JEE Exam > JEE Questions > One mole of an ideal monoatomic gas expands i...
Start Learning for Free
One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is :
[R = 0.082 L atm mol–1 K–1 = 8.3 J mol–1K–1].
[R = 0.082 L atm mol–1 K–1 = 8.3 J mol–1K–1].
- a)0
- b)Rln (24.6)
- c)R ln (2490)
- d)3/2 Rln(24.6)
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
One mole of an ideal monoatomic gas expands isothermally against const...
Most Upvoted Answer
One mole of an ideal monoatomic gas expands isothermally against const...
The entropy change of an ideal gas in an isothermal process can be calculated using the formula:
ΔS = nR ln(Vf/Vi)
where ΔS is the entropy change, n is the number of moles of gas, R is the gas constant, and Vf and Vi are the final and initial volumes of the gas, respectively.
In this case, n = 1 mole, R = 0.082 L atm mol^(-1) K^(-1), Vi = 1 L, and Vf is the final volume of the gas.
Since the final pressure of the gas becomes equal to the external pressure (1 atm), we can use the ideal gas law to find the final volume:
PfVf = nRT
1 atm * Vf = 1 mole * 0.082 L atm mol^(-1) K^(-1) * 300 K
Vf = 24.6 L
Substituting the values into the entropy change formula:
ΔS = (1 mole) * (0.082 L atm mol^(-1) K^(-1)) * ln(24.6 L / 1 L)
ΔS ≈ 2.02 J/K
Therefore, the total entropy change of the system in the given process is approximately 2.02 J/K.
ΔS = nR ln(Vf/Vi)
where ΔS is the entropy change, n is the number of moles of gas, R is the gas constant, and Vf and Vi are the final and initial volumes of the gas, respectively.
In this case, n = 1 mole, R = 0.082 L atm mol^(-1) K^(-1), Vi = 1 L, and Vf is the final volume of the gas.
Since the final pressure of the gas becomes equal to the external pressure (1 atm), we can use the ideal gas law to find the final volume:
PfVf = nRT
1 atm * Vf = 1 mole * 0.082 L atm mol^(-1) K^(-1) * 300 K
Vf = 24.6 L
Substituting the values into the entropy change formula:
ΔS = (1 mole) * (0.082 L atm mol^(-1) K^(-1)) * ln(24.6 L / 1 L)
ΔS ≈ 2.02 J/K
Therefore, the total entropy change of the system in the given process is approximately 2.02 J/K.
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is :[R = 0.082 L atm mol–1 K–1 = 8.3 J mol–1K–1].a)0b)Rln (24.6)c)R ln (2490)d)3/2 Rln(24.6)Correct answer is option 'B'. Can you explain this answer?
Question Description
One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is :[R = 0.082 L atm mol–1 K–1 = 8.3 J mol–1K–1].a)0b)Rln (24.6)c)R ln (2490)d)3/2 Rln(24.6)Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is :[R = 0.082 L atm mol–1 K–1 = 8.3 J mol–1K–1].a)0b)Rln (24.6)c)R ln (2490)d)3/2 Rln(24.6)Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is :[R = 0.082 L atm mol–1 K–1 = 8.3 J mol–1K–1].a)0b)Rln (24.6)c)R ln (2490)d)3/2 Rln(24.6)Correct answer is option 'B'. Can you explain this answer?.
One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is :[R = 0.082 L atm mol–1 K–1 = 8.3 J mol–1K–1].a)0b)Rln (24.6)c)R ln (2490)d)3/2 Rln(24.6)Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is :[R = 0.082 L atm mol–1 K–1 = 8.3 J mol–1K–1].a)0b)Rln (24.6)c)R ln (2490)d)3/2 Rln(24.6)Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is :[R = 0.082 L atm mol–1 K–1 = 8.3 J mol–1K–1].a)0b)Rln (24.6)c)R ln (2490)d)3/2 Rln(24.6)Correct answer is option 'B'. Can you explain this answer?.
Solutions for One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is :[R = 0.082 L atm mol–1 K–1 = 8.3 J mol–1K–1].a)0b)Rln (24.6)c)R ln (2490)d)3/2 Rln(24.6)Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is :[R = 0.082 L atm mol–1 K–1 = 8.3 J mol–1K–1].a)0b)Rln (24.6)c)R ln (2490)d)3/2 Rln(24.6)Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is :[R = 0.082 L atm mol–1 K–1 = 8.3 J mol–1K–1].a)0b)Rln (24.6)c)R ln (2490)d)3/2 Rln(24.6)Correct answer is option 'B'. Can you explain this answer?, a detailed solution for One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is :[R = 0.082 L atm mol–1 K–1 = 8.3 J mol–1K–1].a)0b)Rln (24.6)c)R ln (2490)d)3/2 Rln(24.6)Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is :[R = 0.082 L atm mol–1 K–1 = 8.3 J mol–1K–1].a)0b)Rln (24.6)c)R ln (2490)d)3/2 Rln(24.6)Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is :[R = 0.082 L atm mol–1 K–1 = 8.3 J mol–1K–1].a)0b)Rln (24.6)c)R ln (2490)d)3/2 Rln(24.6)Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE 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