JEE Exam > JEE Questions > Latent heat of vaporisation of a liquid at 50...
Start Learning for Free
Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0 kcal/mol. What will be the change in internal energy (ΔE) of 3 mol of liquid at same temperature?
- a)13.0 kcal
- b)-13.0 kcal
- c)27.0 kcal
- d)-27.0 kcal
Correct answer is option 'C'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is...
Vaporization of 3 moles of H2O vapors is:
3H2O(l) → 3H2O(g)
Δn = 3 - 0 = 3
Therefore,
ΔU = ΔH - ΔnRT
= (3 x10) - 3 (0.002) (500) = 27
change in internal energy is:
ΔU = 27 Kcal
3H2O(l) → 3H2O(g)
Δn = 3 - 0 = 3
Therefore,
ΔU = ΔH - ΔnRT
= (3 x10) - 3 (0.002) (500) = 27
change in internal energy is:
ΔU = 27 Kcal
Most Upvoted Answer
Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is...
Given data:
- Latent heat of vaporisation at 500 K and 1 atm pressure = 10.0 kcal/mol
- Number of moles of liquid = 3 mol
Calculating Change in Internal Energy:
- The change in internal energy (ΔE) when a substance changes from liquid to vapor phase is given by the latent heat of vaporisation.
- Therefore, the change in internal energy for 1 mole of liquid at 500 K and 1 atm pressure is 10.0 kcal.
- For 3 moles of liquid, the total change in internal energy will be 3 times the change for 1 mole.
- ΔE = 3 * 10.0 kcal = 30.0 kcal
Final Answer:
Therefore, the change in internal energy of 3 moles of liquid at the same temperature is 30.0 kcal.
- Latent heat of vaporisation at 500 K and 1 atm pressure = 10.0 kcal/mol
- Number of moles of liquid = 3 mol
Calculating Change in Internal Energy:
- The change in internal energy (ΔE) when a substance changes from liquid to vapor phase is given by the latent heat of vaporisation.
- Therefore, the change in internal energy for 1 mole of liquid at 500 K and 1 atm pressure is 10.0 kcal.
- For 3 moles of liquid, the total change in internal energy will be 3 times the change for 1 mole.
- ΔE = 3 * 10.0 kcal = 30.0 kcal
Final Answer:
Therefore, the change in internal energy of 3 moles of liquid at the same temperature is 30.0 kcal.
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
Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0 kcal/mol. What will be the change in internal energy (ΔE) of 3 mol of liquid at same temperature?a)13.0 kcalb)-13.0 kcalc)27.0 kcald)-27.0 kcalCorrect answer is option 'C'. Can you explain this answer?
Question Description
Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0 kcal/mol. What will be the change in internal energy (ΔE) of 3 mol of liquid at same temperature?a)13.0 kcalb)-13.0 kcalc)27.0 kcald)-27.0 kcalCorrect answer is option 'C'. 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 Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0 kcal/mol. What will be the change in internal energy (ΔE) of 3 mol of liquid at same temperature?a)13.0 kcalb)-13.0 kcalc)27.0 kcald)-27.0 kcalCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0 kcal/mol. What will be the change in internal energy (ΔE) of 3 mol of liquid at same temperature?a)13.0 kcalb)-13.0 kcalc)27.0 kcald)-27.0 kcalCorrect answer is option 'C'. Can you explain this answer?.
Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0 kcal/mol. What will be the change in internal energy (ΔE) of 3 mol of liquid at same temperature?a)13.0 kcalb)-13.0 kcalc)27.0 kcald)-27.0 kcalCorrect answer is option 'C'. 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 Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0 kcal/mol. What will be the change in internal energy (ΔE) of 3 mol of liquid at same temperature?a)13.0 kcalb)-13.0 kcalc)27.0 kcald)-27.0 kcalCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0 kcal/mol. What will be the change in internal energy (ΔE) of 3 mol of liquid at same temperature?a)13.0 kcalb)-13.0 kcalc)27.0 kcald)-27.0 kcalCorrect answer is option 'C'. Can you explain this answer?.
Solutions for Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0 kcal/mol. What will be the change in internal energy (ΔE) of 3 mol of liquid at same temperature?a)13.0 kcalb)-13.0 kcalc)27.0 kcald)-27.0 kcalCorrect answer is option 'C'. 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 Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0 kcal/mol. What will be the change in internal energy (ΔE) of 3 mol of liquid at same temperature?a)13.0 kcalb)-13.0 kcalc)27.0 kcald)-27.0 kcalCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0 kcal/mol. What will be the change in internal energy (ΔE) of 3 mol of liquid at same temperature?a)13.0 kcalb)-13.0 kcalc)27.0 kcald)-27.0 kcalCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0 kcal/mol. What will be the change in internal energy (ΔE) of 3 mol of liquid at same temperature?a)13.0 kcalb)-13.0 kcalc)27.0 kcald)-27.0 kcalCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0 kcal/mol. What will be the change in internal energy (ΔE) of 3 mol of liquid at same temperature?a)13.0 kcalb)-13.0 kcalc)27.0 kcald)-27.0 kcalCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0 kcal/mol. What will be the change in internal energy (ΔE) of 3 mol of liquid at same temperature?a)13.0 kcalb)-13.0 kcalc)27.0 kcald)-27.0 kcalCorrect answer is option 'C'. 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
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup