NEET Exam  >  NEET 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 10.0Kcal/mol. What will be the change in internal energy(∆E) of 3 mol of liquid at same temperature. Ansr :27.0 Kcal?
Most Upvoted Answer
Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is...
3 H2O ( L) is converted into 3 H2O(g)...then according to formula.. delta H = delta E + delta ngRT..and in this question we have to find delta E ..then delta E = delta H - delta ng RT..so...by putting values we get the answer... 3×10^-3 ×500 × 0.002 = 27.0 kcal...I hope that helps you...
Attention NEET Students!
To make sure you are not studying endlessly, EduRev has designed NEET study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in NEET.
Explore Courses for NEET exam

Top Courses for NEET

Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0Kcal/mol. What will be the change in internal energy(∆E) of 3 mol of liquid at same temperature. Ansr :27.0 Kcal?
Question Description
Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0Kcal/mol. What will be the change in internal energy(∆E) of 3 mol of liquid at same temperature. Ansr :27.0 Kcal? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0Kcal/mol. What will be the change in internal energy(∆E) of 3 mol of liquid at same temperature. Ansr :27.0 Kcal? covers all topics & solutions for NEET 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 10.0Kcal/mol. What will be the change in internal energy(∆E) of 3 mol of liquid at same temperature. Ansr :27.0 Kcal?.
Solutions for Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0Kcal/mol. What will be the change in internal energy(∆E) of 3 mol of liquid at same temperature. Ansr :27.0 Kcal? in English & in Hindi are available as part of our courses for NEET. Download more important topics, notes, lectures and mock test series for NEET 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 10.0Kcal/mol. What will be the change in internal energy(∆E) of 3 mol of liquid at same temperature. Ansr :27.0 Kcal? 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 10.0Kcal/mol. What will be the change in internal energy(∆E) of 3 mol of liquid at same temperature. Ansr :27.0 Kcal?, a detailed solution for Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0Kcal/mol. What will be the change in internal energy(∆E) of 3 mol of liquid at same temperature. Ansr :27.0 Kcal? has been provided alongside types of Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is 10.0Kcal/mol. What will be the change in internal energy(∆E) of 3 mol of liquid at same temperature. Ansr :27.0 Kcal? 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 10.0Kcal/mol. What will be the change in internal energy(∆E) of 3 mol of liquid at same temperature. Ansr :27.0 Kcal? tests, examples and also practice NEET tests.
Explore Courses for NEET exam

Top Courses for NEET

Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev