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Assuming that water vapour is an ideal gas, internal energy change (ΔE)when 1 mole of water is vaporised at 1 bar pressure and 100°C will be(Given, molar enthalpy of vaporisation of water at 1 bar and 373 K = 41 kJ mol-1)[IIT JEE 2007]a)4.100 kJ mol-1b)3.7904 kJ mol-1c)37.904 k J mol-1d)41.00 kJ mol-1Correct answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Assuming that water vapour is an ideal gas, internal energy change (ΔE)when 1 mole of water is vaporised at 1 bar pressure and 100°C will be(Given, molar enthalpy of vaporisation of water at 1 bar and 373 K = 41 kJ mol-1)[IIT JEE 2007]a)4.100 kJ mol-1b)3.7904 kJ mol-1c)37.904 k J mol-1d)41.00 kJ mol-1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Assuming that water vapour is an ideal gas, internal energy change (ΔE)when 1 mole of water is vaporised at 1 bar pressure and 100°C will be(Given, molar enthalpy of vaporisation of water at 1 bar and 373 K = 41 kJ mol-1)[IIT JEE 2007]a)4.100 kJ mol-1b)3.7904 kJ mol-1c)37.904 k J mol-1d)41.00 kJ mol-1Correct answer is option 'C'. Can you explain this answer?.

Solutions for Assuming that water vapour is an ideal gas, internal energy change (ΔE)when 1 mole of water is vaporised at 1 bar pressure and 100°C will be(Given, molar enthalpy of vaporisation of water at 1 bar and 373 K = 41 kJ mol-1)[IIT JEE 2007]a)4.100 kJ mol-1b)3.7904 kJ mol-1c)37.904 k J mol-1d)41.00 kJ mol-1Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11. Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.

Here you can find the meaning of Assuming that water vapour is an ideal gas, internal energy change (ΔE)when 1 mole of water is vaporised at 1 bar pressure and 100°C will be(Given, molar enthalpy of vaporisation of water at 1 bar and 373 K = 41 kJ mol-1)[IIT JEE 2007]a)4.100 kJ mol-1b)3.7904 kJ mol-1c)37.904 k J mol-1d)41.00 kJ mol-1Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Assuming that water vapour is an ideal gas, internal energy change (ΔE)when 1 mole of water is vaporised at 1 bar pressure and 100°C will be(Given, molar enthalpy of vaporisation of water at 1 bar and 373 K = 41 kJ mol-1)[IIT JEE 2007]a)4.100 kJ mol-1b)3.7904 kJ mol-1c)37.904 k J mol-1d)41.00 kJ mol-1Correct answer is option 'C'. Can you explain this answer?, a detailed solution for Assuming that water vapour is an ideal gas, internal energy change (ΔE)when 1 mole of water is vaporised at 1 bar pressure and 100°C will be(Given, molar enthalpy of vaporisation of water at 1 bar and 373 K = 41 kJ mol-1)[IIT JEE 2007]a)4.100 kJ mol-1b)3.7904 kJ mol-1c)37.904 k J mol-1d)41.00 kJ mol-1Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of Assuming that water vapour is an ideal gas, internal energy change (ΔE)when 1 mole of water is vaporised at 1 bar pressure and 100°C will be(Given, molar enthalpy of vaporisation of water at 1 bar and 373 K = 41 kJ mol-1)[IIT JEE 2007]a)4.100 kJ mol-1b)3.7904 kJ mol-1c)37.904 k J mol-1d)41.00 kJ mol-1Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Assuming that water vapour is an ideal gas, internal energy change (ΔE)when 1 mole of water is vaporised at 1 bar pressure and 100°C will be(Given, molar enthalpy of vaporisation of water at 1 bar and 373 K = 41 kJ mol-1)[IIT JEE 2007]a)4.100 kJ mol-1b)3.7904 kJ mol-1c)37.904 k J mol-1d)41.00 kJ mol-1Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice Class 11 tests.