- #1
patrickmoloney
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Homework Statement
[/B]Phospine exist in three forms. known as the \alpha, \beta and \gamma forms. The \alpha and \beta forms are in equilibrium with each other at 49.43 \, K, and the \alpha and \gamma forms are in equilibrium at 30.29 \, K. Obtain the molar heat of transformation for the \gamma form changing to the \alpha form at 30.29 \, K from the following data:
(i) The entropy of the \alpha at 49.43 \, K is 34.03 \, \text{J/mol K}.
(ii) The change in entropy heating the \gamma from 0\, K to 30.29 \, K is 11.22 \, \text{J/mol K}.
(iii)The change in entropy in heating the \alpha form from 30.29 \, K to 49.43 \, K is 20.10 \, \text{J/mol K}
Homework Equations
The relationship between latent heat Q_T and molar entropy \Delta S is
Q_T = \Delta S \cdot T
The total entropy is calculated by adding the individual entropies of the system.
The Attempt at a Solution
Using the data given we know that S_{\alpha} = 34.03\, \text{J/mol K} at 49.43\, K
We also know that the entropy change \Delta S_{{\alpha}_2}= 20.10\, \text{J/mol K} from 30.29 \, K \to 49.43 \, K
Hence,
\Delta S_{\alpha_{1}}= S_{alpha}- \Delta S_{\alpha_{2}} = 34.03 - 20.10 = 13.93 \, \text{J/mol K}
is the entropy change in heating the \alpha form from 0 \, K \to 30.29\, K
We know the entropy change for \gamma is \Delta S_{\gamma}= 11.22
There for the total entropy \Delta S_{tot} is given by
\Delta S_{tot} = \Delta S_{alpha_{1}} + \Delta S_{\gamma}= 13.93 + 11.22 = \, 25.15 \text{J/mol K}
Our latent heat is
Q = \Delta S_{tot}\cdot T = (25.15\, \text{J/mol K})(30.29 \, K)= 761.79 \, \text{J/mol}is this a correct way to tackle this problem. I struggled for a while cause it's a bit of a weird problem but is it really that straight forward or am I missing something. As they say if physics seems easy - you're doing it wrong.