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Calculate the molar specific heat at constant volume. Given : specific heat of hydrogen at constant pressure is 6.85 Cal /mol/K and density of hydrogen = 0.8999 g/dm^3. One mole of gas = 2.016 g, J= 4.2 *10 ^7 erg/Cal and 1 atmosphere = 10^6dyne /cm^2. Answer is 4.89 cal/mol/K. How???!!!!? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Calculate the molar specific heat at constant volume. Given : specific heat of hydrogen at constant pressure is 6.85 Cal /mol/K and density of hydrogen = 0.8999 g/dm^3. One mole of gas = 2.016 g, J= 4.2 *10 ^7 erg/Cal and 1 atmosphere = 10^6dyne /cm^2. Answer is 4.89 cal/mol/K. How???!!!!? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the molar specific heat at constant volume. Given : specific heat of hydrogen at constant pressure is 6.85 Cal /mol/K and density of hydrogen = 0.8999 g/dm^3. One mole of gas = 2.016 g, J= 4.2 *10 ^7 erg/Cal and 1 atmosphere = 10^6dyne /cm^2. Answer is 4.89 cal/mol/K. How???!!!!?.

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Here you can find the meaning of Calculate the molar specific heat at constant volume. Given : specific heat of hydrogen at constant pressure is 6.85 Cal /mol/K and density of hydrogen = 0.8999 g/dm^3. One mole of gas = 2.016 g, J= 4.2 *10 ^7 erg/Cal and 1 atmosphere = 10^6dyne /cm^2. Answer is 4.89 cal/mol/K. How???!!!!? defined & explained in the simplest way possible. Besides giving the explanation of Calculate the molar specific heat at constant volume. Given : specific heat of hydrogen at constant pressure is 6.85 Cal /mol/K and density of hydrogen = 0.8999 g/dm^3. One mole of gas = 2.016 g, J= 4.2 *10 ^7 erg/Cal and 1 atmosphere = 10^6dyne /cm^2. Answer is 4.89 cal/mol/K. How???!!!!?, a detailed solution for Calculate the molar specific heat at constant volume. Given : specific heat of hydrogen at constant pressure is 6.85 Cal /mol/K and density of hydrogen = 0.8999 g/dm^3. One mole of gas = 2.016 g, J= 4.2 *10 ^7 erg/Cal and 1 atmosphere = 10^6dyne /cm^2. Answer is 4.89 cal/mol/K. How???!!!!? has been provided alongside types of Calculate the molar specific heat at constant volume. Given : specific heat of hydrogen at constant pressure is 6.85 Cal /mol/K and density of hydrogen = 0.8999 g/dm^3. One mole of gas = 2.016 g, J= 4.2 *10 ^7 erg/Cal and 1 atmosphere = 10^6dyne /cm^2. Answer is 4.89 cal/mol/K. How???!!!!? theory, EduRev gives you an ample number of questions to practice Calculate the molar specific heat at constant volume. Given : specific heat of hydrogen at constant pressure is 6.85 Cal /mol/K and density of hydrogen = 0.8999 g/dm^3. One mole of gas = 2.016 g, J= 4.2 *10 ^7 erg/Cal and 1 atmosphere = 10^6dyne /cm^2. Answer is 4.89 cal/mol/K. How???!!!!? tests, examples and also practice NEET tests.