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25 grams of water vaporises at 373 K against a constant external pressure of 1 atm. The molar enthalpy of vaporisation is 9.72 kcal/mol. If steam obeys perfect gas laws, then work done against the atmosphere and change of internal energy in the above process, respectively, area)-1036.194 cal and 12,464.886 calb)921.4 cal and 11,074 calc)1029.4 cal and 12,470.6 cald)1129.3 cal and 10,207 calCorrect answer is option 'A'. 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 25 grams of water vaporises at 373 K against a constant external pressure of 1 atm. The molar enthalpy of vaporisation is 9.72 kcal/mol. If steam obeys perfect gas laws, then work done against the atmosphere and change of internal energy in the above process, respectively, area)-1036.194 cal and 12,464.886 calb)921.4 cal and 11,074 calc)1029.4 cal and 12,470.6 cald)1129.3 cal and 10,207 calCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 25 grams of water vaporises at 373 K against a constant external pressure of 1 atm. The molar enthalpy of vaporisation is 9.72 kcal/mol. If steam obeys perfect gas laws, then work done against the atmosphere and change of internal energy in the above process, respectively, area)-1036.194 cal and 12,464.886 calb)921.4 cal and 11,074 calc)1029.4 cal and 12,470.6 cald)1129.3 cal and 10,207 calCorrect answer is option 'A'. Can you explain this answer?.

Solutions for 25 grams of water vaporises at 373 K against a constant external pressure of 1 atm. The molar enthalpy of vaporisation is 9.72 kcal/mol. If steam obeys perfect gas laws, then work done against the atmosphere and change of internal energy in the above process, respectively, area)-1036.194 cal and 12,464.886 calb)921.4 cal and 11,074 calc)1029.4 cal and 12,470.6 cald)1129.3 cal and 10,207 calCorrect answer is option 'A'. 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 25 grams of water vaporises at 373 K against a constant external pressure of 1 atm. The molar enthalpy of vaporisation is 9.72 kcal/mol. If steam obeys perfect gas laws, then work done against the atmosphere and change of internal energy in the above process, respectively, area)-1036.194 cal and 12,464.886 calb)921.4 cal and 11,074 calc)1029.4 cal and 12,470.6 cald)1129.3 cal and 10,207 calCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of 25 grams of water vaporises at 373 K against a constant external pressure of 1 atm. The molar enthalpy of vaporisation is 9.72 kcal/mol. If steam obeys perfect gas laws, then work done against the atmosphere and change of internal energy in the above process, respectively, area)-1036.194 cal and 12,464.886 calb)921.4 cal and 11,074 calc)1029.4 cal and 12,470.6 cald)1129.3 cal and 10,207 calCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for 25 grams of water vaporises at 373 K against a constant external pressure of 1 atm. The molar enthalpy of vaporisation is 9.72 kcal/mol. If steam obeys perfect gas laws, then work done against the atmosphere and change of internal energy in the above process, respectively, area)-1036.194 cal and 12,464.886 calb)921.4 cal and 11,074 calc)1029.4 cal and 12,470.6 cald)1129.3 cal and 10,207 calCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of 25 grams of water vaporises at 373 K against a constant external pressure of 1 atm. The molar enthalpy of vaporisation is 9.72 kcal/mol. If steam obeys perfect gas laws, then work done against the atmosphere and change of internal energy in the above process, respectively, area)-1036.194 cal and 12,464.886 calb)921.4 cal and 11,074 calc)1029.4 cal and 12,470.6 cald)1129.3 cal and 10,207 calCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice 25 grams of water vaporises at 373 K against a constant external pressure of 1 atm. The molar enthalpy of vaporisation is 9.72 kcal/mol. If steam obeys perfect gas laws, then work done against the atmosphere and change of internal energy in the above process, respectively, area)-1036.194 cal and 12,464.886 calb)921.4 cal and 11,074 calc)1029.4 cal and 12,470.6 cald)1129.3 cal and 10,207 calCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.