Class 11 Exam > Class 11 Questions > One mole of an ideal monatomic gas is at an i...
Start Learning for Free
One mole of an ideal monatomic gas is at an initial temperature of 300 K. The gas undergoes an isovolumetric process, acquiring 500 J of energy by heat. It then undergoes an isobaric process, losing this same amount of energy by heat. Determine the work done on the gas.
- a)123 J
- b)231 J
- c)333 J
- d)200 J
Correct answer is option 'D'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
One mole of an ideal monatomic gas is at an initial temperature of 300...
Explanation:
At const volume,
Q = 500 J
Q=nCPΔT
500=1×2.5×8.31ΔT
ΔT=24.06
W=nRΔT=1×8.31×24.06=200J
Most Upvoted Answer
One mole of an ideal monatomic gas is at an initial temperature of 300...
Given information:
- Initial temperature of the gas (T1) = 300 K
- Heat added to the gas in the isovolumetric process (Q1) = 500 J
- Heat removed from the gas in the isobaric process (Q2) = -500 J
To find: The work done on the gas
Formula:
The first law of thermodynamics states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W):
ΔU = Q - W
Approach:
1. Calculate the change in internal energy (ΔU) of the gas in the isovolumetric process.
2. Calculate the work done on the gas in the isovolumetric process.
3. Calculate the change in internal energy (ΔU) of the gas in the isobaric process.
4. Calculate the work done on the gas in the isobaric process.
5. Add the work done in both processes to find the total work done on the gas.
Calculations:
1. Change in internal energy (ΔU) in the isovolumetric process:
ΔU1 = Q1
ΔU1 = 500 J
2. Work done on the gas in the isovolumetric process:
In an isovolumetric process, the volume (V) remains constant, so the work done (W1) is zero.
W1 = 0
3. Change in internal energy (ΔU) in the isobaric process:
Using the ideal gas law, we can relate the change in internal energy to the change in temperature (ΔT):
ΔU2 = nCvΔT
Since the gas is monatomic, the molar specific heat at constant volume (Cv) is given by:
Cv = (3/2)R, where R is the ideal gas constant.
ΔT = T2 - T1
ΔT = 0 - 300
ΔT = -300 K
ΔU2 = (1 mol)(3/2)(8.314 J/mol·K)(-300 K)
ΔU2 = -3737 J
4. Work done on the gas in the isobaric process:
In an isobaric process, the pressure (P) remains constant, so the work done (W2) is given by:
W2 = -PΔV
Since the volume (V) remains constant, the change in volume (ΔV) is zero.
W2 = 0
5. Total work done on the gas:
Total work done = W1 + W2
Total work done = 0 + 0
Total work done = 0 J
Therefore, the correct answer is option D) 200 J.
- Initial temperature of the gas (T1) = 300 K
- Heat added to the gas in the isovolumetric process (Q1) = 500 J
- Heat removed from the gas in the isobaric process (Q2) = -500 J
To find: The work done on the gas
Formula:
The first law of thermodynamics states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W):
ΔU = Q - W
Approach:
1. Calculate the change in internal energy (ΔU) of the gas in the isovolumetric process.
2. Calculate the work done on the gas in the isovolumetric process.
3. Calculate the change in internal energy (ΔU) of the gas in the isobaric process.
4. Calculate the work done on the gas in the isobaric process.
5. Add the work done in both processes to find the total work done on the gas.
Calculations:
1. Change in internal energy (ΔU) in the isovolumetric process:
ΔU1 = Q1
ΔU1 = 500 J
2. Work done on the gas in the isovolumetric process:
In an isovolumetric process, the volume (V) remains constant, so the work done (W1) is zero.
W1 = 0
3. Change in internal energy (ΔU) in the isobaric process:
Using the ideal gas law, we can relate the change in internal energy to the change in temperature (ΔT):
ΔU2 = nCvΔT
Since the gas is monatomic, the molar specific heat at constant volume (Cv) is given by:
Cv = (3/2)R, where R is the ideal gas constant.
ΔT = T2 - T1
ΔT = 0 - 300
ΔT = -300 K
ΔU2 = (1 mol)(3/2)(8.314 J/mol·K)(-300 K)
ΔU2 = -3737 J
4. Work done on the gas in the isobaric process:
In an isobaric process, the pressure (P) remains constant, so the work done (W2) is given by:
W2 = -PΔV
Since the volume (V) remains constant, the change in volume (ΔV) is zero.
W2 = 0
5. Total work done on the gas:
Total work done = W1 + W2
Total work done = 0 + 0
Total work done = 0 J
Therefore, the correct answer is option D) 200 J.
