n1 + n2
As the gas is assumed to be ideal, we can use the ideal gas equation to calculate the volume of the gas in the first part of the tank.
PV = nRT
V1 = n1RT1 / P1
V1 = 5 × 8.314 × 673.15 / 500
V1 = 5.629 m³
As both parts of the tank are equal in volume, V2 = V1 = 5.629 m³
The total volume of the tank after removal of the partition is V = 2V1 = 2V2 = 11.258 m³
The final pressure and temperature of the gas in the tank can be calculated using the ideal gas equation and the conservation of mass.
n1RT1 / V1 = (n1 + n2)RT2 / V
T2 = (n1 + n2)T1 / n1
T2 = (5 + n2 / 5) × 673.15 / 5
T2 = (n2 / 5) × 134.63 + 673.15
P2 = nRT / V
P2 = (n1 + n2)RT2 / V
P2 = (5 + n2 / 5) × 8.314 × [(n2 / 5) × 134.63 + 673.15] / 5.629
P2 = (n2 / 5) × 327.7 + 206.3 kPa
The change in entropy of the gas can be calculated using the ideal gas equation and the definition of entropy.
ΔS = ∫(dQ / T)
For an ideal gas, dQ = nCv dT
ΔS = nCv ln(T2 / T1)
Cv for a diatomic gas = 5R / 2
ΔS = (5 + n2 / 5) × (5R / 2) ln[(n2 / 5) × 134.63 + 673.15 / 673.15]
ΔS = (5 + n2 / 5) × (5R / 2) ln(n2 / 5 + 1.013)
As the total number of moles in the tank after removal of the partition is not known, we can assume a small change in the number of moles, dn2, and calculate the corresponding change in entropy, dS.
dS = (5 + (n2 + dn2) / 5) × (5R / 2) ln[(n2 + dn2) / 5 + 1.013] - (5 + n2 / 5) × (5R / 2) ln(n2 / 5 + 1.013)
dS = (5R / 2) ln[(n2 + dn2) / n
Attention Mechanical Engineering Students!
To make sure you are not studying endlessly, EduRev has designed Mechanical Engineering study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Mechanical Engineering.
|
Explore Courses for Mechanical Engineering exam
|
|
Similar Mechanical Engineering Doubts
Top Courses for Mechanical EngineeringView all
A rigid insulated tank has two equal parts by partition. First part initially contains 5k mol of an ideal gas at500 kPa, 400C and other part is evacuated. If partition is removed, then the gas fills the entire tank, what will be the total entropy change during this process.a)28.81 kJ/Kb)14.4 kJ/Kc)7.2 kJ/Kd)21.6 kJ/KCorrect answer is option 'A'. Can you explain this answer?
Question Description
A rigid insulated tank has two equal parts by partition. First part initially contains 5k mol of an ideal gas at500 kPa, 400C and other part is evacuated. If partition is removed, then the gas fills the entire tank, what will be the total entropy change during this process.a)28.81 kJ/Kb)14.4 kJ/Kc)7.2 kJ/Kd)21.6 kJ/KCorrect answer is option 'A'. Can you explain this answer? 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 has two equal parts by partition. First part initially contains 5k mol of an ideal gas at500 kPa, 400C and other part is evacuated. If partition is removed, then the gas fills the entire tank, what will be the total entropy change during this process.a)28.81 kJ/Kb)14.4 kJ/Kc)7.2 kJ/Kd)21.6 kJ/KCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rigid insulated tank has two equal parts by partition. First part initially contains 5k mol of an ideal gas at500 kPa, 400C and other part is evacuated. If partition is removed, then the gas fills the entire tank, what will be the total entropy change during this process.a)28.81 kJ/Kb)14.4 kJ/Kc)7.2 kJ/Kd)21.6 kJ/KCorrect answer is option 'A'. Can you explain this answer?.
A rigid insulated tank has two equal parts by partition. First part initially contains 5k mol of an ideal gas at500 kPa, 400C and other part is evacuated. If partition is removed, then the gas fills the entire tank, what will be the total entropy change during this process.a)28.81 kJ/Kb)14.4 kJ/Kc)7.2 kJ/Kd)21.6 kJ/KCorrect answer is option 'A'. Can you explain this answer? 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 has two equal parts by partition. First part initially contains 5k mol of an ideal gas at500 kPa, 400C and other part is evacuated. If partition is removed, then the gas fills the entire tank, what will be the total entropy change during this process.a)28.81 kJ/Kb)14.4 kJ/Kc)7.2 kJ/Kd)21.6 kJ/KCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rigid insulated tank has two equal parts by partition. First part initially contains 5k mol of an ideal gas at500 kPa, 400C and other part is evacuated. If partition is removed, then the gas fills the entire tank, what will be the total entropy change during this process.a)28.81 kJ/Kb)14.4 kJ/Kc)7.2 kJ/Kd)21.6 kJ/KCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A rigid insulated tank has two equal parts by partition. First part initially contains 5k mol of an ideal gas at500 kPa, 400C and other part is evacuated. If partition is removed, then the gas fills the entire tank, what will be the total entropy change during this process.a)28.81 kJ/Kb)14.4 kJ/Kc)7.2 kJ/Kd)21.6 kJ/KCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free.
Here you can find the meaning of A rigid insulated tank has two equal parts by partition. First part initially contains 5k mol of an ideal gas at500 kPa, 400C and other part is evacuated. If partition is removed, then the gas fills the entire tank, what will be the total entropy change during this process.a)28.81 kJ/Kb)14.4 kJ/Kc)7.2 kJ/Kd)21.6 kJ/KCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A rigid insulated tank has two equal parts by partition. First part initially contains 5k mol of an ideal gas at500 kPa, 400C and other part is evacuated. If partition is removed, then the gas fills the entire tank, what will be the total entropy change during this process.a)28.81 kJ/Kb)14.4 kJ/Kc)7.2 kJ/Kd)21.6 kJ/KCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A rigid insulated tank has two equal parts by partition. First part initially contains 5k mol of an ideal gas at500 kPa, 400C and other part is evacuated. If partition is removed, then the gas fills the entire tank, what will be the total entropy change during this process.a)28.81 kJ/Kb)14.4 kJ/Kc)7.2 kJ/Kd)21.6 kJ/KCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A rigid insulated tank has two equal parts by partition. First part initially contains 5k mol of an ideal gas at500 kPa, 400C and other part is evacuated. If partition is removed, then the gas fills the entire tank, what will be the total entropy change during this process.a)28.81 kJ/Kb)14.4 kJ/Kc)7.2 kJ/Kd)21.6 kJ/KCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A rigid insulated tank has two equal parts by partition. First part initially contains 5k mol of an ideal gas at500 kPa, 400C and other part is evacuated. If partition is removed, then the gas fills the entire tank, what will be the total entropy change during this process.a)28.81 kJ/Kb)14.4 kJ/Kc)7.2 kJ/Kd)21.6 kJ/KCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
|
|
Explore Courses for Mechanical Engineering exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup