When two mole of an ideal gas ( C p , m = 2 5 R ) heated from 300 K to 600 K at constant pressure. The change in entropy of gas ( Δ S ) is :

Δ S = nC p , m In T 1 T 2 = 2 × 2 5 R In 300 600 = 5 R In 2

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When two mole of an ideal gas $(C_{p, m} = \frac{5}{2} R)$ heated from 300 K to 600 K at constant pressure. The change in entropy of gas $(Δ S)$ is :

Answer:

Solution:

$Δ S = nC_{p, m} In \frac{T _{2}}{T _{1}} = 2 \times \frac{5}{2} R In \frac{600}{300} = 5 R In 2$

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

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