The value of entropy change, S β - S α (in Jmol - 1 K - 1 ), at 300 K is [Use: l | TestHub

Passage / Comprehension

The entropy versus temperature plot for phases $α$ and $β$ at $1$ bar pressure is given. $S_{T}$ and $S_{0}$ 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 $C_{p, β} - C_{p, α} = 1 J {mol}^{- 1} K^{- 1}$ . Assume $(C_{p, β} - C_{p, α})$ is independent of temperature in the range of $200$ to $700 K . C_{p, α}$ and $C_{p, β}$ are heat capacities of $α$ and $β$ phases, respectively.

The value of entropy change, $S_{β} - S_{α}$ (in ${Jmol}^{- 1} K^{- 1}$ ), at $300 K$ is [Use: $\ln 2 = 0.69$ ].

Given: $S_{β} - S_{α} = 0$ at $0 K$ ]

Question diagram: The value of entropy change, S β - S α (in Jmol - 1 K - 1 ),

Answer:

0.31

Solution:

$S = S_{0} + \int_{} C_{p} \frac{d T}{T}$
$S_{α} = S_{0} + \int_{} {[C_{p}]}_{α} \frac{d T}{T}$
$S_{β} = S_{0} + \int_{} {[C_{p}]}_{β} \frac{d T}{T}$
Now,
$S_{β} - S_{α} = S_{0} + \int [{[C_{p}]}_{β} - {[C_{p}]}_{α}] \frac{d T}{T}$
Given ${[C_{p}]}_{β} - {[C_{p}]}_{α} = 1$
$S_{β} - S_{α} = lnT + C$ at any temperature of $T$
${[S_{β} - S_{α}]}_{T_{2}} - {[S_{β} - S_{α}]}_{T_{1}} = {lnT}_{2} - {lnT}_{1}$
$T_{2} = 600 K, T_{1} = 300 K and S_{β} - S_{α} = 1 at 600 K$
$1 - {[S_{β} - S_{α}]}_{300} = \ln 600 - \ln 300 \Rightarrow S_{β} - S_{α} = 1 - 0.69 = 0.31$

Chemistry - Thermodynamics - II Question with Solution | TestHub

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