Attention Class 11 Students!
To make sure you are not studying endlessly, EduRev has designed Class 11 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 11.
|
Explore Courses for Class 11 exam
|
|
Similar Class 11 Doubts
Top Courses for Class 11View all
One mole of an ideal monatomic gas is at an initial temperature of 300 K. The gas undergoes an isovolumetric process, acquiring 500 J of energy by heat. It then undergoes an isobaric process, losing this same amount of energy by heat. Determine the work done on the gas.a)123 Jb)231 Jc)333 Jd)200 JCorrect answer is option 'D'. Can you explain this answer?
Question Description
One mole of an ideal monatomic gas is at an initial temperature of 300 K. The gas undergoes an isovolumetric process, acquiring 500 J of energy by heat. It then undergoes an isobaric process, losing this same amount of energy by heat. Determine the work done on the gas.a)123 Jb)231 Jc)333 Jd)200 JCorrect answer is option 'D'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about One mole of an ideal monatomic gas is at an initial temperature of 300 K. The gas undergoes an isovolumetric process, acquiring 500 J of energy by heat. It then undergoes an isobaric process, losing this same amount of energy by heat. Determine the work done on the gas.a)123 Jb)231 Jc)333 Jd)200 JCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of an ideal monatomic gas is at an initial temperature of 300 K. The gas undergoes an isovolumetric process, acquiring 500 J of energy by heat. It then undergoes an isobaric process, losing this same amount of energy by heat. Determine the work done on the gas.a)123 Jb)231 Jc)333 Jd)200 JCorrect answer is option 'D'. Can you explain this answer?.
One mole of an ideal monatomic gas is at an initial temperature of 300 K. The gas undergoes an isovolumetric process, acquiring 500 J of energy by heat. It then undergoes an isobaric process, losing this same amount of energy by heat. Determine the work done on the gas.a)123 Jb)231 Jc)333 Jd)200 JCorrect answer is option 'D'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about One mole of an ideal monatomic gas is at an initial temperature of 300 K. The gas undergoes an isovolumetric process, acquiring 500 J of energy by heat. It then undergoes an isobaric process, losing this same amount of energy by heat. Determine the work done on the gas.a)123 Jb)231 Jc)333 Jd)200 JCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of an ideal monatomic gas is at an initial temperature of 300 K. The gas undergoes an isovolumetric process, acquiring 500 J of energy by heat. It then undergoes an isobaric process, losing this same amount of energy by heat. Determine the work done on the gas.a)123 Jb)231 Jc)333 Jd)200 JCorrect answer is option 'D'. Can you explain this answer?.
Solutions for One mole of an ideal monatomic gas is at an initial temperature of 300 K. The gas undergoes an isovolumetric process, acquiring 500 J of energy by heat. It then undergoes an isobaric process, losing this same amount of energy by heat. Determine the work done on the gas.a)123 Jb)231 Jc)333 Jd)200 JCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.
Here you can find the meaning of One mole of an ideal monatomic gas is at an initial temperature of 300 K. The gas undergoes an isovolumetric process, acquiring 500 J of energy by heat. It then undergoes an isobaric process, losing this same amount of energy by heat. Determine the work done on the gas.a)123 Jb)231 Jc)333 Jd)200 JCorrect answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
One mole of an ideal monatomic gas is at an initial temperature of 300 K. The gas undergoes an isovolumetric process, acquiring 500 J of energy by heat. It then undergoes an isobaric process, losing this same amount of energy by heat. Determine the work done on the gas.a)123 Jb)231 Jc)333 Jd)200 JCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for One mole of an ideal monatomic gas is at an initial temperature of 300 K. The gas undergoes an isovolumetric process, acquiring 500 J of energy by heat. It then undergoes an isobaric process, losing this same amount of energy by heat. Determine the work done on the gas.a)123 Jb)231 Jc)333 Jd)200 JCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of One mole of an ideal monatomic gas is at an initial temperature of 300 K. The gas undergoes an isovolumetric process, acquiring 500 J of energy by heat. It then undergoes an isobaric process, losing this same amount of energy by heat. Determine the work done on the gas.a)123 Jb)231 Jc)333 Jd)200 JCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice One mole of an ideal monatomic gas is at an initial temperature of 300 K. The gas undergoes an isovolumetric process, acquiring 500 J of energy by heat. It then undergoes an isobaric process, losing this same amount of energy by heat. Determine the work done on the gas.a)123 Jb)231 Jc)333 Jd)200 JCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice Class 11 tests.
|
|
Explore Courses for Class 11 exam
|
|
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