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Chemistry - Thermodynamics - II Question with Solution | TestHub

ChemistryThermodynamics - IICalculation of DSEasy2 minPYQ_2012

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The entropy of a sample of a certain substance increases by $0.836 J K^{- 1}$ on adding reversibly $0.3344 J$ of heat at constant temperature. The temperature of the sample is:

Options:

Answer:

Solution:

$Δ S = \frac{Δ H}{T}; T = \frac{Δ H}{Δ S} = \frac{0.3344}{0.836} = 0.4 K$

Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Calculation of DS

⏱ 2mℹ️ Source: PYQ_2012

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