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  • Calculate the change in entropy  when 100 g of liquid water at 100°

Calculate the change in entropy \mathrm{(\Delta S)} when 100 g of liquid water at 100°C is converted into steam at 100°C. Given the heat of vaporization\mathrm{(\Delta H_{vap})} of water is 40.79 kJ/mol.

Option: 1

109.35J/K
 


Option: 2

108.50 J/L
 


Option: 3

100.60 J/L
 


Option: 4

 8090.3 J/L


Answers (1)

best_answer

The change in entropy (?S) during a phase transition can be calculated using the equation:

\mathrm{ \Delta S=\frac{\Delta H_{\mathrm{vap}}}{T} }

where \mathrm{H_{\mathrm{vap}} is the heat of vaporization and T is the temperature in Kelvin.
Given:

\mathrm{ \begin{aligned} \Delta H_{\text {vap }} & =40.79 \mathrm{~kJ} / \mathrm{mol} \\ T & =100^{\circ} \mathrm{C}=373.15 \mathrm{~K} \end{aligned} }

Converting heat of vaporization to J/mol:

\mathrm{ \Delta H_{\text {vap }}=40.79 \times 10^3 \mathrm{~J} / \mathrm{mol} }

Substituting the values and calculating the change in entropy (?S):

\mathrm{ \Delta S=\frac{40.79 \times 10^3 \mathrm{~J} / \mathrm{mol}}{373.15 \mathrm{~K}} }

Therefore, the change in entropy (?S) when 100 g of liquid water at 100°C is converted into steam at 100°C is approximately \mathrm{ \Delta S \approx 109.35 \mathrm{~J} / \mathrm{K} . } J/K. Therefore, the correct option is A.

Posted by

Gautam harsolia

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