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Heat and Thermodynamics question

2021 · 20 Jul · Shift 1 · Q50
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Heat and Thermodynamics question

2021 · 20 Jul · Shift 1 · Q50

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
The amount of heat needed to raise the temperature of 4 moles of rigid diatomic gas from 0 ∘^\circ∘ C to 50 ∘^\circ∘ C when no work is done is ‾\underline{\hspace{2cm}}​. (R is the universal gas constant).
  1. A
    500 R
  2. B
    250 R
  3. C
    750 R
  4. D
    175 R
View written solutionFree

Correct answer: A

  1. Identify the process

    The gas is in a rigid container, so volume is constant.

    Hence, W=intP dV=0W = \\int P\,dV = 0W=intPdV=0 and the heat supplied goes entirely into changing internal energy: Q=nCVΔTQ = n C_V \Delta TQ=nCV​ΔT

  2. Heat capacity of a diatomic gas

    For a rigid diatomic ideal gas (ignoring vibrational modes in this temperature range), CV=52RC_V = \frac{5}{2}RCV​=25​R

  3. Given data

    • Number of moles: n=4n = 4n=4
    • Temperature rise: ΔT=50∘C−0∘C=50 K\Delta T = 50^\circ C - 0^\circ C = 50\,KΔT=50∘C−0∘C=50K
  4. Compute heat required

    Q=nCVΔTQ = n C_V \Delta TQ=nCV​ΔT Q=4×52R×50Q = 4 \times \frac{5}{2}R \times 50Q=4×25​R×50

    Simplifying, Q=4×125R=500RQ = 4 \times 125R = 500RQ=4×125R=500R

  5. Match with options

    Q=500RQ = 500RQ=500R

    So the correct option is A.

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