AIIMS2019Physics-Thermodynamics

AIIMS 2019 Physics Entropy MCQ Question

Type: MCQ-numerical-Medium-Class 11

N2\text{N}_2 gas is heated from 300 K temperature to 600 K through an isobaric process. Then find the change in entropy of the gas. (n=1mole)

A

10 J/k10\text{ J/k}

B

20 J/k20\text{ J/k}

C

30 J/k30\text{ J/k}

D

40 J/k40\text{ J/k}

Correct Answer

Option B

Detailed Explanation

The calculation of the change in entropy (ΔS) for the isobaric process yields approximately 20 J/K, which aligns with option B. In this scenario, the values for the number of moles (n = 1), heat capacity at constant pressure (Cₚ = 7/2 R), and the temperatures (T₁ = 300 K and T₂ = 600 K) were correctly substituted into the formula ΔS = nCₚ ln(T₂/T₁). Other options are not applicable as they do not provide relevant values or calculations related to the entropy change in this context. Understanding this calculation is crucial for grasping the thermodynamic principles governing entropy in isobaric processes.

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