Chemical Thermodynamics Question 49
Question: One mole of water at $ 100^{o}C $ is converted into steam at $ 100^{o}C $ at a constant pressure of 1 atm. The change in entropy is [heat of vaporisation of water at $ 100^{o}C=540,cal/gm $ ] [Pb. PMT 2004]
Options:
A) 8.74
B) 18.76
C) 24.06
D) 26.06
Show Answer
Answer:
Correct Answer: D
Solution:
The entropy change $ =\frac{\text{heat of vaporisation}}{temperature} $ Here, heat of vaporisation $ =540,cal/gm $
$ =540\times 18,cal,mo{{l}^{-1}} $ Temperature of water $ =100+273=373K $
$ \therefore $ entropy change $ =\frac{540\times 18}{373}=26.06,cal,mo{{l}^{-1}}{{K}^{-1}} $
