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Chemistry - Thermodynamics - II Question with Solution | TestHub

ChemistryThermodynamics - IICalculation of DSMedium2 minPYQ_2023
ChemistryMediumnumerical
Passage / Comprehension

The entropy versus temperature plot for phases α and β at 1 bar pressure is given. ST and S0 are entropies of the phases at temperatures T and 0 K, respectively.

The transition temperature for α to β phase change is 600 K and Cp,β-Cp,α=1 J mol-1 K-1. Assume Cp,β-Cp,α is independent of temperature in the range of 200 to 700 K.Cp,α and Cp,β are heat capacities of α and β phases, respectively.

The value of enthalpy change,Hβ-Hα(inJmol-1), at300 Kis

Question diagram: The value of enthalpy change, H β - H α (in Jmol - 1 ), at 3
Answer:
300.00
Solution:

According to Kirchhoff's law 

ΔHT2-ΔHT1=n×Cp,β-Cp,α(T2-T1)

ΔH600-ΔH300=1×Cp,β-Cp,α(600-300)

Now, at transition temperature, G=0

ΔH600=TΔS600

=600×(6-5)

600-ΔH300=1×1×300

ΔH300=600-300

=300 J mol-1

Stream:JEE_ADVSubject:ChemistryTopic:Thermodynamics - IISubtopic:Calculation of DS
2mℹ️ Source: PYQ_2023

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