• What amount of heat must be supplied to 2.0 x 10-2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N2 = 28; R = 8.3 J mol-1 K-1.)

  • What amount of heat must be supplied to | Class 11 Physics Chapter Thermodynamics, Thermodynamics NCERT Solutions

    Q2.

    What amount of heat must be supplied to 2.0 x 10-2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N2 = 28; R = 8.3 J mol-1 K-1.)

    Mass of nitrogen, m = 2.0 × 10-2 kg = 20 g

    Rise in temperature, ΔT = 45°C

    Molecular mass of N2, M = 28

    Universal gas constant, R = 8.3 J mol-1 K-1

    Number of moles, n  =  m / M

    = 2.0  x 10-2 x 103  /  28  =  0.714

    Molar specific heat at constant pressure for nitrogen, CP   =   7/2R

    = 7/2 x 8.3

    = 29.05 J mol-1 K-1

    The total amount of heat to be supplied is given by the relation:

    ΔQ = nCP ΔT

    = 0.714 × 29.05 × 45

    = 933.38 J

    Therefore, the amount of heat to be supplied is 933.38 J.

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    What is the correct answer to: What amount of heat must be supplied to 2.0 x 10-2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N2 = 28; R = 8.3 J mol-1 K-1.)?

    Mass of nitrogen, m = 2.0 × 10-2 kg = 20 g

    Rise in temperature, ΔT = 45°C

    Molecular mass of N2, M = 28

    Universal gas constant, R = 8.3 J mol-1 K-1

    Number of moles, n  =  m / M

    = ...

    How do you solve What amount of heat must be supplied to 2.0 x 10-2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N2 = 28; R = 8.3 J mol-1 K-1.) step by step?

    Step-by-step explanation:
    • Mass of nitrogen, m = 2
    • 0 × 10-2 kg = 20 g

    • Rise in temperature, ΔT = 45°C

    What common mistakes should be avoided in this question?

    Students often lose marks by skipping steps, writing incomplete explanations, or misunderstanding keywords used in the question.

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