IIT JAM Exam > IIT JAM Questions > A di-atomic gas undergoes adiabatic expansion...
Start Learning for Free
A di-atomic gas undergoes adiabatic expansion against the piston of a cylinder. As a result, the temperature of the gas drops from 1150 K to 400 K. The number of moles of the gas required to obtain 2300 J of work from the expansion is ________.
(The gas constant R = 8.314 J mol–1K–1.)
(Round off to 2 decimal places)
(The gas constant R = 8.314 J mol–1K–1.)
(Round off to 2 decimal places)
Correct answer is between '0.14,0.16'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
A di-atomic gas undergoes adiabatic expansion against the piston of a ...
To solve this problem, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:
ΔU = Q - W
In an adiabatic process, no heat is transferred, so Q = 0. Therefore, the equation becomes:
ΔU = -W
The change in internal energy can be expressed as:
ΔU = nCvΔT
Where n is the number of moles of the gas, Cv is the molar specific heat at constant volume, and ΔT is the change in temperature. For a diatomic gas, Cv = (5/2)R.
Plugging in the given values:
ΔU = nCvΔT
ΔU = n(5/2)R(400 - 1150)
We also know that ΔU = -W, so:
n(5/2)R(400 - 1150) = -2300
Simplifying:
n(5/2)(-750) = -2300
n = -2300 / ((5/2)(-750))
n = 0.12267
Therefore, the number of moles of the gas required to obtain 2300 J of work from the expansion is approximately 0.12267 moles.
ΔU = Q - W
In an adiabatic process, no heat is transferred, so Q = 0. Therefore, the equation becomes:
ΔU = -W
The change in internal energy can be expressed as:
ΔU = nCvΔT
Where n is the number of moles of the gas, Cv is the molar specific heat at constant volume, and ΔT is the change in temperature. For a diatomic gas, Cv = (5/2)R.
Plugging in the given values:
ΔU = nCvΔT
ΔU = n(5/2)R(400 - 1150)
We also know that ΔU = -W, so:
n(5/2)R(400 - 1150) = -2300
Simplifying:
n(5/2)(-750) = -2300
n = -2300 / ((5/2)(-750))
n = 0.12267
Therefore, the number of moles of the gas required to obtain 2300 J of work from the expansion is approximately 0.12267 moles.
|
Explore Courses for IIT JAM exam
|
|
Similar IIT JAM Doubts
A di-atomic gas undergoes adiabatic expansion against the piston of a cylinder. As a result, the temperature of the gas drops from 1150 K to 400 K. The number of moles of the gas required to obtain 2300 J of work from the expansion is ________.(The gas constant R = 8.314 J mol–1K–1.)(Round off to 2 decimal places)Correct answer is between '0.14,0.16'. Can you explain this answer?
Question Description
A di-atomic gas undergoes adiabatic expansion against the piston of a cylinder. As a result, the temperature of the gas drops from 1150 K to 400 K. The number of moles of the gas required to obtain 2300 J of work from the expansion is ________.(The gas constant R = 8.314 J mol–1K–1.)(Round off to 2 decimal places)Correct answer is between '0.14,0.16'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A di-atomic gas undergoes adiabatic expansion against the piston of a cylinder. As a result, the temperature of the gas drops from 1150 K to 400 K. The number of moles of the gas required to obtain 2300 J of work from the expansion is ________.(The gas constant R = 8.314 J mol–1K–1.)(Round off to 2 decimal places)Correct answer is between '0.14,0.16'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A di-atomic gas undergoes adiabatic expansion against the piston of a cylinder. As a result, the temperature of the gas drops from 1150 K to 400 K. The number of moles of the gas required to obtain 2300 J of work from the expansion is ________.(The gas constant R = 8.314 J mol–1K–1.)(Round off to 2 decimal places)Correct answer is between '0.14,0.16'. Can you explain this answer?.
A di-atomic gas undergoes adiabatic expansion against the piston of a cylinder. As a result, the temperature of the gas drops from 1150 K to 400 K. The number of moles of the gas required to obtain 2300 J of work from the expansion is ________.(The gas constant R = 8.314 J mol–1K–1.)(Round off to 2 decimal places)Correct answer is between '0.14,0.16'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A di-atomic gas undergoes adiabatic expansion against the piston of a cylinder. As a result, the temperature of the gas drops from 1150 K to 400 K. The number of moles of the gas required to obtain 2300 J of work from the expansion is ________.(The gas constant R = 8.314 J mol–1K–1.)(Round off to 2 decimal places)Correct answer is between '0.14,0.16'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A di-atomic gas undergoes adiabatic expansion against the piston of a cylinder. As a result, the temperature of the gas drops from 1150 K to 400 K. The number of moles of the gas required to obtain 2300 J of work from the expansion is ________.(The gas constant R = 8.314 J mol–1K–1.)(Round off to 2 decimal places)Correct answer is between '0.14,0.16'. Can you explain this answer?.
Solutions for A di-atomic gas undergoes adiabatic expansion against the piston of a cylinder. As a result, the temperature of the gas drops from 1150 K to 400 K. The number of moles of the gas required to obtain 2300 J of work from the expansion is ________.(The gas constant R = 8.314 J mol–1K–1.)(Round off to 2 decimal places)Correct answer is between '0.14,0.16'. Can you explain this answer? in English & in Hindi are available as part of our courses for IIT JAM.
Download more important topics, notes, lectures and mock test series for IIT JAM Exam by signing up for free.
Here you can find the meaning of A di-atomic gas undergoes adiabatic expansion against the piston of a cylinder. As a result, the temperature of the gas drops from 1150 K to 400 K. The number of moles of the gas required to obtain 2300 J of work from the expansion is ________.(The gas constant R = 8.314 J mol–1K–1.)(Round off to 2 decimal places)Correct answer is between '0.14,0.16'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A di-atomic gas undergoes adiabatic expansion against the piston of a cylinder. As a result, the temperature of the gas drops from 1150 K to 400 K. The number of moles of the gas required to obtain 2300 J of work from the expansion is ________.(The gas constant R = 8.314 J mol–1K–1.)(Round off to 2 decimal places)Correct answer is between '0.14,0.16'. Can you explain this answer?, a detailed solution for A di-atomic gas undergoes adiabatic expansion against the piston of a cylinder. As a result, the temperature of the gas drops from 1150 K to 400 K. The number of moles of the gas required to obtain 2300 J of work from the expansion is ________.(The gas constant R = 8.314 J mol–1K–1.)(Round off to 2 decimal places)Correct answer is between '0.14,0.16'. Can you explain this answer? has been provided alongside types of A di-atomic gas undergoes adiabatic expansion against the piston of a cylinder. As a result, the temperature of the gas drops from 1150 K to 400 K. The number of moles of the gas required to obtain 2300 J of work from the expansion is ________.(The gas constant R = 8.314 J mol–1K–1.)(Round off to 2 decimal places)Correct answer is between '0.14,0.16'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A di-atomic gas undergoes adiabatic expansion against the piston of a cylinder. As a result, the temperature of the gas drops from 1150 K to 400 K. The number of moles of the gas required to obtain 2300 J of work from the expansion is ________.(The gas constant R = 8.314 J mol–1K–1.)(Round off to 2 decimal places)Correct answer is between '0.14,0.16'. Can you explain this answer? tests, examples and also practice IIT JAM tests.
|
|
Explore Courses for IIT JAM 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
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup