Class 11 Exam  >  Class 11 Questions  >  In a constant volume calorimeter, 3.5 g of a ... Start Learning for Free
In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is
[IIT JEE 2009]
    Correct answer is '9'. Can you explain this answer?
    Verified Answer
    In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mo...
    Energy released by combustion of 3.5 g gas = 2.5×(298.45-298) kJ
    Energy released by 1 mole of gas = 2.5×0.45/3.5/28 = 9 kJ mol-1
    View all questions of this test
    Most Upvoted Answer
    In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mo...
    Given:
    - Mass of gas = 3.5 g
    - Molar mass of gas = 28 g/mol
    - Temperature change in calorimeter = 0.45 K
    - Heat capacity of calorimeter = 2.5 kJ/K

    To Find:
    The heat given out in kJ/mol

    Solution:

    Step 1: Calculate the moles of the gas:
    Moles of gas = Mass of gas / Molar mass of gas
    Moles of gas = 3.5 g / 28 g/mol
    Moles of gas = 0.125 mol

    Step 2: Calculate the heat absorbed by the calorimeter:
    Heat absorbed by the calorimeter = Heat capacity of calorimeter * Temperature change
    Heat absorbed by the calorimeter = 2.5 kJ/K * 0.45 K
    Heat absorbed by the calorimeter = 1.125 kJ

    Step 3: Calculate the heat given out by the gas:
    Since the reaction is exothermic (heat is given out), the magnitude of heat given out by the gas will be equal to the magnitude of heat absorbed by the calorimeter.
    Heat given out by the gas = Heat absorbed by the calorimeter
    Heat given out by the gas = 1.125 kJ

    Step 4: Calculate the heat given out in kJ/mol:
    Heat given out in kJ/mol = Heat given out by the gas / Moles of gas
    Heat given out in kJ/mol = 1.125 kJ / 0.125 mol
    Heat given out in kJ/mol = 9 kJ/mol

    Answer:
    The numerical value of the heat given out in kJ/mol is 9.
    Free Test
    Community Answer
    In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mo...
    Attention Class 11 Students!
    To make sure you are not studying endlessly, EduRev has designed Class 11 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 11.
    Explore Courses for Class 11 exam

    Similar Class 11 Doubts

    Top Courses for Class 11

    In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer?
    Question Description
    In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. 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 In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. 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 In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer?.
    Solutions for In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. 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 In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer?, a detailed solution for In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer? has been provided alongside types of In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K-1, the numerical value of the heat given out in kJ mol-1 is[IITJEE2009]Correct answer is '9'. Can you explain this answer? tests, examples and also practice Class 11 tests.
    Explore Courses for Class 11 exam

    Top Courses for Class 11

    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